Ch 5  ·  Q–
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Chapter 5 Exercise 5.3 Solutions

Continuity and Differentiability

Step-by-Step Solutions to the NCERT Class 12 Mathematics Chapter 5 Exercise 5.3

Class 12 Mathematics Exercise 5.3 NCERT Solutions Continuity and Differentiability Class 12 Mathematics Chapter 5 CBSE Board Exam JEE Main CUET Continuity Differentiability Derivatives Differentiation Implicit Differentiation Chain Rule Inverse Trigonometric Functions Derivatives of Inverse Trigonometric Functions
15 Questions
35–50 min Ideal time
Q1 Now at
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Q1
NUMERIC3 marks

Find \(\dfrac{dy}{dx}\) in the following:

\[ 2x+3y=\sin x \]

📘 Concept & Theory
Concept/Theory
›

This question illustrates the use of implicit differentiation. When \(y\) is given explicitly as a function of \(x\), such as \(y=f(x)\), we can differentiate directly. However, in this question, \(x\) and \(y\) are related by an equation rather than by an explicit expression for \(y\).

Since \(y\) is ultimately a function of \(x\), we differentiate both sides of the equation with respect to \(x\). While differentiating a term containing \(y\), we must remember that \(y=y(x)\). Therefore,

\[ \frac{d}{dx}(y)=\frac{dy}{dx} \]

The important differentiation rules used here are:

  • \[ \frac{d}{dx}(x)=1 \]
  • \[ \frac{d}{dx}(kx)=k \]
    where \(k\) is a constant.
  • \[ \frac{d}{dx}(ky)=k\frac{dy}{dx} \]
  • \[ \frac{d}{dx}(\sin x)=\cos x \]
🗺️ Solution Roadmap
Step-by-step Plan
›

  1. Start with the given relation between \(x\) and \(y\).

  2. Differentiate both sides with respect to \(x\).

  3. Differentiate every term carefully, treating \(y\) as a function of \(x\).

  4. Collect the terms containing \(\dfrac{dy}{dx}\) on one side.

  5. Solve for \(\dfrac{dy}{dx}\).

  6. Present the final derivative in its simplest form.

✏️ Solution
Complete Solution
›
Step-by-step Solution  ·  9 steps
  1. Given — relation:
    \[2x+3y=\sin x\]
  2. Since \(y\) depends on \(x\), differentiate both sides with respect to \(x\):
    \[\frac{d}{dx}(2x+3y)=\frac{d}{dx}(\sin x)\]
  3. Applying the derivative operator term by term, we get:
    \[ \frac{d}{dx}(2x)+\frac{d}{dx}(3y) = \frac{d}{dx}(\sin x) \]
  4. For the first term,
    \[\frac{d}{dx}(2x)=2\]
  5. For the second term, \(3\) is a constant and \(y\) is a function of \(x\):
    \[\frac{d}{dx}(3y)=3\frac{dy}{dx}\]
  6. For the right-hand side,
    \[\frac{d}{dx}(\sin x)=\cos x\]
  7. Therefore,
    \[2+3\frac{dy}{dx}=\cos x\]
  8. Subtract \(2\) from both sides:
    \[3\frac{dy}{dx}=\cos x-2\]
  9. Divide both sides by \(3\):
    \[\frac{dy}{dx}=\frac{\cos x-2}{3}\]
  10. Hence, the required derivative is
    \[\boxed{\frac{dy}{dx}=\frac{\cos x-2}{3}}\]
🎯 Exam Significance
Exam Significance
›

This is a fundamental application of implicit differentiation, an important concept in Class 12 Mathematics. The question tests whether a student can differentiate an equation containing both \(x\) and \(y\) without first solving explicitly for \(y\).

For board examinations, particular attention should be given to the step

\[ \frac{d}{dx}(3y)=3\frac{dy}{dx} \]
because omitting \(\dfrac{dy}{dx}\) is a common error. Writing the differentiation steps clearly also helps demonstrate the complete method and reduces the possibility of losing marks for an unexplained jump in working.

Significance for Competitive Entrance Exam Aspirants

Implicit differentiation is frequently used as a foundational technique in calculus problems involving curves, tangent slopes, related rates, and equations in which \(y\) cannot be conveniently isolated. A strong command of this basic procedure allows students to handle more complicated implicit functions efficiently.

The key competitive-exam skill tested here is recognizing immediately that every occurrence of \(y\) must be differentiated with respect to \(x\), producing a factor of \(\dfrac{dy}{dx}\).

🔑 Key Takeaways
Key Takeaways
›
Key Takeaways  ·  6 points

  1. When \(x\) and \(y\) are related implicitly, differentiate the entire equation with respect to \(x\).

  2. Always treat \(y\) as a function of \(x\) during implicit differentiation.

  3. The derivative of \(y\) with respect to \(x\) is \(\dfrac{dy}{dx}\).

  4. For a constant \(k\), \(\dfrac{d}{dx}(ky)=k\dfrac{dy}{dx}\).

  5. After differentiation, collect all terms containing \(\dfrac{dy}{dx}\) and solve algebraically.

  6. The final result is:

    \[ \boxed{\frac{dy}{dx}=\frac{\cos x-2}{3}} \]
↑ Top
1 / 15  ·  7%
Q2 →
Q2
NUMERIC3 marks
Find \(\dfrac{dy}{dx}\) in the following: \[2x+3y=\sin y\]
📘 Concept & Theory
Concept/Theory
›

This question is another application of implicit differentiation. Here, both sides of the equation contain \(y\), and \(y\) is considered a function of \(x\). Therefore, while differentiating with respect to \(x\), every term involving \(y\) must produce a factor of \(\dfrac{dy}{dx}\).

The most important step is the differentiation of \(\sin y\). Since \(y\) is a function of \(x\), the chain rule is required:

\[ \frac{d}{dx}(\sin y) = \cos y\frac{dy}{dx} \]

Similarly,

\[ \frac{d}{dx}(3y)=3\frac{dy}{dx} \]

Thus, after differentiating, the resulting equation contains \(\dfrac{dy}{dx}\) on both sides. The next task is to collect these terms and solve algebraically for \(\dfrac{dy}{dx}\).

🗺️ Solution Roadmap
Step-by-step Plan
›

  1. Write the given implicit relation.

  2. Differentiate both sides with respect to \(x\).

  3. Use the chain rule while differentiating \(\sin y\).

  4. Obtain an equation containing \(\dfrac{dy}{dx}\) on both sides.

  5. Bring all terms containing \(\dfrac{dy}{dx}\) to one side.

  6. Factor out \(\dfrac{dy}{dx}\).

  7. Divide by the remaining factor to obtain the required derivative.

✏️ Solution
Complete Solution
›
Step-by-step Solution  ·  12 steps
  1. Given —
    \[2x+3y=\sin y\]
  2. Since \(y\) is a function of \(x\), differentiate both sides with respect to \(x\):
    \[\frac{d}{dx}(2x+3y)=\frac{d}{dx}(\sin y)\]
  3. Differentiate each term on the left-hand side:
    \[ \frac{d}{dx}(2x)+\frac{d}{dx}(3y) = \frac{d}{dx}(\sin y) \]
  4. For the first term,
    \[\frac{d}{dx}(2x)=2\]
  5. For the second term, \(3\) is a constant and \(y\) is a function of \(x\):
    \[\frac{d}{dx}(3y)=3\frac{dy}{dx}\]
  6. For the right-hand side, apply the chain rule:
    \[ \frac{d}{dx}(\sin y) = \cos y\frac{dy}{dx} \]
  7. Therefore,
    \[ 2+3\frac{dy}{dx} = \cos y\frac{dy}{dx} \]
  8. Move \(3\dfrac{dy}{dx}\) to the right-hand side:
    \[ 2 = \cos y\frac{dy}{dx} - 3\frac{dy}{dx} \]
  9. Equivalently,
    \[ -3\frac{dy}{dx} + \cos y\frac{dy}{dx} = 2 \]
  10. Take \(\dfrac{dy}{dx}\) as the common factor:
    \[\frac{dy}{dx}\left(-3+\cos y\right)=2\]
  11. Rearranging the factor gives:
    \[\frac{dy}{dx}\left(\cos y-3\right)=2\]
  12. Divide both sides by \(\cos y-3\):
    \[\frac{dy}{dx}=\frac{2}{\cos y-3}\]
  13. Hence, the required derivative is
    \[\boxed{\frac{dy}{dx}=\frac{2}{\cos y-3}}\]
🎯 Exam Significance
Exam Significance
›

This problem tests two important Class 12 calculus skills: implicit differentiation and the chain rule. In particular, students must recognize that \(y\) is dependent on \(x\), so the derivative of \(\sin y\) is not simply \(\cos y\). It is

\[ \frac{d}{dx}(\sin y)=\cos y\frac{dy}{dx} \]

For board examinations, every algebraic step should be written clearly. Moving the \(\dfrac{dy}{dx}\) terms to one side and factoring them makes the method easy to follow and minimizes sign errors.

Significance for Competitive Entrance Exam Aspirants

The same technique forms the basis of more advanced problems involving implicitly defined curves, tangent and normal equations, related rates, and differentiation of composite functions. Competitive examinations often require students to identify the chain-rule factor immediately and isolate the derivative efficiently.

A useful recognition pattern is:

\[ \frac{d}{dx}\bigl[f(y)\bigr] = f'(y)\frac{dy}{dx} \]

Thus, whenever a function of \(y\) appears while differentiating with respect to \(x\), the factor \(\dfrac{dy}{dx}\) must be included.

🔑 Key Takeaways
Key Takeaways
›
Key Takeaways  ·  7 points

  1. In implicit differentiation, \(y\) is treated as a function of \(x\).

  2. The derivative of \(3y\) is \(3\dfrac{dy}{dx}\).

  3. The chain rule is essential when differentiating a function of \(y\).

  4. \[ \frac{d}{dx}(\sin y)=\cos y\frac{dy}{dx} \]

  5. Collect all terms containing \(\dfrac{dy}{dx}\) before solving for the derivative.

  6. Be particularly careful with signs while transposing terms.

  7. The final result is:

    \[ \boxed{\frac{dy}{dx}=\frac{2}{\cos y-3}} \]
← Q1
2 / 15  ·  13%
Q3 →
Q3
NUMERIC3 marks
Find \(\dfrac{dy}{dx}\) in the following: \[ax+by^2=\cos y\]
📘 Concept & Theory
Concept/Theory
›

This question is an application of implicit differentiation combined with the chain rule. Here, \(x\) and \(y\) are related implicitly, and \(y\) is considered a function of \(x\).

The constants \(a\) and \(b\) are treated as constants during differentiation. The important derivatives required are:

  • \[ \frac{d}{dx}(ax)=a \]
  • \[ \frac{d}{dx}(by^2) = b\frac{d}{dx}(y^2) = 2by\frac{dy}{dx} \]
  • \[ \frac{d}{dx}(\cos y) = -\sin y\frac{dy}{dx} \]

The factor \(\dfrac{dy}{dx}\) appears when differentiating \(y^2\) and \(\cos y\) because \(y\) depends on \(x\). After differentiation, the terms containing \(\dfrac{dy}{dx}\) must be collected and factored.

🗺️ Solution Roadmap
Step-by-step Plan
›

  1. Write the given implicit equation.

  2. Differentiate both sides with respect to \(x\).

  3. Differentiate \(ax\) using the constant-multiple rule.

  4. Differentiate \(by^2\) using the chain rule.

  5. Differentiate \(\cos y\) using the chain rule.

  6. Collect all terms containing \(\dfrac{dy}{dx}\) on one side.

  7. Factor out \(\dfrac{dy}{dx}\).

  8. Solve for \(\dfrac{dy}{dx}\).

✏️ Solution
Complete Solution
›
Step-by-step Solution  ·  14 steps
  1. Given —
    \[ax+by^2=\cos y\]
  2. Since \(y\) is a function of \(x\), differentiate both sides with respect to \(x\):
    \[\frac{d}{dx}(ax+by^2)=\frac{d}{dx}(\cos y)\]
  3. Differentiate each term on the left-hand side:
    \[\frac{d}{dx}(ax)+\frac{d}{dx}(by^2)=\frac{d}{dx}(\cos y)\]
  4. Since \(a\) is a constant,
    \[\frac{d}{dx}(ax)=a\]
  5. For the second term, \(b\) is a constant. Therefore,
    \[\frac{d}{dx}(by^2)=b\frac{d}{dx}(y^2)\]
  6. Since \(y\) is a function of \(x\), apply the chain rule:
    \[\frac{d}{dx}(y^2)=2y\frac{dy}{dx}\]
  7. Hence,
    \[\frac{d}{dx}(by^2)=2by\frac{dy}{dx}\]
  8. On the right-hand side, apply the chain rule to \(\cos y\):
    \[\frac{d}{dx}(\cos y)=-\sin y\frac{dy}{dx}\]
  9. Therefore, after differentiating both sides, we obtain
    \[a+2by\frac{dy}{dx}=-\sin y\frac{dy}{dx}\]
  10. Move \(2by\dfrac{dy}{dx}\) to the right-hand side:
    \[a=-\sin y\frac{dy}{dx}-2by\frac{dy}{dx}\]
  11. Take \(\dfrac{dy}{dx}\) as the common factor on the right-hand side:
    \[a=\frac{dy}{dx}\left(-\sin y-2by\right)\]
  12. Multiply both sides by \(-1\):
    \[-a=\frac{dy}{dx}\left(\sin y+2by\right)\]
  13. Therefore,
    \[\frac{dy}{dx}\left(2by+\sin y\right)=-a\]
  14. Divide both sides by \(2by+\sin y\):
    \[\frac{dy}{dx}=\frac{-a}{2by+\sin y}\]
  15. Hence, the required derivative is
    \[\boxed{\frac{dy}{dx}=-\frac{a}{2by+\sin y}}\]
🎯 Exam Significance
Exam Significance
›

This problem combines implicit differentiation with the chain rule and tests whether students can correctly differentiate algebraic and trigonometric expressions involving \(y\). It is particularly important to remember that \(y\) is not treated as a constant when differentiation is performed with respect to \(x\).

The two critical steps are

\[ \frac{d}{dx}(y^2)=2y\frac{dy}{dx} \]

and

\[ \frac{d}{dx}(\cos y) = -\sin y\frac{dy}{dx} \]

Writing these steps explicitly is useful in board examinations because it demonstrates the application of the chain rule and makes the subsequent algebraic isolation of \(\dfrac{dy}{dx}\) clear.

Significance for Competitive Entrance Exam Aspirants

This problem develops a standard pattern used in more advanced calculus: differentiating an implicitly defined function and isolating its derivative. The same procedure is applicable when algebraic powers, exponential functions, logarithmic functions, and trigonometric functions of \(y\) occur together.

A useful general pattern is

\[ \frac{d}{dx}\bigl[f(y)\bigr] = f'(y)\frac{dy}{dx} \]

Recognising this pattern quickly can significantly reduce the time required for implicit differentiation questions in entrance examinations.

🔑 Key Takeaways
Key Takeaways
›
Key Takeaways  ·  7 points

  1. \(a\) and \(b\) are constants and remain unchanged during differentiation.

  2. When differentiating \(y\) with respect to \(x\), include the factor \(\dfrac{dy}{dx}\).

  3. \[ \frac{d}{dx}(y^2)=2y\frac{dy}{dx} \]

  4. \[ \frac{d}{dx}(\cos y)=-\sin y\frac{dy}{dx} \]

  5. After differentiation, collect all terms containing \(\dfrac{dy}{dx}\).

  6. Careful handling of negative signs is essential when transposing and factoring terms.

  7. The final derivative is:

    \[ \boxed{\frac{dy}{dx}=-\frac{a}{2by+\sin y}} \]
← Q2
3 / 15  ·  20%
Q4 →
Q4
NUMERIC3 marks
Find \(\dfrac{dy}{dx}\) in the following: \[xy+y^2=\tan^2x+y\]
📘 Concept & Theory
Concept/Theory
›

This question requires implicit differentiation, the product rule, the chain rule, and standard trigonometric differentiation.

Since \(x\) and \(y\) occur together in the equation, \(y\) is treated as a function of \(x\). Therefore, whenever \(y\) is differentiated with respect to \(x\), the factor \(\dfrac{dy}{dx}\) must be included.

The product \(xy\) requires the product rule:

\[ \frac{d}{dx}(xy) = x\frac{dy}{dx}+y\frac{dx}{dx} = x\frac{dy}{dx}+y \]

For \(y^2\), the chain rule gives:

\[ \frac{d}{dx}(y^2) = 2y\frac{dy}{dx} \]

The right-hand side contains \(\tan^2x\), not merely \(\tan x\). Therefore, the chain rule must be applied carefully:

\[ \frac{d}{dx}(\tan^2x) = 2\tan x\sec^2x \]

Also,

\[ \frac{d}{dx}(y)=\frac{dy}{dx} \]

Important Correction to the Given Working

There are two errors in the supplied solution that need to be corrected.

  1. The derivative of \(\tan^2x\) is
    \[ 2\tan x\sec^2x, \]
    not \(\sec^2x\). The latter is the derivative of \(\tan x\).
  2. The term \(y\) on the right-hand side contributes
    \[ \frac{dy}{dx}, \]
    which must be retained while solving for the derivative.
🗺️ Solution Roadmap
Step-by-step Plan
›

  1. Write the given implicit equation.

  2. Differentiate both sides with respect to \(x\).

  3. Apply the product rule to \(xy\).

  4. Apply the chain rule to \(y^2\).

  5. Apply the chain rule to \(\tan^2x\).

  6. Differentiate the \(y\) term on the right-hand side.

  7. Collect all terms containing \(\dfrac{dy}{dx}\) on one side.

  8. Factor out \(\dfrac{dy}{dx}\).

  9. Solve for \(\dfrac{dy}{dx}\).

✏️ Solution
Complete Solution
›
Step-by-step Solution  ·  16 steps
  1. Given —
    \[xy+y^2=\tan^2x+y\]
  2. Differentiate both sides with respect to \(x\):
    \[\frac{d}{dx}(xy+y^2)=\frac{d}{dx}(\tan^2x+y)\]
  3. Differentiating term by term, we get
    \[\frac{d}{dx}(xy)+\frac{d}{dx}(y^2)=\frac{d}{dx}(\tan^2x)+\frac{d}{dx}(y)\]
  4. For the product \(xy\), apply the product rule:
    \[\frac{d}{dx}(xy)=x\frac{dy}{dx}+y\frac{dx}{dx}\]
  5. Since
    \[\frac{dx}{dx}=1,\]
  6. we obtain
    \[\frac{d}{dx}(xy)=x\frac{dy}{dx}+y\]
  7. For \(y^2\), apply the chain rule:
    \[\frac{d}{dx}(y^2)=2y\frac{dy}{dx}\]
  8. For \(\tan^2x=(\tan x)^2\), apply the chain rule:
    \[\frac{d}{dx}(\tan^2x)=2\tan x\frac{d}{dx}(\tan x)\]
  9. Since
    \[\frac{d}{dx}(\tan x)=\sec^2x,\]
  10. therefore,
    \[\frac{d}{dx}(\tan^2x)=2\tan x\sec^2x\]
  11. Finally,
    \[\frac{d}{dx}(y)=\frac{dy}{dx}\]
  12. Substituting all these derivatives into the differentiated equation gives
    \[ x\frac{dy}{dx} +y +2y\frac{dy}{dx} = 2\tan x\sec^2x + \frac{dy}{dx} \]
  13. Bring the term \(\dfrac{dy}{dx}\) on the right-hand side to the left-hand side:
    \[x\frac{dy}{dx}+2y\frac{dy}{dx}-\frac{dy}{dx}+y=2\tan x\sec^2x\]
  14. Collect the terms containing \(\dfrac{dy}{dx}\):
    \[\frac{dy}{dx}\left(x+2y-1\right)+y=2\tan x\sec^2x\]
  15. Subtract \(y\) from both sides:
    \[\frac{dy}{dx}\left(x+2y-1\right)=2\tan x\sec^2x-y\]
  16. Divide both sides by \(x+2y-1\):
    \[\frac{dy}{dx}=\frac{2\tan x\sec^2x-y}{x+2y-1}\]
  17. Hence, the required derivative is
    \[\boxed{\frac{dy}{dx}=\frac{2\tan x\sec^2x-y}{x+2y-1}}\]
🎯 Exam Significance
Exam Significance
›

This question is particularly important because it combines several differentiation rules in a single implicit equation. A student must correctly identify when to use the product rule and when to use the chain rule.

The most important point is that

\[ \tan^2x \]
means
\[ (\tan x)^2, \]
so its derivative is
\[ 2\tan x\sec^2x. \]
Confusing it with the derivative of \(\tan x\) can lead to an incorrect answer even when the remaining algebra is correct.

This is also a good board-examination question for practising complete, logically connected working. The product-rule and chain-rule steps should not be skipped.

Significance for Competitive Entrance Exam Aspirants

This problem develops the ability to recognize multiple differentiation rules within a single expression. Such recognition is essential in competitive calculus problems involving implicit curves and composite functions.

A useful pattern to remember is

\[ \frac{d}{dx}\left[f(x)^n\right] = nf(x)^{n-1}f'(x) \]

Therefore, for \(f(x)=\tan x\) and \(n=2\),

\[ \frac{d}{dx}(\tan^2x) = 2\tan x\sec^2x \]

Competitive-exam aspirants should also develop the habit of checking whether every term on both sides of an implicit equation has been differentiated. Missing a term such as \(\dfrac{dy}{dx}\) from the derivative of \(y\) changes the final result.

🔑 Key Takeaways
Key Takeaways
›
Key Takeaways  ·  9 points

  1. Use the product rule for \(xy\).

  2. \[ \frac{d}{dx}(xy) = x\frac{dy}{dx}+y \]

  3. \[ \frac{d}{dx}(y^2) = 2y\frac{dy}{dx} \]

  4. \[ \frac{d}{dx}(\tan^2x) = 2\tan x\sec^2x \]

  5. Do not confuse the derivative of \(\tan^2x\) with the derivative of \(\tan x\).

  6. The derivative of \(y\) with respect to \(x\) is \(\dfrac{dy}{dx}\).

  7. After differentiation, collect all \(\dfrac{dy}{dx}\) terms before solving.

  8. Do not omit the term \(y\) produced by differentiating \(xy\).

  9. The final result is

    \[ \boxed{ \frac{dy}{dx} = \frac{2\tan x\sec^2x-y}{x+2y-1} } \]
← Q3
4 / 15  ·  27%
Q5 →
Q5
NUMERIC3 marks
Find \(\dfrac{dy}{dx}\) in the following: \[x^2+xy+y^2=100\]
📘 Concept & Theory
Concept/Theory
›

This problem is an application of implicit differentiation. Since \(x\) and \(y\) occur together in the given equation, \(y\) is regarded as a function of \(x\). Therefore, while differentiating with respect to \(x\), every occurrence of \(y\) must be differentiated accordingly.

Three differentiation rules are particularly important in this question:

  • Power rule:
    \[ \frac{d}{dx}(x^2)=2x \]
  • Product rule:
    \[ \frac{d}{dx}(xy) = x\frac{dy}{dx}+y\frac{dx}{dx} = x\frac{dy}{dx}+y \]
  • Chain rule:
    \[ \frac{d}{dx}(y^2) = 2y\frac{dy}{dx} \]

Since \(100\) is a constant, its derivative with respect to \(x\) is zero:

\[ \frac{d}{dx}(100)=0 \]

🗺️ Solution Roadmap
Step-by-step Plan
›

  1. Write the given implicit equation.

  2. Differentiate both sides with respect to \(x\).

  3. Differentiate \(x^2\) using the power rule.

  4. Differentiate \(xy\) using the product rule.

  5. Differentiate \(y^2\) using the chain rule.

  6. Differentiate the constant \(100\).

  7. Collect all terms containing \(\dfrac{dy}{dx}\) on one side.

  8. Factor out \(\dfrac{dy}{dx}\).

  9. Solve for \(\dfrac{dy}{dx}\).

✏️ Solution
Complete Solution
›
Step-by-step Solution  ·  13 steps
  1. Given —
    \[x^2+xy+y^2=100\]
  2. Differentiate both sides with respect to \(x\):
    \[\frac{d}{dx}(x^2+xy+y^2)=\frac{d}{dx}(100)\]
  3. Differentiating term by term,
    \[\frac{d}{dx}(x^2)+\frac{d}{dx}(xy)+\frac{d}{dx}(y^2)=\frac{d}{dx}(100)\]
  4. Using the power rule,
    \[\frac{d}{dx}(x^2)=2x\]
  5. For the product \(xy\), use the product rule:
    \[\frac{d}{dx}(xy)=x\frac{dy}{dx}+y\frac{dx}{dx}\]
  6. Since
    \[\frac{dx}{dx}=1\]
  7. we get
    \[\frac{d}{dx}(xy)=x\frac{dy}{dx}+y\]
  8. For \(y^2\), apply the chain rule because \(y\) is a function of \(x\):
    \[\frac{d}{dx}(y^2)=2y\frac{dy}{dx}\]
  9. Finally, since \(100\) is a constant,
    \[\frac{d}{dx}(100)=0\]
  10. Substituting these derivatives into the equation gives
    \[2x+x\frac{dy}{dx}+y+2y\frac{dy}{dx}=0\]
  11. Collect the terms containing \(\dfrac{dy}{dx}\):
    \[x\frac{dy}{dx}+2y\frac{dy}{dx}=-2x-y\]
  12. Take \(\dfrac{dy}{dx}\) as the common factor:
    \[\frac{dy}{dx}(x+2y)=-2x-y\]
  13. Divide both sides by \(x+2y\):
    \[\frac{dy}{dx}=\frac{-2x-y}{x+2y}\]
  14. Therefore,
    \[\boxed{\frac{dy}{dx}=-\frac{2x+y}{x+2y}}\]
🎯 Exam Significance
Exam Significance
›

This is a standard implicit differentiation problem that tests the correct application of the power rule, product rule, and chain rule in a single equation.

The most important step is differentiating the product \(xy\). Students must not write its derivative as \(x\dfrac{dy}{dx}\) alone. The complete product rule gives

\[ \frac{d}{dx}(xy) = x\frac{dy}{dx}+y \]

Similarly, the term \(y^2\) produces

\[ 2y\frac{dy}{dx} \]

These are common points at which marks can be lost. Writing the intermediate differentiation steps clearly is therefore important in a board examination.

Significance for Competitive Entrance Exam Aspirants

The question develops a fundamental skill required for differentiating implicitly defined curves. The same technique is used in questions involving tangents, normals, slopes of curves, and higher-order derivatives.

For competitive examinations, students should quickly recognize products and powers involving \(y\). A useful pattern is

\[ \frac{d}{dx}(xy) = x\frac{dy}{dx}+y \]

and

\[ \frac{d}{dx}(y^n) = ny^{n-1}\frac{dy}{dx} \]

Mastering these patterns helps reduce calculation time while maintaining accuracy.

🔑 Key Takeaways
Key Takeaways
›
Key Takeaways  ·  9 points

  1. In implicit differentiation, \(y\) is treated as a function of \(x\).

  2. \[ \frac{d}{dx}(x^2)=2x \]

  3. \[ \frac{d}{dx}(xy)=x\frac{dy}{dx}+y \]

  4. \[ \frac{d}{dx}(y^2)=2y\frac{dy}{dx} \]

  5. \[ \frac{d}{dx}(100)=0 \]

  6. After differentiation, collect all terms containing \(\dfrac{dy}{dx}\).

  7. Factor out \(\dfrac{dy}{dx}\) before isolating it.

  8. Be careful not to omit the \(y\) term generated by differentiating \(xy\).

  9. The final result is

    \[ \boxed{ \frac{dy}{dx} = -\frac{2x+y}{x+2y} } \]
← Q4
5 / 15  ·  33%
Q6 →
Q6
NUMERIC3 marks
Find \(\dfrac{dy}{dx}\) in the following: \[x^3+x^2y+xy^2+y^3=81\]
📘 Concept & Theory
Concept/Theory
›

This question is an application of implicit differentiation involving the product rule and the chain rule. Since \(x\) and \(y\) occur together in the equation, \(y\) is treated as a function of \(x\).

The terms \(x^2y\) and \(xy^2\) are products, so the product rule must be used. The terms containing powers of \(y\) also require the chain rule.

The important rules are:

  • \[ \frac{d}{dx}(x^n)=nx^{n-1} \]
  • \[ \frac{d}{dx}(x^2y) = x^2\frac{dy}{dx}+2xy \]
  • \[ \frac{d}{dx}(xy^2) = x\left(2y\frac{dy}{dx}\right)+y^2 = 2xy\frac{dy}{dx}+y^2 \]
  • \[ \frac{d}{dx}(y^3) = 3y^2\frac{dy}{dx} \]
  • \[ \frac{d}{dx}(81)=0 \]
🗺️ Solution Roadmap
Step-by-step Plan
›

  1. Write the given implicit equation.

  2. Differentiate both sides with respect to \(x\).

  3. Differentiate \(x^3\) using the power rule.

  4. Differentiate \(x^2y\) using the product rule.

  5. Differentiate \(xy^2\) using the product rule and chain rule.

  6. Differentiate \(y^3\) using the chain rule.

  7. Differentiate the constant \(81\).

  8. Collect all terms containing \(\dfrac{dy}{dx}\).

  9. Factor out \(\dfrac{dy}{dx}\).

  10. Solve for \(\dfrac{dy}{dx}\) and simplify.

✏️ Solution
Complete Solution
›
Step-by-step Solution  ·  18 steps
  1. Given —
    \[x^3+x^2y+xy^2+y^3=81\]
  2. Differentiate both sides with respect to \(x\):
    \[\frac{d}{dx}\left(x^3+x^2y+xy^2+y^3\right)=\frac{d}{dx}(81)\]
  3. Differentiating term by term,
    \[\frac{d}{dx}(x^3)+\frac{d}{dx}(x^2y)+\frac{d}{dx}(xy^2)+\frac{d}{dx}(y^3)=\frac{d}{dx}(81)\]
  4. Differentiate \(x^3\)
  5. Using the power rule,
    \[\frac{d}{dx}(x^3)=3x^2\]
  6. Differentiate \(x^2y\)
  7. Since \(x^2y\) is a product of \(x^2\) and \(y\), apply the product rule:
    \[\frac{d}{dx}(x^2y)=x^2\frac{dy}{dx}+y\frac{d}{dx}(x^2)\]
  8. Since
    \[\frac{d}{dx}(x^2)=2x,\]
  9. we obtain
    \[\frac{d}{dx}(x^2y)=x^2\frac{dy}{dx}+2xy\]
  10. Differentiate \(xy^2\)
  11. Again, \(xy^2\) is a product. Therefore, use the product rule:
    \[\frac{d}{dx}(xy^2)=x\frac{d}{dx}(y^2)+y^2\frac{d}{dx}(x)\]
  12. Since \(y\) is a function of \(x\), apply the chain rule to \(y^2\):
    \[\frac{d}{dx}(y^2)=2y\frac{dy}{dx}\]
  13. Also,
    \[\frac{d}{dx}(x)=1\]
  14. Therefore,
    \[\frac{d}{dx}(xy^2)=x\left(2y\frac{dy}{dx}\right)+y^2(1)\]
  15. Hence,
    \[\frac{d}{dx}(xy^2)=2xy\frac{dy}{dx}+y^2\]
  16. Differentiate \(y^3\)
  17. Since \(y\) is a function of \(x\), apply the chain rule:
    \[\frac{d}{dx}(y^3)=3y^2\frac{dy}{dx}\]
  18. Differentiate the constant \(81\)
  19. Since \(81\) is a constant,
    \[frac{d}{dx}(81)=0\]
  20. Substitute all derivatives
  21. Substituting the derivatives into the differentiated equation gives
    \[3x^2+x^2\frac{dy}{dx}+2xy+2xy\frac{dy}{dx}+y^2+3y^2\frac{dy}{dx}=0\]
  22. Collect the terms containing \(\dfrac{dy}{dx}\)
  23. Move the terms that do not contain \(\dfrac{dy}{dx}\) to the right-hand side:
    \[x^2\frac{dy}{dx}+2xy\frac{dy}{dx}+3y^2\frac{dy}{dx}=-3x^2-2xy-y^2\]
  24. Factor out \(\dfrac{dy}{dx}\)
  25. Taking \(\dfrac{dy}{dx}\) as the common factor,
    \[\frac{dy}{dx}\left(x^2+2xy+3y^2\right)=-\left(3x^2+2xy+y^2\right)\]
  26. Solve for \(\dfrac{dy}{dx}\)
  27. Divide both sides by \(x^2+2xy+3y^2\):
    \[ \frac{dy}{dx} = -\frac{3x^2+2xy+y^2} {x^2+2xy+3y^2} \]
  28. Hence, the required derivative is
    \[\boxed{\frac{dy}{dx}=-\frac{3x^2+2xy+y^2}{x^2+2xy+3y^2}}\]
🎯 Exam Significance
Exam Significance
›

This problem is an important practice question because it combines the power rule, product rule, and chain rule in one implicit differentiation problem.

The two product terms require special attention:

\[ \frac{d}{dx}(x^2y) = x^2\frac{dy}{dx}+2xy \]

and

\[ \frac{d}{dx}(xy^2) = 2xy\frac{dy}{dx}+y^2 \]

In board examinations, writing these derivatives step by step is important. A common mistake is to differentiate a product as though only one factor changes. The product rule requires both resulting terms.

Significance for Competitive Entrance Exam Aspirants

This question builds speed and accuracy in handling polynomial expressions containing both \(x\) and \(y\). Such implicit differentiation techniques are useful in problems involving curves, slopes, tangents, normals, and higher derivatives.

A useful general pattern for competitive examinations is

\[ \frac{d}{dx}\left(x^m y^n\right) = mx^{m-1}y^n + nx^m y^{n-1}\frac{dy}{dx} \]

For example, applying this pattern to \(x^2y\) gives

\[ \frac{d}{dx}(x^2y) = 2xy+x^2\frac{dy}{dx} \]

Recognising such patterns can make lengthy implicit differentiation questions considerably faster without sacrificing accuracy.

🔑 Key Takeaways
Key Takeaways
›
Key Takeaways  ·  9 points

  1. Use the power rule for pure powers of \(x\).

  2. Use the product rule whenever \(x\)- and \(y\)-dependent factors are multiplied.

  3. \[ \frac{d}{dx}(x^2y) = x^2\frac{dy}{dx}+2xy \]

  4. \[ \frac{d}{dx}(xy^2) = 2xy\frac{dy}{dx}+y^2 \]

  5. \[ \frac{d}{dx}(y^3) = 3y^2\frac{dy}{dx} \]

  6. The derivative of a constant such as \(81\) is zero.

  7. Collect all terms containing \(\dfrac{dy}{dx}\) before isolating the derivative.

  8. When using the product rule, do not omit either of the two product terms.

  9. The final result is

    \[ \boxed{ \frac{dy}{dx} = -\frac{3x^2+2xy+y^2} {x^2+2xy+3y^2} } \]
← Q5
6 / 15  ·  40%
Q7 →
Q7
NUMERIC3 marks
Find \(\dfrac{dy}{dx}\) in the following: \[\sin^2 y+\cos(xy)=k\]
📘 Concept & Theory
Concept/Theory
›

This problem involves implicit differentiation together with the chain rule and product rule. Since \(y\) is a function of \(x\), every expression containing \(y\) must be differentiated with respect to \(x\).

There are three important derivatives in this question.

First, for \(\sin^2y=(\sin y)^2\), the chain rule gives

\[ \frac{d}{dx}(\sin^2y) = 2\sin y\cos y\frac{dy}{dx} \]

Using the identity

\[ 2\sin y\cos y=\sin 2y, \]
this can also be written as

\[ \frac{d}{dx}(\sin^2y) = \sin 2y\frac{dy}{dx} \]

Second, for \(\cos(xy)\), the outer function is cosine and the inner function is \(xy\). Therefore, the chain rule gives

\[ \frac{d}{dx}\bigl(\cos(xy)\bigr) = -\sin(xy)\frac{d}{dx}(xy) \]

The derivative of \(xy\) requires the product rule:

\[ \frac{d}{dx}(xy) = x\frac{dy}{dx}+y \]

Third, \(k\) is a constant, so

\[ \frac{d}{dx}(k)=0 \]

🗺️ Solution Roadmap
Step-by-step Plan
›

  1. Write the given implicit relation.

  2. Differentiate both sides with respect to \(x\).

  3. Differentiate \(\sin^2y\) using the chain rule.

  4. Differentiate \(\cos(xy)\) using the chain rule.

  5. Differentiate \(xy\) using the product rule.

  6. Use \(2\sin y\cos y=\sin 2y\) to simplify where appropriate.

  7. Collect all terms containing \(\dfrac{dy}{dx}\).

  8. Factor out \(\dfrac{dy}{dx}\).

  9. Solve for \(\dfrac{dy}{dx}\).

✏️ Solution
Complete Solution
›
Step-by-step Solution  ·  17 steps
  1. Given —
    \[\sin^2y+\cos(xy)=k\]
  2. Differentiate both sides with respect to \(x\):
    \[\frac{d}{dx}(\sin^2y)+\frac{d}{dx}\bigl(\cos(xy)\bigr)=\frac{d}{dx}(k)\]
  3. Differentiate \(\sin^2y\)
  4. Write \(\sin^2y\) as \((\sin y)^2\):
    \[\frac{d}{dx}\left((\sin y)^2\right)=2\sin y\frac{d}{dx}(\sin y)\]
  5. Since \(y\) is a function of \(x\), apply the chain rule:
    \[\frac{d}{dx}(\sin y)=\cos y\frac{dy}{dx}\]
  6. Therefore,
    \[\frac{d}{dx}(\sin^2y)=2\sin y\cos y\frac{dy}{dx}\]
  7. Using
    \[2\sin y\cos y=\sin 2y,\]
  8. we obtain
    \[\frac{d}{dx}(\sin^2y)=\sin 2y\frac{dy}{dx}\]
  9. Differentiate \(\cos(xy)\)
  10. Apply the chain rule:
    \[\frac{d}{dx}\bigl(\cos(xy)\bigr)=-\sin(xy)\frac{d}{dx}(xy)\]
  11. Now differentiate \(xy\) using the product rule:
    \[\frac{d}{dx}(xy)=x\frac{dy}{dx}+y\frac{dx}{dx}\]
  12. Since
    \[\frac{dx}{dx}=1,\]
  13. we get
    \[\frac{d}{dx}(xy)=x\frac{dy}{dx}+y\]
  14. Hence,
    \[\frac{d}{dx}\bigl(\cos(xy)\bigr)=-\sin(xy)\left(x\frac{dy}{dx}+y\right)\]
  15. Differentiate the constant \(k\)
  16. Since \(k\) is a constant,
    \[\frac{d}{dx}(k)=0\]
  17. Substitute all derivatives
  18. Substituting all the derivatives into the original differentiated equation:
    \[\sin 2y\frac{dy}{dx}-\sin(xy)\left(x\frac{dy}{dx}+y\right)=0\]
  19. Expand the bracket
  20. Distributing \(-\sin(xy)\), we get
    \[\sin 2y\frac{dy}{dx}-x\sin(xy)\frac{dy}{dx}-y\sin(xy)=0\]
  21. Collect the terms containing \(\dfrac{dy}{dx}\)
  22. Move the term \(-y\sin(xy)\) to the right-hand side:
    \[\sin 2y\frac{dy}{dx}-x\sin(xy)\frac{dy}{dx}=y\sin(xy)\]
  23. Factor out \(\dfrac{dy}{dx}\)
  24. Taking \(\dfrac{dy}{dx}\) as the common factor:
    \[\frac{dy}{dx}\left[\sin 2y-x\sin(xy)\right]=y\sin(xy)\]
  25. Solve for \(\dfrac{dy}{dx}\)
  26. Divide both sides by
    \[\sin 2y-x\sin(xy).\]
    Therefore,
    \[\frac{dy}{dx}=\frac{y\sin(xy)}{\sin 2y-x\sin(xy)}\]
  27. Hence, the required derivative is
    \[\boxed{\frac{dy}{dx}=\frac{y\sin(xy)}{\sin 2y-x\sin(xy)}}\]
🎯 Exam Significance
Exam Significance
›

This problem is important because it combines the chain rule with the product rule in an implicit differentiation setting. The expression \(\cos(xy)\) requires two levels of differentiation: first differentiating the outer cosine function and then differentiating the inner product \(xy\).

The critical chain-rule step is

\[\frac{d}{dx}\bigl(\cos(xy)\bigr)=-\sin(xy)\left(x\frac{dy}{dx}+y\right)\]

Students should also remember that

\[ \frac{d}{dx}(\sin^2y) \]
is not simply \(2\sin y\). The complete derivative is

\[ 2\sin y\cos y\frac{dy}{dx} = \sin 2y\frac{dy}{dx}. \]

Significance for Competitive Entrance Exam Aspirants

This question develops the ability to differentiate nested functions and products efficiently. Such structures frequently occur in calculus problems involving implicit curves and trigonometric functions.

A useful general chain-rule pattern is

\[ \frac{d}{dx}\bigl[f(g(x))\bigr] = f'(g(x))g'(x) \]

Here, the outer function is \(f(u)=\cos u\) and the inner function is \(u=xy\). Thus,

\[ \frac{d}{dx}\bigl(\cos(xy)\bigr) = -\sin(xy)\frac{d}{dx}(xy). \]

Recognising the outer-function and inner-function structure helps competitive exam aspirants avoid missing chain-rule factors and reduces sign errors.

🔑 Key Takeaways
Key Takeaways
›
Key Takeaways  ·  9 points

  1. In implicit differentiation, \(y\) is treated as a function of \(x\).

  2. \[ \frac{d}{dx}(\sin^2y) = 2\sin y\cos y\frac{dy}{dx} = \sin 2y\frac{dy}{dx} \]

  3. \[ \frac{d}{dx}(xy) = x\frac{dy}{dx}+y \]

  4. \[ \frac{d}{dx}\bigl(\cos(xy)\bigr) = -\sin(xy) \left( x\frac{dy}{dx}+y \right) \]

  5. The constant \(k\) has derivative zero.

  6. Apply the product rule whenever \(x\) and \(y\) occur as a product.

  7. Apply the chain rule whenever a function contains another function as its argument.

  8. Collect and factor all terms containing \(\dfrac{dy}{dx}\) before isolating the derivative.

  9. The final result is

    \[ \boxed{ \frac{dy}{dx} = \frac{y\sin(xy)} {\sin 2y-x\sin(xy)} } \]
← Q6
7 / 15  ·  47%
Q8 →
Q8
NUMERIC3 marks
Find \(\dfrac{dy}{dx}\) in the following: \[\sin^2x+\cos^2y=1\]
📘 Concept & Theory
Concept/Theory
›

This question illustrates implicit differentiation involving trigonometric functions. Since \(y\) is a function of \(x\), the expression \(\cos^2y\) must be differentiated using the chain rule.

For a squared trigonometric function, the chain rule gives

\[\frac{d}{dx}(\sin^2x)=2\sin x\cos x=\sin 2x\]

Similarly,

\[\frac{d}{dx}(\cos^2y)=2\cos y\left(-\sin y\frac{dy}{dx}\right)\]

Therefore,

\[\frac{d}{dx}(\cos^2y)=-2\sin y\cos y\frac{dy}{dx}=-\sin 2y\frac{dy}{dx}\]

The derivative of the constant \(1\) is zero.

🗺️ Solution Roadmap
Step-by-step Plan
›

  1. Write the given trigonometric relation.

  2. Differentiate both sides with respect to \(x\).

  3. Differentiate \(\sin^2x\) using the chain rule.

  4. Differentiate \(\cos^2y\) using the chain rule and include \(\dfrac{dy}{dx}\).

  5. Use double-angle identities to simplify the expression.

  6. Isolate \(\dfrac{dy}{dx}\).

✏️ Solution
Complete Solution
›
Step-by-step Solution  ·  14 steps
  1. Given —
    \[\sin^2x+\cos^2y=1\]
  2. Differentiate both sides with respect to \(x\):
    \[\frac{d}{dx}(\sin^2x)+\frac{d}{dx}(\cos^2y)=\frac{d}{dx}(1)\]
  3. Differentiate \(\sin^2x\)
  4. Write \(\sin^2x\) as \((\sin x)^2\). By the chain rule,
    \[\frac{d}{dx}(\sin^2x)=2\sin x\cos x\]
  5. Using the double-angle identity
    \[2\sin x\cos x=\sin 2x,\]
  6. we obtain
    \[\frac{d}{dx}(\sin^2x)=\sin 2x\]
  7. Differentiate \(\cos^2y\)
  8. Since \(y\) is a function of \(x\), write
    \[\cos^2y=(\cos y)^2.\]
  9. Applying the chain rule,
    \[\frac{d}{dx}(\cos^2y)=2\cos y\frac{d}{dx}(\cos y)\]
  10. Again, because \(y\) depends on \(x\),
    \[\frac{d}{dx}(\cos y)=-\sin y\frac{dy}{dx}\]
  11. Therefore,
    \[\frac{d}{dx}(\cos^2y)=2\cos y\left(-\sin y\frac{dy}{dx}\right)\]
  12. Hence,
    \[\frac{d}{dx}(\cos^2y)=-2\sin y\cos y\frac{dy}{dx}\]
  13. Using
    \[2\sin y\cos y=\sin 2y,\]
  14. we get
    \[\frac{d}{dx}(\cos^2y)=-\sin 2y\frac{dy}{dx}\]
  15. Differentiate the constant \(1\)
  16. Since \(1\) is a constant,
    \[\frac{d}{dx}(1)=0\]
  17. Substitute the derivatives
  18. Substituting the derivatives into the differentiated equation:
    \[\sin 2x-\sin 2y\frac{dy}{dx}=0\]
  19. Isolate \(\dfrac{dy}{dx}\)
  20. Move the term containing \(\dfrac{dy}{dx}\) to the right-hand side:
    \[\sin 2x=\sin 2y\frac{dy}{dx}\]
  21. Divide both sides by \(\sin 2y\):
    \[\frac{dy}{dx}=\frac{\sin 2x}{\sin 2y}\]
  22. Hence, the required derivative is
    \[\boxed{\frac{dy}{dx}=\frac{\sin 2x}{\sin 2y}}\]
🎯 Exam Significance
Exam Significance
›

This problem is useful for practising the differentiation of composite trigonometric functions in an implicit equation. The key point is that \(\cos^2y\) contains \(y\), and therefore its differentiation must produce the factor \(\dfrac{dy}{dx}\).

A frequent mistake is to write

\[ \frac{d}{dx}(\cos^2y) = -2\sin y\cos y \]

without including \(\dfrac{dy}{dx}\). The correct derivative is

\[ \frac{d}{dx}(\cos^2y) = -2\sin y\cos y\frac{dy}{dx}. \]

Students should also be comfortable using the double-angle identities

\[ 2\sin x\cos x=\sin 2x \]
and
\[ 2\sin y\cos y=\sin 2y \]
to obtain the final simplified form.

Significance for Competitive Entrance Exam Aspirants

This problem reinforces the chain rule and helps develop quick recognition of composite trigonometric expressions. The pattern

\[ \frac{d}{dx}\bigl[f(y)\bigr] = f'(y)\frac{dy}{dx} \]

is fundamental in implicit differentiation questions.

Competitive-exam aspirants should also be able to move between equivalent forms such as

\[ 2\sin y\cos y \quad\text{and}\quad \sin 2y. \]

Such algebraic and trigonometric simplification can make the final answer considerably more compact.

🔑 Key Takeaways
Key Takeaways
›
Key Takeaways  ·  8 points

  1. Use implicit differentiation when \(x\) and \(y\) occur together in an equation.

  2. When differentiating a function of \(y\) with respect to \(x\), include \(\dfrac{dy}{dx}\).

  3. \[ \frac{d}{dx}(\sin^2x)=2\sin x\cos x=\sin 2x \]

  4. \[ \frac{d}{dx}(\cos^2y) = -2\sin y\cos y\frac{dy}{dx} = -\sin 2y\frac{dy}{dx} \]

  5. \[ \frac{d}{dx}(1)=0 \]

  6. Use double-angle identities to simplify the differentiated equation.

  7. Do not forget the negative sign arising from the derivative of \(\cos y\).

  8. The final result is

    \[ \boxed{ \frac{dy}{dx} = \frac{\sin 2x}{\sin 2y} } \]
← Q7
8 / 15  ·  53%
Q9 →
Q9
NUMERIC3 marks
Find \(\dfrac{dy}{dx}\) in the following: \[y=\sin^{-1}\left(\frac{2x}{1+x^2}\right)\]
📘 Concept & Theory
Concept/Theory
›

This question involves differentiation of an inverse trigonometric function. Although the given equation can be differentiated directly using the standard formula for \(\sin^{-1}u\), an especially clear method is to first convert the inverse trigonometric relation into an implicit trigonometric equation.

The standard derivative formula is

\[ \frac{d}{dx}\left(\sin^{-1}u\right) = \frac{1}{\sqrt{1-u^2}}\frac{du}{dx} \]

Here,

\[ u=\frac{2x}{1+x^2}. \]

Alternatively, from

\[ y=\sin^{-1}u, \]
we have

\[ \sin y=u. \]

Differentiating this relation gives

\[ \cos y\frac{dy}{dx}=\frac{du}{dx}. \]

Since \(u\) is a quotient, the quotient rule is required.

🗺️ Solution Roadmap
Step-by-step Plan
›

  1. Start with the inverse trigonometric equation.

  2. Convert \(y=\sin^{-1}u\) into \(\sin y=u\).

  3. Differentiate both sides with respect to \(x\).

  4. Use the chain rule on \(\sin y\).

  5. Differentiate \(\dfrac{2x}{1+x^2}\) using the quotient rule.

  6. Simplify the resulting expression.

  7. Use the relation \(\sin y=\dfrac{2x}{1+x^2}\) to determine \(\cos y\).

  8. Substitute \(\cos y\) and simplify to obtain \(\dfrac{dy}{dx}\).

✏️ Solution
Complete Solution
›
Step-by-step Solution  ·  32 steps
  1. Given —
    \[y=\sin^{-1}\left(\frac{2x}{1+x^2}\right)\]
  2. Convert the inverse trigonometric relation
  3. From
    \[y=\sin^{-1}\left(\frac{2x}{1+x^2}\right),\]
  4. taking sine on both sides gives
    \[ \sin y=\frac{2x}{1+x^2}\]
  5. Differentiate both sides with respect to \(x\)
  6. Differentiating,
    \[\frac{d}{dx}(\sin y)=\frac{d}{dx}\left(\frac{2x}{1+x^2}\right)\]
  7. Since \(y\) is a function of \(x\), the chain rule gives
    \[\frac{d}{dx}(\sin y)=\cos y\frac{dy}{dx}\]
  8. Therefore,
    \[\cos y\frac{dy}{dx}=\frac{d}{dx}\left(\frac{2x}{1+x^2}\right)\]
  9. Differentiate the quotient
  10. Let
    \[u=2x,\quad v=1+x^2.\]
  11. By the quotient rule,
    \[\frac{d}{dx}\left(\frac{u}{v}\right)=\frac{v\frac{du}{dx}-u\frac{dv}{dx}}{v^2}\]
  12. Now,
    \[\frac{du}{dx}=\frac{d}{dx}(2x)=2\]
    and
    \[\frac{dv}{dx}=\frac{d}{dx}(1+x^2)=2x\]
  13. Hence,
    \[\frac{d}{dx}\left(\frac{2x}{1+x^2}\right)=\frac{(1+x^2)(2)-(2x)(2x)}{(1+x^2)^2}\]
  14. Simplifying the numerator,
    \[=\frac{2+2x^2-4x^2}{(1+x^2)^2}\]
  15. Therefore,
    \[=\frac{2-2x^2}{(1+x^2)^2}\]
  16. Taking \(2\) common,
    \[=\frac{2(1-x^2)}{(1+x^2)^2}\]
  17. Thus,
    \[\cos y\frac{dy}{dx}=\frac{2(1-x^2)}{(1+x^2)^2}\]
  18. Solve for \(\dfrac{dy}{dx}\)
  19. Dividing both sides by \(\cos y\),
    \[\frac{dy}{dx}=\frac{2(1-x^2)}{(1+x^2)^2\cos y}\]
  20. To obtain an answer entirely in terms of \(x\), we now determine \(\cos y\).
  21. Find \(\cos y\)
  22. We already have
    \[\sin y=\frac{2x}{1+x^2}\]
  23. Using
    \[\cos^2y=1-\sin^2y,\]
  24. we get
    \[\cos^2y=1-\left(\frac{2x}{1+x^2}\right)^2\]
  25. Therefore,
    \[\cos^2y=1-\frac{4x^2}{(1+x^2)^2}\]
  26. Taking the common denominator,
    \[\cos^2y=\frac{(1+x^2)^2-4x^2}{(1+x^2)^2}\]
  27. Expand \((1+x^2)^2\)
    \[(1+x^2)^2=1+2x^2+x^4\]
  28. Hence,
    \[\cos^2y=\frac{1+2x^2+x^4-4x^2}{(1+x^2)^2}\]
  29. Simplifying,
    \[\cos^2y=\frac{1-2x^2+x^4}{(1+x^2)^2}\]
  30. Since
    \[1-2x^2+x^4=(1-x^2)^2,\]
  31. we obtain
    \[\cos^2y=\frac{(1-x^2)^2}{(1+x^2)^2}\]
  32. Thus,
    \[\cos y=\frac{|1-x^2|}{1+x^2}\]
  33. Here, the absolute value is important because \(y=\sin^{-1}(\cdots)\) has its principal value in
    \[ -\frac{\pi}{2}\leq y\leq\frac{\pi}{2}, \]
    so \(\cos y\geq0\).
  34. Substitute \(\cos y\)
  35. From
    \[\frac{dy}{dx}=\frac{2(1-x^2)}{(1+x^2)^2\cos y},\]
  36. substituting
    \[\cos y=\frac{|1-x^2|}{1+x^2},\]
  37. gives
    \[\frac{dy}{dx}=\frac{2(1-x^2)}{(1+x^2)^2\left(\frac{|1-x^2|}{1+x^2}\right)}\]
  38. Simplifying,
    \[\frac{dy}{dx}=\frac{2(1-x^2)}{(1+x^2)|1-x^2|}\]
  39. Therefore, for \(x\neq\pm1\),
    \[\boxed{\frac{dy}{dx}=\frac{2\,\operatorname{sgn}(1-x^2)}{1+x^2}}\]
  40. Equivalently, the derivative can be written piecewise as
    \[ \boxed{ \frac{dy}{dx} = \begin{cases} \dfrac{2}{1+x^2}, & |x|<1,\\[8pt] -\dfrac{2}{1+x^2}, & |x|>1. \end{cases} } \]
  41. Check the points \(x=\pm1\)
  42. At \(x=1\),
    \[y=\sin^{-1}(1)=\frac{\pi}{2}\]
  43. At \(x=-1\),
    \[y=\sin^{-1}(-1)=-\frac{\pi}{2}\]
    At both points, \(\cos y=0\), so the derivative formula obtained by division by \(\cos y\) is not defined. In fact, the function has a cusp at these points, so \(\dfrac{dy}{dx}\) does not exist there.
🎯 Exam Significance
Exam Significance
›

This problem is valuable because it combines inverse trigonometric functions, implicit differentiation, the quotient rule, the chain rule, and trigonometric identities. It also highlights an important issue involving the principal range of an inverse trigonometric function.

A key board-examination skill is knowing that

\[ y=\sin^{-1}u \]
implies
\[ \sin y=u, \]
after which implicit differentiation can be applied.

Students should also be careful when using

\[ \cos y=\sqrt{1-\sin^2y}. \]
The positive square root is appropriate here because the principal range of \(\sin^{-1}u\) is
\[ \left[-\frac{\pi}{2},\frac{\pi}{2}\right], \]
throughout which \(\cos y\geq0\).

Significance for Competitive Entrance Exam Aspirants

This question illustrates why inverse-trigonometric differentiation can sometimes require more than simply applying a memorised formula. The expression inside \(\sin^{-1}\) has a special structure that causes the derivative to change sign at \(x=\pm1\).

Competitive-exam aspirants should be comfortable with

\[1-\left(\frac{2x}{1+x^2}\right)^2=\frac{(1-x^2)^2}{(1+x^2)^2}\]

and should remember that

\[ \sqrt{(1-x^2)^2}=|1-x^2|, \]
not simply \(1-x^2\).

🔑 Key Takeaways
Key Takeaways
›
Key Takeaways  ·  10 points

  1. For an inverse trigonometric relation, an implicit-differentiation approach can simplify the calculation.

  2. \[ y=\sin^{-1}u \quad\Longrightarrow\quad \sin y=u \]

  3. \[ \frac{d}{dx}(\sin y) = \cos y\frac{dy}{dx} \]

  4. The quotient rule is required for \(\dfrac{2x}{1+x^2}\).

  5. \[ \frac{d}{dx} \left( \frac{2x}{1+x^2} \right) = \frac{2(1-x^2)} {(1+x^2)^2} \]

  6. Because \(y\) is the principal value of \(\sin^{-1}\),

    \[ \cos y=\frac{|1-x^2|}{1+x^2}. \]

  7. For \(|x|<1\),

    \[ \frac{dy}{dx}=\frac{2}{1+x^2}. \]

  8. For \(|x|>1\),

    \[ \frac{dy}{dx}=-\frac{2}{1+x^2}. \]

  9. The derivative does not exist at \(x=\pm1\).

  10. The complete result is

    \[\boxed{\frac{dy}{dx}=\begin{cases}\dfrac{2}{1+x^2}, & |x|<1,\\[8pt]-\dfrac{2}{1+x^2}, & |x|>1,\end{cases}}\]
    with \(\dfrac{dy}{dx}\) undefined at \(x=\pm1\).

← Q8
9 / 15  ·  60%
Q10 →
Q10
NUMERIC3 marks
Find \(\dfrac{dy}{dx}\) in the following: \[ y=\tan^{-1}\left(\frac{3x-x^3}{1-3x^2}\right), \qquad -\frac{1}{\sqrt{3}} < x < \frac{1}{\sqrt{3}} \]
📘 Concept & Theory
Concept/Theory
›

This question involves inverse trigonometric functions, the chain rule, and the triple-angle identity for tangent. The given restriction on \(x\) is important because it allows the inverse tangent expression to be simplified without introducing an additional multiple of \(\pi\).

The standard identity

\[ \tan 3\theta = \frac{3\tan\theta-\tan^3\theta} {1-3\tan^2\theta} \]

becomes, on putting \(\tan\theta=x\),

\[ \tan 3\theta = \frac{3x-x^3}{1-3x^2}. \]

Therefore, if

\[ \theta=\tan^{-1}x, \]
then

\[ \frac{3x-x^3}{1-3x^2} = \tan(3\tan^{-1}x). \]

The given restriction

\[ -\frac{1}{\sqrt3} < x < \frac{1}{\sqrt3} \]
implies
\[ -\frac{\pi}{6}<\tan^{-1}x < \frac{\pi}{6}. \]
Consequently,
\[ -\frac{\pi}{2}<3\tan^{-1}x < \frac{\pi}{2}, \]
which lies within the principal range of \(\tan^{-1}\).

Hence,

\[ \tan^{-1}\left(\tan(3\tan^{-1}x)\right) = 3\tan^{-1}x. \]

This observation provides a much shorter and more elegant solution than applying the quotient rule directly to the complicated rational expression.

🗺️ Solution Roadmap
Step-by-step Plan
›

  1. Set \(\theta=\tan^{-1}x\), so that \(x=\tan\theta\).

  2. Use the triple-angle identity for \(\tan 3\theta\).

  3. Recognise the expression inside \(\tan^{-1}\) as \(\tan 3\theta\).

  4. Use the given restriction on \(x\) to determine the correct principal branch.

  5. Simplify \(y\) to \(3\tan^{-1}x\).

  6. Differentiate using the standard derivative of \(\tan^{-1}x\).

✏️ Solution
Complete Solution
›
Step-by-step Solution  ·  18 steps
  1. Given —
    \[y=\tan^{-1}\left(\frac{3x-x^3}{1-3x^2}\right)\]
  2. with
    \[-\frac{1}{\sqrt3} < x < \frac{1}{\sqrt3}\]
  3. Introduce a suitable substitution
  4. Let
    \[\theta=\tan^{-1}x\]
  5. Therefore,
    \[\tan\theta=x.\]
  6. Apply the triple-angle identity
  7. We know that
    \[\tan3\theta=\frac{3\tan\theta-\tan^3\theta}{1-3\tan^2\theta}.\]
  8. Since
    \[\tan\theta=x,\]
    substitute \(x\) for \(\tan\theta\):
    \[\tan3\theta=\frac{3x-x^3}{1-3x^2}\]
  9. Hence,
    \[\frac{3x-x^3}{1-3x^2}=\tan3\theta\]
  10. Since
    \[\theta=\tan^{-1}x,\]
  11. we have
    \[\frac{3x-x^3}{1-3x^2}=\tan\left(3\tan^{-1}x\right)\]
  12. Rewrite the given equation
  13. Substituting this result into the given expression,
    \[y=\tan^{-1}\left[\tan\left(3\tan^{-1}x\right)\right]\]
  14. Use the given restriction on \(x\)
  15. We are given
    \[-\frac{1}{\sqrt3} < x < \frac{1}{\sqrt3}.\]
  16. Since
    \[\tan^{-1}\left(\frac{1}{\sqrt3}\right)=\frac{\pi}{6},\]
    and \(\tan^{-1}x\) is an increasing function,
  17. \[-\frac{\pi}{6}<\tan^{-1}x < \frac{\pi}{6}.\]
  18. Multiplying the entire inequality by \(3\),
    \[-\frac{\pi}{2} < 3\tan^{-1}x < \frac{\pi}{2}\]
  19. The interval
    \[\left(-\frac{\pi}{2},\frac{\pi}{2}\right)\]
    is the principal range of \(\tan^{-1}\).
  20. Therefore,
    \[\tan^{-1}\left[\tan\left(3\tan^{-1}x\right)\right]=3\tan^{-1}x\]
  21. Hence, the given equation simplifies to
    \[y=3\tan^{-1}x\]
  22. Differentiate with respect to \(x\)
  23. Differentiating both sides,
    \[\frac{dy}{dx}=3\frac{d}{dx}(\tan^{-1}x)\]
  24. Using
    \[\frac{d}{dx}(\tan^{-1}x)=\frac{1}{1+x^2},\]
  25. we get
    \[\frac{dy}{dx}=\frac{3}{1+x^2}\]
  26. Therefore,
    \[\boxed{\frac{dy}{dx}=\frac{3}{1+x^2}}\]
🎯 Exam Significance
Exam Significance
›

This question is important because it tests more than routine differentiation. It requires students to recognise a standard trigonometric identity and understand why the restriction on \(x\) has been supplied.

The key identity is

\[ \tan3\theta = \frac{3\tan\theta-\tan^3\theta} {1-3\tan^2\theta}. \]

Recognising this structure converts a complicated-looking inverse trigonometric expression into the simple function

\[ y=3\tan^{-1}x. \]

The given interval is not incidental. It ensures that

\[ 3\tan^{-1}x \]
remains within the principal range of \(\tan^{-1}\), allowing
\[ \tan^{-1}(\tan\theta)=\theta \]
to be used directly.

Significance for Competitive Entrance Exam Aspirants

For competitive examinations, this is a classic example of an identity-recognition problem. Direct differentiation is possible, but identifying the triple-angle pattern is substantially more efficient.

Whenever an expression resembles

\[ \frac{3x-x^3}{1-3x^2}, \]

students should immediately check the identity

\[ \tan3\theta = \frac{3\tan\theta-\tan^3\theta} {1-3\tan^2\theta}. \]

The problem also reinforces an important inverse-function principle:

\[ \tan^{-1}(\tan\theta)=\theta \]
only when \(\theta\) lies within the principal range
\[ \left(-\frac{\pi}{2},\frac{\pi}{2}\right). \]
Recognising domain and range restrictions is therefore essential.

🔑 Key Takeaways
Key Takeaways
›
Key Takeaways  ·  9 points

  1. Recognise the triple-angle identity for tangent:

    \[ \tan3\theta = \frac{3\tan\theta-\tan^3\theta} {1-3\tan^2\theta}. \]

  2. Put \(\theta=\tan^{-1}x\), so that \(\tan\theta=x\).

  3. Then

    \[ \frac{3x-x^3}{1-3x^2} = \tan(3\tan^{-1}x). \]

  4. The restriction

    \[ -\frac{1}{\sqrt3} < x < \frac{1}{\sqrt3} \]
    gives
    \[ -\frac{\pi}{2} < 3\tan^{-1}x<\frac{\pi}{2}. \]

  5. Therefore,

    \[ \tan^{-1}\left(\tan(3\tan^{-1}x)\right) = 3\tan^{-1}x. \]

  6. Hence,

    \[ y=3\tan^{-1}x. \]

  7. Using

    \[ \frac{d}{dx}(\tan^{-1}x)=\frac{1}{1+x^2}, \]
    we obtain
    \[ \frac{dy}{dx}=\frac{3}{1+x^2}. \]

  8. The given restriction on \(x\) is essential for selecting the correct principal branch.

  9. For this question, identity recognition is more efficient than direct quotient-rule differentiation.
← Q9
10 / 15  ·  67%
Q11 →
Q11
NUMERIC3 marks
Find \(\dfrac{dy}{dx}\) in the following: \[y=\cos^{-1}\left(\frac{1-x^2}{1+x^2}\right), \quad 0 < x < 1\]
📘 Concept & Theory
Concept/Theory
›

This question involves implicit differentiation, the quotient rule, the identity

\[ \sin^2y+\cos^2y=1, \]
and the principal range of the inverse cosine function.

Since

\[ y=\cos^{-1}\left(\frac{1-x^2}{1+x^2}\right), \]
we can write

\[ \cos y=\frac{1-x^2}{1+x^2}. \]

Differentiating implicitly gives a term containing \(\sin y\). We therefore need to determine \(\sin y\) in terms of \(x\).

The condition

\[ 0 < x < 1 \]
is important because it ensures \(x>0\). Consequently, when the square root involving \(x^2\) is simplified, we have
\[ \sqrt{x^2}=x. \]

There is also a useful trigonometric identity behind the expression:

\[ \cos(2\theta) = \frac{1-\tan^2\theta}{1+\tan^2\theta}. \]

Thus, if \(x=\tan\theta\), then

\[ \frac{1-x^2}{1+x^2}=\cos2\theta. \]
Because \(0 < x < 1\), we have
\[ 0<\tan^{-1}x<\frac{\pi}{4}, \]
and hence
\[ 0<2\tan^{-1}x<\frac{\pi}{2}. \]
Therefore, the inverse cosine can also be simplified as
\[ y=2\tan^{-1}x. \]
This provides a shorter alternative solution.
🗺️ Solution Roadmap
Step-by-step Plan
›

  1. Rewrite the inverse cosine equation as \(\cos y=\dfrac{1-x^2}{1+x^2}\).

  2. Differentiate both sides implicitly.

  3. Differentiate the rational expression using the quotient rule.

  4. Find \(\sin y\) using \(\sin^2y=1-\cos^2y\).

  5. Use \(0

  6. Substitute \(\sin y\) into the differentiated equation.

  7. Simplify to obtain \(\dfrac{dy}{dx}\).

✏️ Solution
Complete Solution
›
Step-by-step Solution  ·  33 steps
  1. Given —
    \[y=\cos^{-1}\left(\frac{1-x^2}{1+x^2}\right), \quad 0 < x < 1\]
  2. Remove the inverse cosine
  3. From
    \[y=\cos^{-1}\left(\frac{1-x^2}{1+x^2}\right)\]
  4. we obtain
    \[\cos y=\frac{1-x^2}{1+x^2}\]
  5. Differentiate both sides
  6. Differentiating the left-hand side with respect to \(x\),
    \[\frac{d}{dx}(\cos y)=-\sin y\frac{dy}{dx}\]
  7. Therefore,
    \[-\sin y\frac{dy}{dx}=\frac{d}{dx}\left(\frac{1-x^2}{1+x^2}\right)\]
  8. Differentiate the rational expression
  9. Using the quotient rule,
    \[\frac{d}{dx}\left(\frac{1-x^2}{1+x^2}\right)=\frac{(1+x^2)\dfrac{d}{dx}(1-x^2)-(1-x^2)\dfrac{d}{dx}(1+x^2)}{(1+x^2)^2}\]
  10. Now,
    \[\frac{d}{dx}(1-x^2)=-2x\]
    and
    \[\frac{d}{dx}(1+x^2)=2x\]
  11. Hence,
    \[\frac{d}{dx}\left(\frac{1-x^2}{1+x^2}\right)=\frac{(1+x^2)(-2x)-(1-x^2)(2x)}{(1+x^2)^2}\]
  12. Expanding the numerator,
    \[(1+x^2)(-2x)=-2x-2x^3\]
    and
    \[(1-x^2)(2x)=2x-2x^3\]
  13. Therefore,
    \[\frac{d}{dx}\left(\frac{1-x^2}{1+x^2}\right)=\frac{-2x-2x^3-(2x-2x^3)}{(1+x^2)^2}\]
  14. Thus,
    \[\frac{d}{dx}\left(\frac{1-x^2}{1+x^2}\right)=\frac{-2x-2x^3-2x+2x^3}{(1+x^2)^2}\]
  15. Combining like terms,
    \[\frac{d}{dx}\left(\frac{1-x^2}{1+x^2}\right)=\frac{-4x}{(1+x^2)^2}\]
  16. Therefore, the differentiated equation becomes
    \[-\sin y\frac{dy}{dx}=-\frac{4x}{(1+x^2)^2}\]
  17. Find \(\sin y\)
  18. From the identity
    \[\sin^2y+\cos^2y=1\]
  19. we have
    \[\sin y=\sqrt{1-\cos^2y}\]
  20. Since
    \[ y=\cos^{-1}(\cdots), \]
    the principal range of \(y\) is
    \[ 0\leq y\leq\pi. \]
    Hence \(\sin y\geq0\), so the positive square root must be taken.
  21. Using
    \[ \cos y=\frac{1-x^2}{1+x^2}, \]
  22. we get
    \[\sin y=\sqrt{1-\left(\frac{1-x^2}{1+x^2}\right)^2}\]
  23. Taking the common denominator,
    \[\sin y=\sqrt{\frac{(1+x^2)^2-(1-x^2)^2}{(1+x^2)^2}}\]
  24. Alternatively, using
    \[1-a^2=(1+a)(1-a),\]
    with
    \[a=\frac{1-x^2}{1+x^2},\]
  25. we get
    \[\sin y=\sqrt{\left(1+\frac{1-x^2}{1+x^2}\right)\left(1-\frac{1-x^2}{1+x^2}\right)}\]
  26. Simplifying the first factor,
    \[1+\frac{1-x^2}{1+x^2}=\frac{1+x^2+1-x^2}{1+x^2}=\frac{2}{1+x^2}\]
  27. Similarly,
    \[1-\frac{1-x^2}{1+x^2}=\frac{1+x^2-1+x^2}{1+x^2}=\frac{2x^2}{1+x^2}\]
  28. Therefore,
    \[\sin y=\sqrt{\frac{2}{1+x^2}\cdot\frac{2x^2}{1+x^2}}\]
  29. Hence,
    \[\sin y=\sqrt{\frac{4x^2}{(1+x^2)^2}}\]
  30. Thus,
    \[\sin y=\frac{2|x|}{1+x^2}\]
  31. But the question specifies
    \[ 0 < x < 1, \]
    so \(x>0\) and therefore
    \[ |x|=x. \]
  32. Hence,
    \[\boxed{\sin y=\frac{2x}{1+x^2}}\]
  33. Substitute \(\sin y\)
  34. We have
    \[-\sin y\frac{dy}{dx}=-\frac{4x}{(1+x^2)^2}\]
  35. Substituting
    \[\sin y=\frac{2x}{1+x^2},\]
  36. we obtain
    \[-\frac{2x}{1+x^2}\frac{dy}{dx}=-\frac{4x}{(1+x^2)^2}\]
  37. Multiplying both sides by \(-1\),
    \[\frac{2x}{1+x^2}\frac{dy}{dx}=\frac{4x}{(1+x^2)^2}\]
  38. Dividing by
    \[\frac{2x}{1+x^2},\]
    which is non-zero because \(x>0\), we get
    \[\frac{dy}{dx}=\frac{4x}{(1+x^2)^2}\cdot\frac{1+x^2}{2x}\]
  39. Cancelling \(2x\) and one factor of \(1+x^2\),
    \[\boxed{\frac{dy}{dx}=\frac{2}{1+x^2}}\]
🎯 Exam Significance
Exam Significance
›

This question is important because it combines inverse trigonometric functions with differentiation and tests whether the student can correctly handle the sign of a square root.

The most important points are:

  • Use implicit differentiation after writing
    \[ \cos y=\frac{1-x^2}{1+x^2}. \]
  • Remember that
    \[ \frac{d}{dx}(\cos y) = -\sin y\frac{dy}{dx}. \]
  • When calculating \(\sin y\), the expression initially gives
    \[ \frac{2|x|}{1+x^2}. \]
  • The condition \(0 < x < 1\) allows us to write
    \[ |x|=x. \]
  • The domain restriction also permits the elegant simplification
    \[ y=2\tan^{-1}x. \]

For a board examination, the implicit differentiation method is a reliable approach because it follows directly from the standard differentiation rules. The identity-based method is shorter if the student recognises the double-angle structure.

Significance for Competitive Entrance Exam Aspirants

For competitive examinations, the expression

\[ \frac{1-x^2}{1+x^2} \]

should immediately suggest the identity

\[ \cos2\theta = \frac{1-\tan^2\theta}{1+\tan^2\theta}. \]

Putting

\[ \theta=\tan^{-1}x \]
converts the original problem into the differentiation of \(2\tan^{-1}x\).

This avoids the quotient rule and the subsequent calculation of \(\sin y\). Such identity recognition is particularly useful in time-constrained examinations.

However, students must not ignore the given condition

\[ 0 < x < 1. \]
It establishes the correct inverse-trigonometric branch and ensures that the simplification
\[ \cos^{-1}(\cos2\tan^{-1}x) = 2\tan^{-1}x \]
is valid.

🔑 Key Takeaways
Key Takeaways
›
Key Takeaways  ·  8 points

  1. Start with

    \[ \cos y=\frac{1-x^2}{1+x^2}. \]

  2. Differentiating implicitly gives

    \[ -\sin y\frac{dy}{dx} = -\frac{4x}{(1+x^2)^2}. \]

  3. From

    \[ \sin^2y+\cos^2y=1, \]
    we obtain
    \[ \sin y=\frac{2x}{1+x^2} \]
    because \(0 < x < 1\).

  4. Substitution gives

    \[ \frac{dy}{dx} = \frac{2}{1+x^2}. \]

  5. The expression can also be recognised using

    \[ \cos2\theta = \frac{1-\tan^2\theta}{1+\tan^2\theta}. \]

  6. With

    \[ \theta=\tan^{-1}x, \]
    the function becomes
    \[ y=2\tan^{-1}x. \]

  7. The condition \(0 < x < 1\) is essential for the correct branch and for replacing \(|x|\) by \(x\).

  8. Identity recognition provides the shortest solution, while implicit differentiation provides a systematic verification.

← Q10
11 / 15  ·  73%
Q12 →
Q12
NUMERIC3 marks
Find \(\dfrac{dy}{dx}\) in the following: \[y=\sin^{-1}\left(\frac{1-x^2}{1+x^2}\right)\]
📘 Concept & Theory
Concept/Theory
›

This question involves inverse trigonometric functions, implicit differentiation, the quotient rule, and careful consideration of the principal range of \(\sin^{-1}x\).

From

\[ y=\sin^{-1}\left(\frac{1-x^2}{1+x^2}\right), \]

we can write

\[ \sin y=\frac{1-x^2}{1+x^2}. \]

Differentiating implicitly gives

\[ \cos y\frac{dy}{dx} = \frac{d}{dx} \left( \frac{1-x^2}{1+x^2} \right). \]

The derivative of the rational expression is obtained using the quotient rule. We then determine \(\cos y\) from

\[ \cos^2y=1-\sin^2y. \]

A crucial point is that the principal range of \(\sin^{-1}x\) is

\[ -\frac{\pi}{2}\leq y\leq\frac{\pi}{2}, \]

so \(\cos y\geq0\). Therefore, the positive square root must be selected when calculating \(\cos y\).

There is also an important distinction from Question 11. Here no restriction such as \(0

🗺️ Solution Roadmap
Step-by-step Plan
›

  1. Rewrite the equation as \(\sin y=\dfrac{1-x^2}{1+x^2}\).

  2. Differentiate both sides implicitly.

  3. Evaluate the derivative of the rational expression using the quotient rule.

  4. Find \(\cos y\) using \(\cos^2y=1-\sin^2y\).

  5. Use the principal range of \(\sin^{-1}\) to select the positive square root.

  6. Keep \(|x|\) because no sign restriction on \(x\) is provided.

  7. Substitute \(\cos y\) and simplify \(\dfrac{dy}{dx}\).

✏️ Solution
Complete Solution
›
Step-by-step Solution  ·  34 steps
  1. Given —
    \[y=\sin^{-1}\left(\frac{1-x^2}{1+x^2}\right)\]
  2. Remove the inverse sine
  3. From the given equation,
    \[\sin y=\frac{1-x^2}{1+x^2}\]
  4. Differentiate implicitly
  5. Differentiating both sides with respect to \(x\),
    \[\frac{d}{dx}(\sin y)=\frac{d}{dx}\left(\frac{1-x^2}{1+x^2}\right)\]
  6. By the chain rule,
    \[\frac{d}{dx}(\sin y)=\cos y\frac{dy}{dx}\]
  7. Hence,
    \[\cos y\frac{dy}{dx}=\frac{d}{dx}\left(\frac{1-x^2}{1+x^2}\right)\]
  8. Differentiate the rational expression
  9. Using the quotient rule,
    \[\frac{d}{dx}\left(\frac{1-x^2}{1+x^2}\right)=\frac{(1+x^2)\dfrac{d}{dx}(1-x^2)-(1-x^2)\dfrac{d}{dx}(1+x^2)}{(1+x^2)^2}\]
  10. Now,
    \[\frac{d}{dx}(1-x^2)=-2x\]
    and
    \[\frac{d}{dx}(1+x^2)=2x\]
  11. Therefore,
    \[\frac{d}{dx}\left(\frac{1-x^2}{1+x^2}\right)=\frac{(1+x^2)(-2x)-(1-x^2)(2x)}{(1+x^2)^2}\]
  12. Expanding the numerator,
    \[(1+x^2)(-2x)=-2x-2x^3\]
    and
    \[(1-x^2)(2x)=2x-2x^3\]
  13. Hence,
    \[\frac{d}{dx}\left(\frac{1-x^2}{1+x^2}\right)=\frac{-2x-2x^3-(2x-2x^3)}{(1+x^2)^2}\]
  14. Therefore,
    \[=\frac{-2x-2x^3-2x+2x^3}{(1+x^2)^2}\]
  15. Combining like terms,
    \[\boxed{\frac{d}{dx}\left(\frac{1-x^2}{1+x^2}\right)=-\frac{4x}{(1+x^2)^2}}\]
  16. Thus,
    \[\cos y\frac{dy}{dx}=-\frac{4x}{(1+x^2)^2}\]
  17. Find \(\cos y\)
  18. We know that
    \[\sin^2y+\cos^2y=1\]
  19. Therefore,
    \[\cos y=\sqrt{1-\sin^2y}\]
  20. From the original equation,
    \[\sin y=\frac{1-x^2}{1+x^2}\]
  21. Hence,
    \[\cos y=\sqrt{1-\left(\frac{1-x^2}{1+x^2}\right)^2}\]
  22. Using
    \[1-a^2=(1-a)(1+a),\]
  23. we get
    \[\cos y=\sqrt{\left(1-\frac{1-x^2}{1+x^2}\right)\left(1+\frac{1-x^2}{1+x^2}\right)}\]
  24. Simplify the two factors
  25. First,
    \[1-\frac{1-x^2}{1+x^2}=\frac{1+x^2-1+x^2}{1+x^2}\]
  26. Therefore,
    \[1-\frac{1-x^2}{1+x^2}=\frac{2x^2}{1+x^2}\]
  27. Similarly,
    \[1+\frac{1-x^2}{1+x^2}=\frac{1+x^2+1-x^2}{1+x^2}\]
  28. Hence,
    \[1+\frac{1-x^2}{1+x^2}=\frac{2}{1+x^2}\]
  29. Thus,
    \[\cos y=\sqrt{\frac{2x^2}{1+x^2}\cdot\frac{2}{1+x^2}}\]
  30. Therefore,
    \[\cos y=\sqrt{\frac{4x^2}{(1+x^2)^2}}\]
  31. Since the principal range of \(\sin^{-1}\) is
    \[ \left[-\frac{\pi}{2},\frac{\pi}{2}\right], \]
    we have
    \[ \cos y\geq0. \]
  32. Hence the positive square root is required:
    \[\cos y=\frac{2|x|}{1+x^2}\]
    Notice that we must retain \(|x|\), because the question does not specify \(x>0\) or \(x < 0\).
  33. Therefore,
    \[\boxed{\cos y=\frac{2|x|}{1+x^2}}\]
  34. Substitute \(\cos y\)
  35. We already obtained
    \[\cos y\frac{dy}{dx}=-\frac{4x}{(1+x^2)^2}\]
  36. Substituting
    \[\cos y=\frac{2|x|}{1+x^2},\]
  37. we get
    \[\frac{2|x|}{1+x^2}\frac{dy}{dx}=-\frac{4x}{(1+x^2)^2}\]
  38. Therefore,
    \[\frac{dy}{dx}=-\frac{4x}{(1+x^2)^2}\cdot\frac{1+x^2}{2|x|}\]
  39. Cancelling the common factors,
    \[\frac{dy}{dx}=-\frac{2x}{|x|(1+x^2)}\]
  40. Since
    \[\frac{x}{|x|}=\begin{cases}1, & x>0,\\-1, & x < 0,\end{cases}\]
  41. the derivative is piecewise:
    \[ \boxed{ \frac{dy}{dx} = \begin{cases} -\dfrac{2}{1+x^2}, & x>0,\\[8pt] \dfrac{2}{1+x^2}, & x < 0. \end{cases} } \]
  42. The original expression is well-defined for every real \(x\), because
    \[ 1+x^2>0. \]
    However, at \(x=0\), the derivative does not exist.
🎯 Exam Significance
Exam Significance
›

This question is particularly useful because it demonstrates that sign restrictions cannot be ignored when simplifying square roots.

Students should remember

\[ \sqrt{x^2}=|x|, \]

not \(x\) for every real value of \(x\).

The principal range of the inverse sine function is also important:

\[ -\frac{\pi}{2}\leq\sin^{-1}x\leq\frac{\pi}{2}. \]

Consequently, when

\[ y=\sin^{-1}(\cdots), \]
we know that
\[ \cos y\geq0. \]

This determines the correct sign when calculating \(\cos y\) from \(\cos^2y\).

For board examinations, the implicit differentiation method is a robust method because every step follows from standard differentiation and trigonometric identities.

Significance for Competitive Entrance Exam Aspirants

For competitive examinations, the expression

\[ \frac{1-x^2}{1+x^2} \]

should immediately suggest the identity

\[ \cos2\theta = \frac{1-\tan^2\theta}{1+\tan^2\theta}. \]

However, identity recognition alone is not sufficient. The principal branch of \(\sin^{-1}\) must also be considered. This is a good example of why inverse-trigonometric identities cannot always be manipulated as though they were ordinary trigonometric identities.

Competitive examination problems frequently use expressions such as

\[ \sqrt{x^2},\quad \sin^{-1}(\sin x),\quad \cos^{-1}(\cos x) \]
to test awareness of domains, ranges, absolute values, and principal branches.

🔑 Key Takeaways
Key Takeaways
›
Key Takeaways  ·  9 points

  1. From

    \[ y=\sin^{-1}\left(\frac{1-x^2}{1+x^2}\right), \]
    write
    \[ \sin y=\frac{1-x^2}{1+x^2}. \]

  2. Quotient-rule differentiation gives

    \[ \frac{d}{dx} \left( \frac{1-x^2}{1+x^2} \right) = -\frac{4x}{(1+x^2)^2}. \]

  3. Therefore,

    \[ \cos y\frac{dy}{dx} = -\frac{4x}{(1+x^2)^2}. \]

  4. Since

    \[ \cos^2y=1-\sin^2y, \]
    we obtain
    \[ \cos y=\frac{2|x|}{1+x^2}. \]

  5. The absolute value is essential because the question does not restrict the sign of \(x\).

  6. Hence,

    \[ \frac{dy}{dx} = -\frac{2x}{|x|(1+x^2)}. \]

  7. In piecewise form,

    \[ \frac{dy}{dx} = \begin{cases} -\dfrac{2}{1+x^2}, & x>0,\\[8pt] \dfrac{2}{1+x^2}, & x<0. \end{cases} \]

  8. The function is not differentiable at \(x=0\), because the left-hand and right-hand derivatives are \(2\) and \(-2\), respectively.

  9. The identity

    \[ \cos2\theta = \frac{1-\tan^2\theta}{1+\tan^2\theta} \]
    provides an alternative route to the same result.

← Q11
12 / 15  ·  80%
Q13 →
Q13
NUMERIC3 marks
Find \(\dfrac{dy}{dx}\) in the following: \[ y=\cos^{-1}\left(\frac{2x}{1+x^2}\right), \qquad -1 < x < 1 \]
📘 Concept & Theory
Concept/Theory
›

This question involves implicit differentiation, the quotient rule, the identity

\[ \sin^2y+\cos^2y=1, \]
and careful handling of the sign of a square root.

Starting with

\[ y=\cos^{-1}\left(\frac{2x}{1+x^2}\right), \]
we write

\[ \cos y=\frac{2x}{1+x^2}. \]

Differentiating implicitly,

\[ -\sin y\frac{dy}{dx} = \frac{d}{dx} \left( \frac{2x}{1+x^2} \right). \]

We then determine \(\sin y\) using

\[ \sin^2y=1-\cos^2y. \]

The restriction

\[ -1 < x < 1 \]
is essential. It ensures that both \(1+x\) and \(1-x\) are positive. Therefore, when a square root produces
\[ \sqrt{(1+x)^2(1-x)^2}, \]
we can correctly simplify it to
\[ (1+x)(1-x)=1-x^2. \]

🗺️ Solution Roadmap
Step-by-step Plan
›

  1. Rewrite the inverse cosine equation as \(\cos y=\dfrac{2x}{1+x^2}\).

  2. Differentiate both sides implicitly.

  3. Use the quotient rule to differentiate \(\dfrac{2x}{1+x^2}\).

  4. Find \(\sin y\) from \(\sin^2y=1-\cos^2y\).

  5. Use \(-1 < x < 1\) to determine the correct sign of \(\sin y\).

  6. Substitute \(\sin y\) into the differentiated equation.

  7. Simplify to obtain \(\dfrac{dy}{dx}\).

✏️ Solution
Complete Solution
›
Step-by-step Solution  ·  38 steps
  1. Given —
    \[ y=\cos^{-1}\left(\frac{2x}{1+x^2}\right), \qquad -1 < x < 1. \]
  2. Remove the inverse cosine
  3. From the given equation,
    \[\cos y=\frac{2x}{1+x^2}\]
  4. Differentiate implicitly
  5. Differentiating both sides with respect to \(x\),
    \[\frac{d}{dx}(\cos y)=\frac{d}{dx}\left(\frac{2x}{1+x^2}\right)\]
  6. By the chain rule,
    \[\frac{d}{dx}(\cos y)=-\sin y\frac{dy}{dx}\]
  7. Hence,
    \[-\sin y\frac{dy}{dx}=\frac{d}{dx}\left(\frac{2x}{1+x^2}\right)\]
  9. Using the quotient rule,
    \[\frac{d}{dx}\left(\frac{2x}{1+x^2}\right)=\frac{(1+x^2)\dfrac{d}{dx}(2x)-(2x)\dfrac{d}{dx}(1+x^2)}{(1+x^2)^2}\]
  10. Now,
    \[\frac{d}{dx}(2x)=2\]
    and
    \[\frac{d}{dx}(1+x^2)=2x\]
  11. Therefore,
    \[\frac{d}{dx}\left(\frac{2x}{1+x^2}\right)=\frac{2(1+x^2)-(2x)(2x)}{(1+x^2)^2}\]
  12. Expanding the numerator,
    \[=\frac{2+2x^2-4x^2}{(1+x^2)^2}\]
  13. Combining like terms,
    \[=\frac{2-2x^2}{(1+x^2)^2}\]
  14. Taking \(2\) common,
    \[\boxed{\frac{d}{dx}\left(\frac{2x}{1+x^2}\right)=\frac{2(1-x^2)}{(1+x^2)^2}}\]
  15. Therefore,
    \[-\sin y\frac{dy}{dx}=\frac{2(1-x^2)}{(1+x^2)^2}\]
  16. Find \(\sin y\)
  17. Using
    \[\sin^2y+\cos^2y=1\]
  18. we have
    \[\sin y=\sqrt{1-\cos^2y}\]
  19. Since
    \[y=\cos^{-1}(\cdots),\]
    the principal range of \(y\) is
    \[0\leq y\leq\pi.\]
    Thus \(\sin y\geq0\), so the positive square root is required.
  20. Since
    \[\cos y=\frac{2x}{1+x^2},\]
  21. we obtain
    \[\sin y=\sqrt{1-\left(\frac{2x}{1+x^2}\right)^2}\]
  22. Using \(1-a^2=(1-a)(1+a)\),
  23. \[\sin y=\sqrt{\left(1-\frac{2x}{1+x^2}\right)\left(1+\frac{2x}{1+x^2}\right)}\]
  24. Simplifying the first factor,
    \[1-\frac{2x}{1+x^2}=\frac{1+x^2-2x}{1+x^2}\]
  25. Since
    \[1+x^2-2x=(1-x)^2,\]
  26. we get
    \[1-\frac{2x}{1+x^2}=\frac{(1-x)^2}{1+x^2}\]
  27. Similarly,
    \[1+\frac{2x}{1+x^2}=\frac{1+x^2+2x}{1+x^2}\]
  28. Since
    \[1+x^2+2x=(1+x)^2,\]
  29. we obtain
    \[1+\frac{2x}{1+x^2}=\frac{(1+x)^2}{1+x^2}\]
  30. Hence,
    \[\sin y=\sqrt{\frac{(1-x)^2}{1+x^2}\cdot\frac{(1+x)^2}{1+x^2}}\]
  31. Therefore,
    \[\sin y=\sqrt{\frac{(1-x)^2(1+x)^2}{(1+x^2)^2}}\]
  32. Thus,
    \[\sin y=\frac{|1-x||1+x|}{1+x^2}\]
  33. Now the given restriction is
    \[-1 < x < 1\]
  34. Therefore,
    \[1-x>0\]
    and
    \[1+x>0.\]
    Hence,
    \[|1-x|=1-x\]
    and
    \[|1+x|=1+x.\]
  35. Consequently,
    \[\sin y=\frac{(1-x)(1+x)}{1+x^2}\]
  36. Using
    \[(1-x)(1+x)=1-x^2,\]
  37. we obtain
    \[\boxed{\sin y=\frac{1-x^2}{1+x^2}}\]
  38. Substitute \(\sin y\)
  39. We have
    \[-\sin y\frac{dy}{dx}=\frac{2(1-x^2)}{(1+x^2)^2}\]
  40. Substituting
    \[\sin y=\frac{1-x^2}{1+x^2},\]
    gives
    \[-\frac{1-x^2}{1+x^2}\frac{dy}{dx}=\frac{2(1-x^2)}{(1+x^2)^2}\]
  41. Since \(-1 < x < 1\), we have
    \[1-x^2>0,\]
    so cancellation is valid.
  42. Therefore,
    \[-\frac{dy}{dx}=\frac{2}{1+x^2}\]
  43. Hence,
    \[\boxed{\frac{dy}{dx}=-\frac{2}{1+x^2}}\]
🎯 Exam Significance
Exam Significance
›

This question is important because it tests several fundamental differentiation skills together:

  • Implicit differentiation of an inverse trigonometric function.
  • Application of the quotient rule.
  • Use of the identity \(\sin^2y+\cos^2y=1\).
  • Correct handling of square roots and absolute values.
  • Use of a given interval to determine the correct sign.

A particularly important examination point is

\[ \sqrt{x^2}=|x|, \]

rather than simply \(x\). Here, the condition

\[ -1 allows us to determine the signs of \(1+x\) and \(1-x\) and hence remove the absolute-value symbols correctly.

The question can also be solved much more quickly by recognising the double-angle identity. Both approaches are useful for board preparation.

Significance for Competitive Entrance Exam Aspirants

For competitive examinations, the expression

\[ \frac{2x}{1+x^2} \]

should immediately suggest

\[ \sin2\theta = \frac{2\tan\theta}{1+\tan^2\theta}. \]

Taking

\[ \theta=\tan^{-1}x \]
gives

\[ \frac{2x}{1+x^2}=\sin2\tan^{-1}x. \]

The domain restriction then determines the correct principal branch of the inverse cosine. This reduces the problem to differentiating \(2\tan^{-1}x\).

This problem is therefore an excellent example of how identity recognition, domain analysis, and principal-value reasoning can replace lengthy algebraic manipulation.

🔑 Key Takeaways
Key Takeaways
›
Key Takeaways  ·  8 points

  1. Start with

    \[ \cos y=\frac{2x}{1+x^2}. \]

  2. Implicit differentiation gives

    \[ -\sin y\frac{dy}{dx} = \frac{2(1-x^2)}{(1+x^2)^2}. \]

  3. Using

    \[ \sin^2y+\cos^2y=1, \]
    we obtain
    \[ \sin y = \frac{2|1-x^2|^{1/2}}{\text{appropriate form}} \]
    and, more directly after factorisation,
    \[ \sin y=\frac{|1-x||1+x|}{1+x^2}. \]

  4. Since

    \[ -1 < x < 1, \]
    both \(1-x\) and \(1+x\) are positive. Hence
    \[ \sin y=\frac{1-x^2}{1+x^2}. \]

  5. Substitution gives

    \[ \frac{dy}{dx} = -\frac{2}{1+x^2}. \]

  6. The expression

    \[ \frac{2x}{1+x^2} \]
    can also be recognised as
    \[ \sin(2\tan^{-1}x). \]

  7. The interval restriction is essential for selecting the correct principal branch of \(\cos^{-1}\).

  8. Always retain absolute values when simplifying square roots unless the given domain provides sufficient information to remove them.

← Q12
13 / 15  ·  87%
Q14 →
Q14
NUMERIC3 marks
Find \(\dfrac{dy}{dx}\) in the following: \[ y=\sin^{-1}\left(2x\sqrt{1-x^2}\right), \quad -\frac{1}{\sqrt2} < x < \frac{1}{\sqrt2} \]
📘 Concept & Theory
Concept/Theory
›

This question involves implicit differentiation, the product rule, the chain rule, and careful treatment of the sign of \(\cos y\).

Starting with

\[ y=\sin^{-1}\left(2x\sqrt{1-x^2}\right), \]

we can write

\[ \sin y=2x\sqrt{1-x^2}. \]

Differentiating implicitly gives

\[ \cos y\frac{dy}{dx} = \frac{d}{dx} \left(2x\sqrt{1-x^2}\right). \]

The right-hand side requires both the product rule and the chain rule.

We also need \(\cos y\). Using

\[ \cos^2y=1-\sin^2y, \]

the expression simplifies to

\[ \cos^2y=(1-2x^2)^2. \]

However, we must be careful: this gives

\[ \cos y=|1-2x^2|, \]
not automatically \(1-2x^2\).

The given interval

\[ -\frac{1}{\sqrt2} < x < \frac{1}{\sqrt2} \]
does not ensure that \(1-2x^2\) is always positive. It changes sign at
\[ x=\pm\frac{1}{\sqrt2}. \]
Since the endpoints are excluded, we have
\[ x^2 < \frac12, \]
and therefore
\[ 1-2x^2>0. \]
Hence, within the specified interval,
\[ \cos y=1-2x^2. \]

This domain condition is therefore essential to the simplification.

🗺️ Solution Roadmap
Step-by-step Plan
›

  1. Rewrite the equation as \(\sin y=2x\sqrt{1-x^2}\).

  2. Differentiate both sides implicitly.

  3. Apply the product rule to \(2x\sqrt{1-x^2}\).

  4. Use the chain rule to differentiate \(\sqrt{1-x^2}\).

  5. Find \(\cos y\) using \(\cos^2y=1-\sin^2y\).

  6. Use the given interval to determine that \(\cos y=1-2x^2>0\).

  7. Substitute \(\cos y\) and simplify.

✏️ Solution
Complete Solution
›
Step-by-step Solution  ·  34 steps
  1. Given —
    \[ y=\sin^{-1}\left(2x\sqrt{1-x^2}\right), \qquad -\frac{1}{\sqrt2} < x < \frac{1}{\sqrt2}. \]
  2. Remove the inverse sine
  3. From the given equation,
    \[\sin y=2x\sqrt{1-x^2}.\]
  4. Differentiate implicitly
  5. Differentiating both sides with respect to \(x\),
    \[\frac{d}{dx}(\sin y)=\frac{d}{dx}\left(2x\sqrt{1-x^2}\right)\]
  6. Applying the chain rule to the left-hand side,
    \[\cos y\frac{dy}{dx}=\frac{d}{dx}\left(2x\sqrt{1-x^2}\right)\]
  7. Therefore,
    \[\frac{dy}{dx}=\frac{1}{\cos y}\frac{d}{dx}\left(2x\sqrt{1-x^2}\right)\]
  8. Differentiate \(2x\sqrt{1-x^2}\)
  9. Write
    \[2x\sqrt{1-x^2}=2x(1-x^2)^{1/2}\]
  10. Applying the product rule,
    \[\frac{d}{dx}\left[2x(1-x^2)^{1/2}\right]=(1-x^2)^{1/2}\frac{d}{dx}(2x)+2x\frac{d}{dx}(1-x^2)^{1/2}\]
  11. We have
    \[\frac{d}{dx}(2x)=2\]
  12. For the second derivative, using the chain rule,
    \[\frac{d}{dx}(1-x^2)^{1/2}=\frac12(1-x^2)^{-1/2}(-2x)\]
  13. Hence,
    \[\frac{d}{dx}(1-x^2)^{1/2}=-\frac{x}{\sqrt{1-x^2}}\]
  14. Therefore,
    \[\frac{d}{dx}\left(2x\sqrt{1-x^2}\right)=2\sqrt{1-x^2}-\frac{2x^2}{\sqrt{1-x^2}}\]
  15. Taking the common denominator,
    \[=\frac{2(1-x^2)-2x^2}{\sqrt{1-x^2}}\]
  16. Simplifying the numerator,
    \[=\frac{2-2x^2-2x^2}{\sqrt{1-x^2}}\]
  17. Therefore,
    \[\boxed{\frac{d}{dx}\left(2x\sqrt{1-x^2}\right)=\frac{2(1-2x^2)}{\sqrt{1-x^2}}}\]
  18. Thus, the differentiated equation becomes
    \[\cos y\frac{dy}{dx}=\frac{2(1-2x^2)}{\sqrt{1-x^2}}\]
  19. Find \(\cos y\)
  20. From
    \[\sin y=2x\sqrt{1-x^2},\]
  21. we have
    \[\cos^2y=1-\sin^2y\]
  22. Therefore,
    \[\cos^2y=1-\left(2x\sqrt{1-x^2}\right)^2\]
  23. Squaring the expression,
    \[\cos^2y=1-4x^2(1-x^2)\]
  24. Expanding,
    \[\cos^2y=1-4x^2+4x^4\]
  25. Rearranging in descending powers,
    \[\cos^2y=4x^4-4x^2+1\]
  26. Recognising the perfect square,
    \[4x^4-4x^2+1=(1-2x^2)^2\]
  27. Hence,
    \[\cos^2y=(1-2x^2)^2\]
  28. Therefore,
    \[\cos y=|1-2x^2|\]
  29. Use the given interval to determine the sign
  30. We are given
    \[-\frac{1}{\sqrt2} < x < \frac{1}{\sqrt2}\]
  31. Squaring both sides appropriately gives
    \[x^2 < \frac12\]
  32. Hence,
    \[2x^2 < 1\]
  33. and therefore
    \[1-2x^2>0\]
  34. Thus,
    \[|1-2x^2|=1-2x^2\]
  35. Therefore,
    \[\boxed{\cos y=1-2x^2}\]
  36. Substitute \(\cos y\)
  37. We have
    \[\cos y\frac{dy}{dx}=\frac{2(1-2x^2)}{\sqrt{1-x^2}}\]
  38. Substituting
    \[\cos y=1-2x^2,\]
  39. we obtain
    \[(1-2x^2)\frac{dy}{dx}=\frac{2(1-2x^2)}{\sqrt{1-x^2}}\]
  40. Since
    \[1-2x^2>0\]
    throughout the given interval, it is non-zero and can be cancelled:
    \[\frac{dy}{dx}=\frac{2}{\sqrt{1-x^2}}\]
  41. Hence,
    \[\boxed{\frac{dy}{dx}=\frac{2}{\sqrt{1-x^2}}}\]
🎯 Exam Significance
Exam Significance
›

This question is important for board examinations because it combines several differentiation techniques in a single problem:

  • Implicit differentiation.
  • Product rule.
  • Chain rule.
  • Trigonometric identities.
  • Manipulation of perfect squares.
  • Domain and principal-range analysis.

A common mistake is to write

\[ \sqrt{(1-2x^2)^2}=1-2x^2 \]

without checking its sign. The correct general result is

\[ \sqrt{(1-2x^2)^2}=|1-2x^2|. \]

Here, the given interval makes \(1-2x^2\) positive, which justifies removing the absolute-value sign.

Students should also recognise that the given interval is deliberately chosen to ensure

\[ -\frac{\pi}{2}<2\sin^{-1}x < \frac{\pi}{2}, \]
allowing the inverse-sine identity to be used directly.

Significance for Competitive Entrance Exam Aspirants

For competitive examinations, the expression

\[ 2x\sqrt{1-x^2} \]

strongly suggests the double-angle identity

\[ \sin2\theta=2\sin\theta\cos\theta. \]

Substituting

\[ x=\sin\theta \]
immediately gives

\[ 2x\sqrt{1-x^2}=\sin2\theta. \]

The given interval then establishes the correct principal branch, leading directly to

\[ y=2\sin^{-1}x. \]

This is considerably faster than performing product-rule differentiation followed by the calculation of \(\cos y\).

The problem therefore illustrates an important competitive-examination strategy: look for a trigonometric identity before beginning lengthy differentiation.

🔑 Key Takeaways
Key Takeaways
›
Key Takeaways  ·  10 points

  1. Start with

    \[ \sin y=2x\sqrt{1-x^2}. \]

  2. Implicit differentiation gives

    \[ \cos y\frac{dy}{dx} = \frac{2(1-2x^2)} {\sqrt{1-x^2}}. \]

  3. From

    \[ \cos^2y=1-\sin^2y, \]
    we obtain
    \[ \cos^2y=(1-2x^2)^2. \]

  4. Therefore,

    \[ \cos y=|1-2x^2|. \]

  5. The given interval implies

    \[ x^2 < \frac12, \]
    so
    \[ 1-2x^2>0. \]

  6. Hence,

    \[ \cos y=1-2x^2. \]

  7. Substitution and cancellation yield

    \[ \frac{dy}{dx} = \frac{2}{\sqrt{1-x^2}}. \]

  8. A shorter identity-based approach uses

    \[ 2x\sqrt{1-x^2} = \sin(2\sin^{-1}x). \]

  9. The given interval ensures that the principal branch gives

    \[ y=2\sin^{-1}x. \]

  10. Always examine the domain before removing an absolute value or applying an inverse-trigonometric identity.

← Q13
14 / 15  ·  93%
Q15 →
Q15
NUMERIC3 marks
Find \(\dfrac{dy}{dx}\) in the following: \[ y=\sec^{-1}\left(\frac{1}{2x^2-1}\right), \qquad 0 < x < \frac{1}{\sqrt2} \]
📘 Concept & Theory
Concept/Theory
›

This question involves inverse trigonometric functions, implicit differentiation, the Pythagorean identity

\[ \sin^2y+\cos^2y=1, \]
and careful analysis of the sign of \(\sin y\).

Since

\[ y=\sec^{-1}\left(\frac{1}{2x^2-1}\right), \]

we have

\[ \sec y=\frac{1}{2x^2-1}. \]

Taking reciprocals,

\[ \cos y=2x^2-1. \]

This transformation makes implicit differentiation particularly simple.

Differentiating,

\[ -\sin y\frac{dy}{dx}=4x. \]

Therefore,

\[ \frac{dy}{dx} = -\frac{4x}{\sin y}. \]

We then determine \(\sin y\) from

\[ \sin^2y=1-\cos^2y. \]

The given interval

\[ 0 < x < \frac{1}{\sqrt2} \]
implies
\[ 0 < x^2 < \frac12. \]
Hence
\[ -1 < 2x^2-1<0. \]
Thus \(\cos y < 0\). For the principal range of \(\sec^{-1}\), the corresponding angle lies in the second quadrant, where \(\sin y>0\). Therefore, the positive square root must be chosen when finding \(\sin y\).

🗺️ Solution Roadmap
Step-by-step Plan
›

  1. Rewrite the equation as \(\sec y=\dfrac{1}{2x^2-1}\).

  2. Take reciprocals to obtain \(\cos y=2x^2-1\).

  3. Differentiate implicitly.

  4. Find \(\sin y\) using \(\sin^2y=1-\cos^2y\).

  5. Use the given interval to determine the correct sign of \(\sin y\).

  6. Substitute \(\sin y\) into the differentiated equation.

  7. Simplify to obtain \(\dfrac{dy}{dx}\).

✏️ Solution
Complete Solution
›
Step-by-step Solution  ·  23 steps
  1. Given —
    \[ y=\sec^{-1}\left(\frac{1}{2x^2-1}\right), \quad 0 < x < \frac{1}{\sqrt2}. \]
  2. Remove the inverse secant
  3. From
    \[y=\sec^{-1}\left(\frac{1}{2x^2-1}\right),\]
  4. we obtain
    \[\sec y=\frac{1}{2x^2-1}\]
  5. Taking reciprocals,
    \[\cos y=2x^2-1\]
  6. Differentiate implicitly
  7. Differentiating both sides with respect to \(x\),
    \[\frac{d}{dx}(\cos y)=\frac{d}{dx}(2x^2-1)\]
  8. Applying the chain rule to the left-hand side,
    \[-\sin y\frac{dy}{dx}=4x\]
  9. Therefore,
    \[\frac{dy}{dx}=-\frac{4x}{\sin y}\]
  10. Find \(\sin y\)
  11. Using the Pythagorean identity,
    \[\sin^2y+\cos^2y=1\]
  12. Hence,
    \[\sin y=\sqrt{1-\cos^2y},\]
  13. where the sign will be determined from the given interval. Since
    \[\cos y=2x^2-1,\]
  14. we have
    \[\sin y=\sqrt{1-(2x^2-1)^2}\]
  15. Simplify the expression under the square root
  16. Expanding the square,
    \[(2x^2-1)^2=(2x^2)^2-2(2x^2)(1)+1^2\]
  17. Therefore,
    \[(2x^2-1)^2=4x^4-4x^2+1\]
  18. Hence,
    \[\sin y=\sqrt{1-(4x^4-4x^2+1)}\]
  19. Simplifying,
    \[\sin y=\sqrt{1-4x^4+4x^2-1}\]
  20. Therefore,
    \[\sin y=\sqrt{4x^2-4x^4}\]
  21. Taking \(4x^2\) common,
    \[\sin y=\sqrt{4x^2(1-x^2)}\]
  22. Hence,
    \[\sin y=2|x|\sqrt{1-x^2}\]
  23. Since the given condition is
    \[ 0 < x < \frac{1}{\sqrt2}, \]
    we have \(x>0\), and therefore
    \[ |x|=x. \]
  24. Thus,
    \[\boxed{\sin y=2x\sqrt{1-x^2}}\]
  25. Substitute \(\sin y\)
  26. We have already obtained
    \[\frac{dy}{dx}=-\frac{4x}{\sin y}\]
  27. Substituting
    \[\sin y=2x\sqrt{1-x^2},\]
  28. we get
    \[\frac{dy}{dx}=-\frac{4x}{2x\sqrt{1-x^2}}\]
  29. Since \(x>0\), \(x\neq0\), so cancellation is valid:
    \[\frac{dy}{dx}=-\frac{2}{\sqrt{1-x^2}}\]
  30. Therefore,
    \[\boxed{\frac{dy}{dx}=-\frac{2}{\sqrt{1-x^2}}}\]
🎯 Exam Significance
Exam Significance
›

This question is important because it combines inverse trigonometric functions with implicit differentiation and tests algebraic accuracy.

The most important steps are:

  • Convert the inverse secant equation into an ordinary trigonometric equation.
  • Take the reciprocal carefully:
    \[ \sec y=\frac{1}{2x^2-1} \quad\Longrightarrow\quad \cos y=2x^2-1. \]
  • Apply the chain rule:
    \[ \frac{d}{dx}(\cos y) = -\sin y\frac{dy}{dx}. \]
  • Be extremely careful when expanding
    \[ (2x^2-1)^2. \]
  • Use the given interval to determine the correct sign of the square root.

A common examination error is to lose the negative sign in

\[ -\sin y\frac{dy}{dx}=4x. \]
The negative sign must remain when solving for \(\dfrac{dy}{dx}\).

Significance for Competitive Entrance Exam Aspirants

For competitive examinations, the quickest route is to recognise that taking the reciprocal transforms the inverse secant expression into

\[ \cos y=2x^2-1. \]

This immediately suggests using

\[ \sin^2y=1-\cos^2y. \]

The resulting expression

\[ 1-(2x^2-1)^2 \]
should be simplified as

\[ 1-(2x^2-1)^2 = 4x^2(1-x^2). \]

This factorisation is much more useful than expanding without a clear target.

The question also reinforces the importance of checking the sign of a square root and using the supplied domain before cancelling factors.

🔑 Key Takeaways
Key Takeaways
›
Key Takeaways  ·  8 points

  1. From

    \[ y=\sec^{-1}\left(\frac{1}{2x^2-1}\right), \]
    we obtain
    \[ \cos y=2x^2-1. \]

  2. Implicit differentiation gives

    \[ -\sin y\frac{dy}{dx}=4x. \]

  3. Hence,

    \[ \frac{dy}{dx} = -\frac{4x}{\sin y}. \]

  4. Using

    \[ \sin^2y=1-\cos^2y, \]
    we obtain
    \[ \sin^2y = 1-(2x^2-1)^2. \]

  5. The correct algebra is

    \[ 1-(2x^2-1)^2 = 4x^2(1-x^2). \]

  6. Since

    \[ 0 < x < \frac{1}{\sqrt2}, \]
    we have
    \[ \sin y=2x\sqrt{1-x^2}. \]

  7. Substitution gives

    \[ \frac{dy}{dx} = -\frac{4x}{2x\sqrt{1-x^2}}. \]

  8. Therefore,

    \[ \boxed{ \frac{dy}{dx} = -\frac{2}{\sqrt{1-x^2}} }. \]
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Chapter Complete!

All 15 solutions for Continuity and Differentiability covered.

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ACADEMIA AETERNUM तमसो मा ज्योतिर्गमय · Est. 2025
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NCERT Class 12 Maths Ex 5.3 Q15 Solution
NCERT Class 12 Maths Ex 5.3 Q15 Solution — Complete Notes & Solutions · academia-aeternum.com
NCERT Class 12 Mathematics Chapter 5: Continuity and Differentiability – Exercise 5.3, Question 15 focuses on finding the derivative of an inverse trigonometric function using implicit differentiation. In this problem, the function is given as \[ y=\sec^{-1}\left(\frac{1}{2x^2-1}\right), \qquad 0
🎓 Class 12 📐 Mathematics 📖 NCERT ✅ Free Access 🏆 CBSE · JEE
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    Frequently Asked Questions

    The function is y = sec?¹(1/(2x² - 1)), where 0 < x < 1/v2.

    The problem is solved using implicit differentiation after converting the inverse secant equation into the equivalent relation cos y = 2x² - 1.

    Differentiating cos y = 2x² - 1 gives -sin y(dy/dx) = 4x, so dy/dx = -4x/sin y.

    Using sin²y + cos²y = 1 and cos y = 2x² - 1, we get sin y = 2xv(1 - x²) for the given domain.

    The condition 0 < x < 1/v2 helps determine the correct sign of the square root and ensures that x is positive when simplifying |x| to x.

    For 0 < x < 1/v2, the correct value is sin y = 2xv(1 - x²).

    The required derivative is dy/dx = -2/v(1 - x²).

    A common mistake is simplifying 1 - (4x4 - 4x² + 1) incorrectly. The correct result is 4x² - 4x4 = 4x²(1 - x²).

    Since v(x²) = |x|, we initially obtain sin y = 2|x|v(1 - x²). The given condition x > 0 then allows |x| = x.

    Yes. It reinforces inverse trigonometric functions, implicit differentiation, chain rule, domain-based sign selection, and algebraic simplification, which are important for Class 12 Mathematics examinations and entrance tests.

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