Ch 5  ·  Q–
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Chapter 5 Exercise 5.4 Solutions

Continuity and Differentiability

Step-by-Step Solutions to the NCERT Class 12 Mathematics Chapter 5 Exercise 5.4

Class 12 Mathematics Exercise 5.4 NCERT Solutions Continuity and Differentiability Class 12 Mathematics Chapter 5 CBSE Board Exam JEE Main CUET Continuity Differentiability Derivative Differentiation Chain Rule Product Rule Quotient Rule Logarithmic Differentiation Exponential Functions Trigonometric Functions
10 Questions
20–30 min Ideal time
Q1 Now at
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Q1
NUMERIC3 marks
Differentiate the following with respect to \(x\): \[y=\frac{e^x}{\sin x}\]
📘 Concept & Theory
Concept/Theory
›

The given function is a quotient of two differentiable functions. Therefore, the quotient rule of differentiation is the most direct method.

If

\[ y=\frac{u}{v}, \]
where \(u\) and \(v\) are differentiable functions of \(x\), then
\[ \frac{dy}{dx} = \frac{v\frac{du}{dx}-u\frac{dv}{dx}}{v^2}. \]

Here,

\[ u=e^x \]
and
\[ v=\sin x. \]

We use the standard derivatives

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

🗺️ Solution Roadmap
Step-by-step Plan
›

  1. Write the given function in the form \(y=\dfrac{u}{v}\).

  2. Identify \(u=e^x\) and \(v=\sin x\).

  3. Differentiate \(u\) and \(v\) separately.

  4. Apply the quotient rule carefully.

  5. Substitute the derivatives and simplify the numerator by taking \(e^x\) common.

  6. Write the final derivative in its simplest form.

✏️ Solution
Complete Solution
›
Step-by-step Solution  ·  9 steps
  1. Given —
    \[y=\frac{e^x}{\sin x}\]
  2. Let
    \[u=e^x\]
    and
    \[v=\sin x\]
  3. Differentiate \(u\) with respect to \(x\):
    \[\frac{du}{dx}=\frac{d}{dx}(e^x)=e^x\]
  4. Differentiate \(v\) with respect to \(x\):
    \[\frac{dv}{dx}=\frac{d}{dx}(\sin x)=\cos x\]
  5. Using the quotient rule,
    \[\frac{dy}{dx}=\frac{v\frac{du}{dx}-u\frac{dv}{dx}}{v^2}\]
  6. Substituting
    \[u=e^x,\quad v=\sin x,\quad \frac{du}{dx}=e^x,\quad \frac{dv}{dx}=\cos x,\]
  7. we get
    \[\frac{dy}{dx}=\frac{\sin x\left(e^x\right)-e^x\left(\cos x\right)}{\sin^2x}\]
  8. Therefore,
    \[\frac{dy}{dx}=\frac{e^x\sin x-e^x\cos x}{\sin^2x}\]
  9. Taking \(e^x\) common from the numerator,
    \[\frac{dy}{dx}=\frac{e^x(\sin x-\cos x)}{\sin^2x}\]
  10. Hence, the required derivative is
    \[\boxed{\frac{dy}{dx}=\frac{e^x(\sin x-\cos x)}{\sin^2x}}\]
🎯 Exam Significance
Exam Significance
›

This question tests the direct application of the quotient rule together with standard derivatives of the exponential and trigonometric functions. Such questions are useful for developing accuracy in differentiation, particularly when the numerator and denominator are both functions of \(x\).

  • It reinforces the correct order of terms in the quotient rule.
  • It tests the standard derivatives of \(e^x\) and \(\sin x\).
  • It develops the habit of identifying \(u\), \(v\), \(\dfrac{du}{dx}\), and \(\dfrac{dv}{dx}\) before substitution.
  • The final factorisation makes the answer concise and easier to verify.
  • Writing every intermediate step reduces sign errors, especially in the term \(v\dfrac{du}{dx}-u\dfrac{dv}{dx}\).
Significance for Competitive Entrance Exam Aspirants

The same differentiation technique forms the basis of more advanced problems involving logarithmic differentiation, maxima and minima, monotonicity, tangents and normals, and differential equations. In objective examinations, recognising the quotient-rule structure quickly can save time while maintaining accuracy.

A useful equivalent form is obtained by rewriting the function as

\[ y=e^x\csc x. \]
This permits the product rule to be used as an alternative verification method. However, the quotient rule provides the most straightforward route for the expression as originally given.

🔑 Key Takeaways
Key Takeaways
›
Key Takeaways  ·  6 points

  1. For \(\displaystyle y=\frac{u}{v}\), use

    \[ \frac{dy}{dx} = \frac{v\frac{du}{dx}-u\frac{dv}{dx}}{v^2}. \]

  2. Remember

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

  3. Remember

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

  4. The order \(v\dfrac{du}{dx}-u\dfrac{dv}{dx}\) must not be reversed.

  5. After applying the quotient rule, factor common terms wherever possible.

  6. The final result is

    \[ \boxed{ \frac{dy}{dx} = \frac{e^x(\sin x-\cos x)}{\sin^2x} }. \]
↑ Top
1 / 10  ·  10%
Q2 →
Q2
NUMERIC3 marks
Differentiate the following with respect to \(x\): \[y=e^{\sin^{-1}x}\]
📘 Concept & Theory
Concept/Theory
›

The given function is an exponential function of an inverse trigonometric function. Since one function is contained inside another, the chain rule of differentiation is required.

The chain rule states that if

\[ y=f(u) \]
and
\[ u=g(x), \]
then
\[ \frac{dy}{dx} = \frac{dy}{du}\frac{du}{dx}. \]

In this question,

\[ u=\sin^{-1}x \]
and
\[ y=e^u. \]

We also use the standard derivatives

\[ \frac{d}{du}(e^u)=e^u \]
and
\[ \frac{d}{dx}(\sin^{-1}x) = \frac{1}{\sqrt{1-x^2}}. \]

Therefore, the derivative can be obtained by multiplying the derivative of the outer function by the derivative of the inner function.

🗺️ Solution Roadmap
Step-by-step Plan
›

  1. Identify the inner function \(u=\sin^{-1}x\).

  2. Rewrite the given function as \(y=e^u\).

  3. Differentiate \(y=e^u\) with respect to \(u\).

  4. Differentiate \(u=\sin^{-1}x\) with respect to \(x\).

  5. Apply the chain rule

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

  6. Substitute \(u=\sin^{-1}x\) back into the result.

  7. Write the final answer in simplified form.

✏️ Solution
Complete Solution
›
Step-by-step Solution  ·  10 steps
  1. Given —
    \[y=e^{\sin^{-1}x}\]
  2. Let
    \[u=\sin^{-1}x\]
  3. Therefore, the given function becomes
    \[y=e^u\]
  4. Differentiate \(y=e^u\) with respect to \(u\):
    \[\frac{dy}{du}=\frac{d}{du}(e^u)=e^u\]
  5. Now differentiate
    \[u=\sin^{-1}x\]
  6. with respect to \(x\):
    \[\frac{du}{dx}=\frac{d}{dx}(\sin^{-1}x)=\frac{1}{\sqrt{1-x^2}}\]
  7. Using the chain rule,
    \[\frac{dy}{dx}=\frac{dy}{du}\frac{du}{dx}\]
  8. Substituting the values of \(\dfrac{dy}{du}\) and \(\dfrac{du}{dx}\),
    \[\frac{dy}{dx}=e^u\cdot\frac{1}{\sqrt{1-x^2}}\]
  9. Hence,
    \[\frac{dy}{dx}=\frac{e^u}{\sqrt{1-x^2}}\]
  10. Since
    \[u=\sin^{-1}x,\]
  11. substituting back, we obtain
    \[\boxed{\frac{dy}{dx}=\frac{e^{\sin^{-1}x}}{\sqrt{1-x^2}}}\]
🎯 Exam Significance
Exam Significance
›

This problem is a direct application of the chain rule and the derivative of an inverse trigonometric function. It is important for Class 12 board examinations because expressions involving exponential, logarithmic and inverse trigonometric functions frequently require nested differentiation.

  • It reinforces the concept of composite functions.
  • It tests the correct use of the chain rule.
  • It requires accurate recall of the derivative of \(\sin^{-1}x\).
  • Introducing \(u\) makes the nested structure easier to understand and reduces errors.
  • The domain restriction helps distinguish the function's domain from the interval on which its derivative is finite.
Significance for Competitive Entrance Exam Aspirants

Composite functions of the form

\[ e^{f(x)},\qquad \ln(f(x)),\qquad \sin(f(x)),\qquad \cos(f(x)) \]
are fundamental patterns in differentiation-based questions. Recognising the outer function and inner function quickly is particularly useful in time-bound objective examinations.

The pattern

\[ \frac{d}{dx}\left(e^{f(x)}\right) = e^{f(x)}f'(x) \]
should be treated as a standard chain-rule form. Here,
\[ f(x)=\sin^{-1}x. \]

🔑 Key Takeaways
Key Takeaways
›
Key Takeaways  ·  7 points

  1. When one function occurs inside another, check for the chain rule.

  2. For \(y=e^{u}\),

    \[ \frac{dy}{du}=e^u. \]

  3. For \(u=\sin^{-1}x\),

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

  4. The chain rule is

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

  5. For the real-valued function, \(\sin^{-1}x\) requires

    \[ -1\leq x\leq1. \]

  6. The derivative is finite for

    \[ -1 < x < 1. \]

  7. The final result is

    \[ \boxed{ \frac{dy}{dx} = \frac{e^{\sin^{-1}x}}{\sqrt{1-x^2}} }. \]
← Q1
2 / 10  ·  20%
Q3 →
Q3
NUMERIC3 marks
Differentiate the following with respect to \(x\): \[y=e^{x^3}\]
📘 Concept & Theory
Concept/Theory
›

The given function is an exponential function with a composite exponent. Since \(x^3\) is itself a function of \(x\), the chain rule of differentiation is required.

The chain rule states that if

\[ y=f(u) \]
and
\[ u=g(x), \]
then
\[ \frac{dy}{dx} = \frac{dy}{du}\cdot\frac{du}{dx}. \]

For an exponential function,

\[ \frac{d}{du}(e^u)=e^u. \]

Also, using the power rule,

\[ \frac{d}{dx}(x^n)=nx^{n-1}, \]
we have
\[ \frac{d}{dx}(x^3)=3x^2. \]

Therefore, for a function of the form

\[ e^{f(x)}, \]
the standard chain-rule result is
\[ \frac{d}{dx}\left(e^{f(x)}\right) = e^{f(x)}f'(x). \]

🗺️ Solution Roadmap
Step-by-step Plan
›

  1. Identify the inner function \(u=x^3\).

  2. Rewrite the given function as \(y=e^u\).

  3. Differentiate \(y=e^u\) with respect to \(u\).

  4. Differentiate \(u=x^3\) with respect to \(x\).

  5. Apply the chain rule.

  6. Substitute \(u=x^3\) back into the expression.

  7. Write the final answer in simplified form.

✏️ Solution
Complete Solution
›
Step-by-step Solution  ·  10 steps
  1. Given —
    \[y=e^{x^3}\]
  2. Let
    \[u=x^3\]
  3. Therefore, the given function becomes
    \[y=e^u\]
  4. Differentiate \(y=e^u\) with respect to \(u\):
    \[\frac{dy}{du}=\frac{d}{du}(e^u)=e^u\]
  5. Now differentiate
    \[u=x^3\]
    with respect to \(x\):
    \[\frac{du}{dx}=\frac{d}{dx}(x^3)\]
  6. Using the power rule,
    \[\frac{du}{dx}=3x^{3-1}=3x^2\]
  7. Using the chain rule,
    \[\frac{dy}{dx}=\frac{dy}{du}\cdot\frac{du}{dx}\]
  8. Substituting the values obtained above,
    \[\frac{dy}{dx}=e^u\cdot3x^2\]
  9. Therefore,
    \[\frac{dy}{dx}=3x^2e^u\]
  10. Since
    \[u=x^3\]
    substituting \(u=x^3\), we get
    \[\frac{dy}{dx}=3x^2e^{x^3}\]
  11. Hence, the required derivative is
    \[\boxed{\frac{dy}{dx}=3x^2e^{x^3}}\]
🎯 Exam Significance
Exam Significance
›

This is a fundamental example of the chain rule. It is important to understand why the factor \(3x^2\) appears in the answer. Differentiating only \(e^{x^3}\) as \(e^{x^3}\) would ignore the derivative of the inner function \(x^3\).

  • It reinforces the differentiation of composite functions.
  • It combines the derivative of an exponential function with the power rule.
  • It demonstrates the standard pattern for differentiating \(e^{f(x)}\).
  • It helps prevent the common mistake of omitting the derivative of the exponent.
  • The step-by-step \(u\)-substitution method provides a reliable approach for longer composite functions.
Significance for Competitive Entrance Exam Aspirants

The expression \(e^{f(x)}\) occurs frequently in objective questions involving differentiation, increasing and decreasing functions, tangents and normals, maxima and minima, and differential equations.

For rapid problem solving, the general result

\[ \frac{d}{dx}\left(e^{f(x)}\right) = e^{f(x)}f'(x) \]
is especially useful. In the present problem,
\[ f(x)=x^3 \]
and
\[ f'(x)=3x^2. \]

Hence,

\[ \frac{d}{dx}\left(e^{x^3}\right) = e^{x^3}(3x^2). \]

🔑 Key Takeaways
Key Takeaways
›
Key Takeaways  ·  7 points

  1. Identify the inner function whenever one function is contained inside another.

  2. For \(y=e^u\),

    \[ \frac{dy}{du}=e^u. \]

  3. For \(u=x^3\),

    \[ \frac{du}{dx}=3x^2. \]

  4. The chain rule is

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

  5. The standard result

    \[ \frac{d}{dx}\left(e^{f(x)}\right) = e^{f(x)}f'(x) \]
    should be recognised quickly in competitive examinations.

  6. Never omit the derivative of the exponent in an exponential composite function.

  7. The final answer is

    \[ \boxed{ \frac{dy}{dx}=3x^2e^{x^3} }. \]
← Q2
3 / 10  ·  30%
Q4 →
Q4
NUMERIC3 marks
Differentiate the following with respect to \(x\): \[y=\sin\left(\tan^{-1}e^{-x}\right)\]
📘 Concept & Theory
Concept/Theory
›

The given function contains three nested functions: an exponential function inside an inverse tangent function, which is itself inside a sine function. Therefore, the chain rule must be applied successively.

The structure is

\[ x \longrightarrow e^{-x} \longrightarrow \tan^{-1}(e^{-x}) \longrightarrow \sin\left(\tan^{-1}e^{-x}\right). \]

If

\[ y=f(u),\qquad u=g(v),\qquad v=h(x), \]
then the chain rule gives
\[ \frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dv} \cdot \frac{dv}{dx}. \]

The required standard derivatives are

\[ \frac{d}{du}(\sin u)=\cos u, \]
\[ \frac{d}{dv}(\tan^{-1}v) = \frac{1}{1+v^2}, \]
and
\[ \frac{d}{dx}(e^{-x}) = -e^{-x}. \]

Notice that the derivative of \(\tan^{-1}v\) has denominator \(1+v^2\), not \(\sqrt{1+v^2}\). This is an important correction to the supplied working.

🗺️ Solution Roadmap
Step-by-step Plan
›

  1. Take \(v=e^{-x}\).

  2. Take \(u=\tan^{-1}v\).

  3. Write the given function as \(y=\sin u\).

  4. Find \(\dfrac{dy}{du}\).

  5. Find \(\dfrac{du}{dv}\).

  6. Find \(\dfrac{dv}{dx}\).

  7. Multiply the three derivatives using the chain rule.

  8. Substitute \(u=\tan^{-1}e^{-x}\) and \(v=e^{-x}\).

  9. Optionally simplify the trigonometric factor using a right-triangle identity.

✏️ Solution
Complete Solution
›
Step-by-step Solution  ·  19 steps
  1. Given —
    \[y=\sin\left(\tan^{-1}e^{-x}\right)\]
  2. Let
    \[v=e^{-x}\]
  3. Then let
    \[u=\tan^{-1}v\]
  4. Therefore,
    \[y=\sin u\]
  5. Now differentiate \(y=\sin u\) with respect to \(u\):
    \[\frac{dy}{du}=\frac{d}{du}(\sin u)=\cos u\]
  6. Next, differentiate
    \[u=\tan^{-1}v\]
  7. with respect to \(v\):
    \[\frac{du}{dv}=\frac{d}{dv}\left(\tan^{-1}v\right)=\frac{1}{1+v^2}\]
  8. Now differentiate
    \[v=e^{-x}\]
    with respect to \(x\):
    \[\frac{dv}{dx}=\frac{d}{dx}(e^{-x})\]
  9. Using the chain rule for the exponential function,
    \[\frac{dv}{dx}=e^{-x}\frac{d}{dx}(-x)=-e^{-x}\]
  10. Using the chain rule for the three nested functions,
    \[\frac{dy}{dx}=\frac{dy}{du}\cdot\frac{du}{dv}\cdot\frac{dv}{dx}\]
  11. Substituting the derivatives obtained above,
    \[\frac{dy}{dx}=\cos u\cdot\frac{1}{1+v^2}\cdot(-e^{-x})\]
  12. Therefore,
    \[\frac{dy}{dx}=-\frac{e^{-x}\cos u}{1+v^2}\]
  13. Since
    \[u=\tan^{-1}e^{-x}\]
    and
    \[v=e^{-x}\]
  14. we obtain
    \[\frac{dy}{dx}=-\frac{e^{-x}\cos\left(\tan^{-1}e^{-x}\right)}{1+e^{-2x}}\]
  15. Hence, the required derivative is
    \[\boxed{\frac{dy}{dx}=-\frac{e^{-x}\cos\left(\tan^{-1}e^{-x}\right)}{1+e^{-2x}}}\]
  16. Further Simplification
  17. The trigonometric factor can be simplified using the identity
    \[ \cos(\tan^{-1}t)=\frac{1}{\sqrt{1+t^2}}, \]
    for real \(t\).
  18. Here,
    \[t=e^{-x}\]
  19. Therefore,
    \[\cos\left(\tan^{-1}e^{-x}\right)=\frac{1}{\sqrt{1+e^{-2x}}}\]
  20. Substituting this into the derivative,
    \[\frac{dy}{dx}=-\frac{e^{-x}}{(1+e^{-2x})\sqrt{1+e^{-2x}}}\]
  21. Hence,
    \[\boxed{\frac{dy}{dx}=-\frac{e^{-x}}{(1+e^{-2x})^{3/2}}}\]
🎯 Exam Significance
Exam Significance
›

This problem is particularly valuable because it combines several differentiation rules in a single expression. It tests whether the student can correctly handle multiple levels of composite functions.

  • It reinforces repeated application of the chain rule.
  • It tests the derivative of \(e^{-x}\), including the negative sign.
  • It tests the standard derivative of \(\tan^{-1}x\).
  • It tests the derivative of \(\sin x\).
  • It demonstrates how a complicated nested function can be broken into simpler intermediate functions.
  • The simplification using \(\cos(\tan^{-1}x)\) strengthens inverse-trigonometric identities useful in board examinations.
Significance for Competitive Entrance Exam Aspirants

Nested composite functions are common in objective questions where speed and accuracy are important. Instead of differentiating the entire expression at once, identifying successive inner functions provides a systematic method.

The general pattern

\[ \frac{d}{dx}f(g(h(x))) = f'(g(h(x)))g'(h(x))h'(x) \]
is directly applicable here.

In addition, the identity

\[ \cos(\tan^{-1}t)=\frac{1}{\sqrt{1+t^2}} \]
can convert the answer into an algebraic expression, which can be useful when comparing answer choices in an objective examination.

🔑 Key Takeaways
Key Takeaways
›
Key Takeaways  ·  7 points

  1. Use intermediate variables such as \(u\) and \(v\) to make the structure clear.

  2. Remember

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

  3. Remember

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

  4. Remember

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

  5. The direct derivative is

    \[ \boxed{ \frac{dy}{dx} = -\frac{ e^{-x}\cos(\tan^{-1}e^{-x}) }{ 1+e^{-2x} } }. \]

  6. An equivalent simplified form is

    \[ \boxed{ \frac{dy}{dx} = -\frac{e^{-x}} {(1+e^{-2x})^{3/2}} }. \]

  7. Another equivalent form using positive powers of \(e^x\) is

    \[ \boxed{ \frac{dy}{dx} = -\frac{e^{2x}} {(1+e^{2x})^{3/2}} }. \]
← Q3
4 / 10  ·  40%
Q5 →
Q5
NUMERIC3 marks
Differentiate the following with respect to \(x\): \[y=\log(\cos e^x)\]
📘 Concept & Theory
Concept/Theory
›

The given expression contains a nested composite function. The exponential function \(e^x\) occurs inside the cosine function, and the cosine function occurs inside the logarithm. Therefore, the chain rule must be applied successively.

The functional structure is

\[x\longrightarrow e^x\longrightarrow\cos(e^x)\longrightarrow\log(\cos e^x)\]

For a composite function with intermediate variables

\[ v=e^x,\qquad u=\cos v,\qquad y=\log u, \]
the chain rule gives
\[\frac{dy}{dx}=\frac{dy}{du}\cdot\frac{du}{dv}\cdot\frac{dv}{dx}\]

The standard derivatives required are

\[ \frac{d}{du}(\log u)=\frac{1}{u}, \]

\[ \frac{d}{dv}(\cos v)=-\sin v, \]

and

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

Therefore, the negative sign in the final answer comes from differentiating the cosine function.

🗺️ Solution Roadmap
Step-by-step Plan
›

  1. Identify the innermost function \(v=e^x\).

  2. Take the next intermediate function \(u=\cos v\).

  3. Write the outer function as \(y=\log u\).

  4. Find \(\dfrac{dy}{du}\).

  5. Find \(\dfrac{du}{dv}\).

  6. Find \(\dfrac{dv}{dx}\).

  7. Multiply the three derivatives using the chain rule.

  8. Substitute \(u=\cos e^x\) and \(v=e^x\).

  9. Simplify the result using the identity \(\tan\theta=\dfrac{\sin\theta}{\cos\theta}\).

✏️ Solution
Complete Solution
›
Step-by-step Solution  ·  15 steps
  1. Given —
    \[y=\log(\cos e^x)\]
  2. Let
    \[v=e^x\]
  3. Then
    \[\cos e^x=\cos v\]
  4. Now let
    \[u=\cos v\]
  5. Therefore,
    \[y=\log u\]
  6. Differentiate \(y=\log u\) with respect to \(u\):
    \[\frac{dy}{du}=\frac{d}{du}(\log u)=\frac{1}{u}\]
  7. Next, differentiate
    \[u=\cos v\]
    with respect to \(v\):
    \[\frac{du}{dv}=\frac{d}{dv}(\cos v)=-\sin v\]
  8. Now differentiate
    \[v=e^x\]
    with respect to \(x\):
    \[\frac{dv}{dx}=\frac{d}{dx}(e^x)=e^x\]
  9. Using the chain rule,
    \[\frac{dy}{dx}=\frac{dy}{du}\cdot\frac{du}{dv}\cdot\frac{dv}{dx}\]
  10. Substituting the derivatives,
    \[\frac{dy}{dx}=\frac{1}{u}\cdot(-\sin v)\cdot e^x\]
  11. Thus,
    \[\frac{dy}{dx}=-\frac{e^x\sin v}{u}\]
  12. Since
    \[u=\cos v\]
    and
    \[v=e^x\]
  13. we get
    \[\frac{dy}{dx}=-\frac{e^x\sin(e^x)}{\cos(e^x)}\]
  14. Using
    \[\tan\theta=\frac{\sin\theta}{\cos\theta}\]
  15. we obtain
    \[\frac{dy}{dx}=-e^x\tan(e^x)\]
  16. Hence, the required derivative is
    \[\boxed{\frac{dy}{dx}=-e^x\tan(e^x)}\]
🎯 Exam Significance
Exam Significance
›

This problem is a useful application of successive chain-rule differentiation. It combines logarithmic, trigonometric and exponential functions in a single expression, making it representative of the type of composite-function differentiation expected from Class 12 students.

  • It tests the derivative of \(\log x\).
  • It tests the derivative of \(\cos x\).
  • It tests the derivative of \(e^x\).
  • It demonstrates how the chain rule is applied through several nested functions.
  • It reinforces careful handling of the negative sign produced by differentiating cosine.
  • The final conversion from a sine-to-cosine ratio to tangent makes the answer concise.
Significance for Competitive Entrance Exam Aspirants

Composite functions involving logarithms, exponentials and trigonometric functions are important in objective examinations. A useful recognition pattern is

\[ \frac{d}{dx}\left[\log f(x)\right] = \frac{f'(x)}{f(x)}. \]

Here,

\[ f(x)=\cos(e^x). \]
Therefore,
\[ f'(x)=-e^x\sin(e^x), \]
giving
\[ \frac{d}{dx}\left[\log(\cos e^x)\right] = -e^x\tan(e^x). \]

Recognising this structure quickly can reduce the number of intermediate steps needed in a time-bound objective examination, while the multi-variable method remains useful for checking the calculation.

🔑 Key Takeaways
Key Takeaways
›
Key Takeaways  ·  7 points

  1. For nested functions, identify each layer before differentiating.

  2. Use

    \[ \frac{d}{dx}(\log f(x)) = \frac{f'(x)}{f(x)}. \]

  3. Remember

    \[ \frac{d}{dx}(\cos x)=-\sin x. \]

  4. Remember

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

  5. The negative sign in the answer comes from the derivative of the cosine function.

  6. The ratio

    \[ \frac{\sin(e^x)}{\cos(e^x)} \]
    can be written as
    \[ \tan(e^x). \]

  7. The final answer is

    \[ \boxed{ \frac{dy}{dx} = -e^x\tan(e^x) }. \]
← Q4
5 / 10  ·  50%
Q6 →
Q6
NUMERIC3 marks
Differentiate the following with respect to \(x\): \[e^x+e^{x^2}+ex^3+\ldots+e^{x^5}\]
📘 Concept & Theory
Concept/Theory
›

The given expression is a sum of functions. Therefore, we use the sum rule of differentiation, which states that the derivative of a sum is the sum of the derivatives of its individual terms.

For example,

\[ \frac{d}{dx}\left[f(x)+g(x)+h(x)\right] = f'(x)+g'(x)+h'(x). \]

The exponential terms also require the chain rule. In general, if \(u=f(x)\), then

\[ \frac{d}{dx}\left(e^{f(x)}\right) = e^{f(x)}f'(x). \]

In particular,

\[ \frac{d}{dx}(e^{x^n}) = e^{x^n}\frac{d}{dx}(x^n) = nx^{n-1}e^{x^n}. \]

Important: The third term in the question is written as \(ex^3\), where \(e\) is the constant Euler's number. It is not \(e^{x^3}\). Therefore,

\[ \frac{d}{dx}(ex^3)=3ex^2. \]
The supplied solution appears to treat this term as \(e^{x^3}\). The differentiation below follows the expression exactly as written.

Also, the final term in the supplied solution contains \(e^{x^5}\), not \(e^{x^s}\). This is corrected below.

🗺️ Solution Roadmap
Step-by-step Plan
›

  1. Differentiate each term separately using the sum rule.

  2. Differentiate \(e^x\) directly.

  3. Differentiate \(e^{x^2}\) using the chain rule.

  4. Differentiate \(ex^3\), treating \(e\) as a constant.

  5. Continue the same pattern for the intermediate exponential terms represented by the ellipsis.

  6. Differentiate the final term \(e^{x^5}\) using the chain rule.

  7. Write the complete derivative without omitting any displayed term.

✏️ Solution
Complete Solution
›
Step-by-step Solution  ·  21 steps
  1. Let
    \[y=e^x+e^{x^2}+ex^3+\ldots+e^{x^5}\]
  2. Using the sum rule of differentiation,
    \[\frac{dy}{dx}=\frac{d}{dx}(e^x)+\frac{d}{dx}(e^{x^2})+\frac{d}{dx}(ex^3)+\ldots+\frac{d}{dx}(e^{x^5})\]
  3. First Term
  4. For
    \[e^x,\]
  5. we have
    \[\frac{d}{dx}(e^x)=e^x\]
  6. Second Term
  7. For
    \[e^{x^2},\]
  8. use the chain rule. Let
    \[u=x^2\]
  9. Then
    \[\frac{du}{dx}=2x\]
  10. Therefore,
    \[\frac{d}{dx}(e^{x^2})=e^{x^2}\frac{d}{dx}(x^2)=2xe^{x^2}\]
  11. Third Term
  12. The third term is
    \[ex^3\]
  13. Since \(e\) is a constant,
    \[\frac{d}{dx}(ex^3)=e\frac{d}{dx}(x^3)\]
  14. Using the power rule,
    \[\frac{d}{dx}(x^3)=3x^2\]
  15. Hence,
    \[\frac{d}{dx}(ex^3)=3ex^2\]
  16. General Pattern for the Exponential Terms
  17. For an exponential term of the form
    \[ e^{x^n}, \]
    let
    \[ u=x^n. \]
  18. Then
    \[\frac{du}{dx}=nx^{n-1}\]
  19. Therefore, by the chain rule,
    \[\frac{d}{dx}(e^{x^n})=e^{x^n}\frac{du}{dx}=nx^{n-1}e^{x^n}\]
  20. Thus, for the final term
    \[e^{x^5},\]
  21. we have
    \[\frac{d}{dx}(e^{x^5})=e^{x^5}\frac{d}{dx}(x^5)\]
  22. Since
    \[frac{d}{dx}(x^5)=5x^4\]
  23. we obtain
    \[ \frac{d}{dx}(e^{x^5}) = 5x^4e^{x^5}. \]
  24. Derivative of the Given Expression
  25. Combining the differentiated terms,
    \[\frac{dy}{dx}=e^x+2xe^{x^2}+3ex^2+\ldots+5x^4e^{x^5}\]
  26. Hence, for the expression exactly as written in the question,
    \[ \boxed{ \frac{dy}{dx} = e^x+2xe^{x^2}+3ex^2+\ldots+5x^4e^{x^5} } \]
🎯 Exam Significance
Exam Significance
›

2-mark standard board question.

🔑 Key Takeaways
Key Takeaways
›
Key Takeaways  ·  6 points

  1. Use the sum rule:

    \[ \frac{d}{dx}(f+g)=f'+g'. \]

  2. Use the standard exponential derivative:

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

  3. For a composite exponential function:

    \[ \frac{d}{dx}(e^{f(x)}) = e^{f(x)}f'(x). \]

  4. For

    \[ e^{x^n}, \]
    the derivative is
    \[ nx^{n-1}e^{x^n}. \]

  5. The notation \(ex^3\) means \(e\times x^3\), whereas \(e^{x^3}\) is an exponential function. They have different derivatives.

  6. If the intended expression is

    \[ e^x+e^{x^2}+e^{x^3}+e^{x^4}+e^{x^5}, \]
    its derivative is
    \[ \boxed{ e^x+2xe^{x^2}+3x^2e^{x^3}+4x^3e^{x^4}+5x^4e^{x^5} }. \]
← Q5
6 / 10  ·  60%
Q7 →
Q7
NUMERIC3 marks
Differentiate the following with respect to \(x\), \(x>0\): \[y=\sqrt{e^{\sqrt{x}}}\]
📘 Concept & Theory
Concept/Theory
›

The given function contains two nested functions: the square-root function applied to the exponential function, and \(\sqrt{x}\) appearing in the exponent. Therefore, the chain rule is required.

The expression can be written as

\[ y=\left(e^{\sqrt{x}}\right)^{1/2}. \]
Since
\[ e^{\sqrt{x}}>0, \]
this can also be simplified to
\[ y=e^{\sqrt{x}/2}. \]

There are two useful approaches:

  1. Use the chain rule directly on the nested expression.
  2. First simplify the square root and then apply the chain rule.

The standard derivatives needed are

\[ \frac{d}{dx}(e^{f(x)}) = e^{f(x)}f'(x) \]
and
\[ \frac{d}{dx}(\sqrt{x}) = \frac{1}{2\sqrt{x}}, \qquad x>0. \]

🗺️ Solution Roadmap
Step-by-step Plan
›

  1. Write the given expression as \(y=\sqrt{e^{\sqrt{x}}}\).

  2. Introduce \(t=\sqrt{x}\) to separate the inner function.

  3. Differentiate \(y=\sqrt{e^t}\) with respect to \(t\), or equivalently use the relation \(y^2=e^t\).

  4. Differentiate \(t=\sqrt{x}\) with respect to \(x\).

  5. Apply the chain rule.

  6. Substitute \(t=\sqrt{x}\) and \(y=\sqrt{e^{\sqrt{x}}}\).

  7. Simplify the result.

✏️ Solution
Complete Solution
›
Step-by-step Solution  ·  17 steps
  1. Given —
    \[y=\sqrt{e^{\sqrt{x}}}\]
  2. Let
    \[t=\sqrt{x}\]
  3. Therefore,
    \[y=\sqrt{e^t}\]
  4. Squaring both sides,
    \[y^2=e^t\]
  5. Differentiate both sides with respect to \(x\):
    \[\frac{d}{dx}(y^2)=\frac{d}{dx}(e^t)\]
  6. Using the chain rule on the left-hand side,
    \[2y\frac{dy}{dx}=\frac{d}{dt}(e^t)\frac{dt}{dx}\]
  7. Now,
    \[\frac{d}{dt}(e^t)=e^t\]
  8. Also, since
    \[t=\sqrt{x}\]
  9. we have
    \[\frac{dt}{dx}=\frac{d}{dx}(\sqrt{x})=\frac{1}{2\sqrt{x}}\]
  10. Therefore,
    \[2y\frac{dy}{dx}=e^t\cdot\frac{1}{2\sqrt{x}}\]
  11. Hence,
    \[2y\frac{dy}{dx}=\frac{e^t}{2\sqrt{x}}\]
  12. Dividing both sides by \(2y\),
    \[\frac{dy}{dx}=\frac{e^t}{4y\sqrt{x}}\]
  13. Now substitute
    \[t=\sqrt{x}\]
    and
    \[y=\sqrt{e^{\sqrt{x}}}\]
  14. Thus,
    \[\frac{dy}{dx}=\frac{e^{\sqrt{x}}}{4\sqrt{e^{\sqrt{x}}}\sqrt{x}}\]
  15. Therefore,
    \[\boxed{\frac{dy}{dx}=\frac{e^{\sqrt{x}}}{4\sqrt{x}\sqrt{e^{\sqrt{x}}}}}\]
  16. Simplification of the Answer
  17. Since
    \[e^{\sqrt{x}}>0,\]
    we can use
    \[\sqrt{x}\sqrt{e^{\sqrt{x}}}=\sqrt{xe^{\sqrt{x}}}\]
  18. Hence,
    \[\frac{dy}{dx}=\frac{1}{4}\frac{e^{\sqrt{x}}}{\sqrt{xe^{\sqrt{x}}}}\]
  19. Therefore, an equivalent form is
    \[\boxed{\frac{dy}{dx}=\frac{e^{\sqrt{x}}}{4\sqrt{xe^{\sqrt{x}}}}}\]
🎯 Exam Significance
Exam Significance
›

This problem is useful because it combines exponential functions, square-root functions and the chain rule. It demonstrates that a complicated-looking expression can often be simplified before differentiation.

  • It reinforces the chain rule for multiple nested functions.
  • It strengthens understanding of the derivative of \(\sqrt{x}\).
  • It demonstrates the useful identity
    \[ \sqrt{e^a}=e^{a/2}. \]
  • It shows that different-looking derivative expressions may be algebraically equivalent.
  • It encourages students to simplify expressions before or after differentiation as appropriate.
  • The condition \(x>0\) ensures that \(\sqrt{x}\) and its derivative are well-defined in the real domain.
Significance for Competitive Entrance Exam Aspirants

Expressions involving several nested functions frequently occur in objective calculus questions. A fast approach is to simplify

\[ \sqrt{e^{\sqrt{x}}} \]
to
\[ e^{\sqrt{x}/2} \]
and then apply the chain rule once more.

The general pattern

\[ \frac{d}{dx}\left(e^{f(x)}\right) = e^{f(x)}f'(x) \]
immediately gives the derivative once
\[ f(x)=\frac{\sqrt{x}}{2} \]
has been identified.

This approach is particularly useful when answer choices contain equivalent forms of the derivative and algebraic simplification is needed to identify the correct option.

🔑 Key Takeaways
Key Takeaways
›
Key Takeaways  ·  6 points

  1. The chain rule is essential for nested functions.

  2. Remember

    \[ \frac{d}{dx}(\sqrt{x})=\frac{1}{2\sqrt{x}}. \]

  3. The expression can be simplified as

    \[ \sqrt{e^{\sqrt{x}}}=e^{\sqrt{x}/2}. \]

  4. After simplification,

    \[ y=e^{\sqrt{x}/2}. \]

  5. Therefore,

    \[ \frac{dy}{dx} = e^{\sqrt{x}/2} \left(\frac{1}{4\sqrt{x}}\right). \]

  6. The most simplified answer is

    \[ \boxed{ \frac{dy}{dx} = \frac{e^{\sqrt{x}/2}}{4\sqrt{x}} } \]
    or equivalently
    \[ \boxed{ \frac{dy}{dx} = \frac{\sqrt{e^{\sqrt{x}}}}{4\sqrt{x}} }. \]
← Q6
7 / 10  ·  70%
Q8 →
Q8
NUMERIC3 marks
Differentiate the following with respect to \(x\), \(x>1\): \[y=\log(\log x)\]
📘 Concept & Theory
Concept/Theory
›

The given function is a composite logarithmic function. The logarithm \(\log x\) occurs inside another logarithm. Hence, the chain rule must be applied.

If

\[ y=f(u),\qquad u=g(x), \]
then
\[ \frac{dy}{dx} = \frac{dy}{du}\cdot\frac{du}{dx}. \]

Here, the inner function is

\[ u=\log x \]
and the outer function is
\[ y=\log u. \]

We use the standard derivative

\[ \frac{d}{dx}(\log x)=\frac{1}{x}, \qquad x>0. \]

Applying the chain rule gives the general result

\[ \frac{d}{dx}\left[\log(f(x))\right] = \frac{f'(x)}{f(x)}. \]

In this problem,

\[ f(x)=\log x \]
and
\[ f'(x)=\frac{1}{x}. \]

🗺️ Solution Roadmap
Step-by-step Plan
›

  1. Identify the inner function \(u=\log x\).

  2. Write the outer function as \(y=\log u\).

  3. Differentiate \(y=\log u\) with respect to \(u\).

  4. Differentiate \(u=\log x\) with respect to \(x\).

  5. Apply the chain rule.

  6. Substitute \(u=\log x\).

  7. Simplify the result.

  8. Verify the domain condition \(x>1\).

✏️ Solution
Complete Solution
›
Step-by-step Solution  ·  10 steps
  1. Given —
    \[y=\log(\log x)\]
  2. Let
    \[u=\log x\]
  3. Therefore,
    \[y=\log u\]
  4. Differentiate \(y=\log u\) with respect to \(u\):
    \[\frac{dy}{du}=\frac{d}{du}(\log u)=\frac{1}{u}\]
  5. Now differentiate
    \[u=\log x\]
    with respect to \(x\):
    \[\frac{du}{dx}=\frac{d}{dx}(\log x)=\frac{1}{x}\]
  6. Using the chain rule,
    \[\frac{dy}{dx}=\frac{dy}{du}\cdot\frac{du}{dx}\]
  7. Substituting the values obtained above,
    \[\frac{dy}{dx}=\frac{1}{u}\cdot\frac{1}{x}\]
  8. Therefore,
    \[\frac{dy}{dx}=\frac{1}{xu}\]
  9. Since
    \[u=\log x\]
  10. we obtain
    \[\frac{dy}{dx}=\frac{1}{x\log x}\]
  11. Hence, the required derivative is
    \[\boxed{\frac{dy}{dx}=\frac{1}{x\log x}}\qquad x>1\]
🎯 Exam Significance
Exam Significance
›

This is a fundamental application of the chain rule to logarithmic functions. It is particularly useful for understanding how the derivative of an outer logarithm combines with the derivative of its inner function.

  • It reinforces the chain rule for composite logarithmic functions.
  • It tests the standard derivative
    \[ \frac{d}{dx}(\log x)=\frac{1}{x}. \]
  • It demonstrates that the derivative of \(\log(f(x))\) is
    \[ \frac{f'(x)}{f(x)}. \]
  • It highlights the importance of checking the domain of a logarithmic expression.
  • The substitution method provides a clear and systematic presentation suitable for written board solutions.
Significance for Competitive Entrance Exam Aspirants

Nested logarithmic functions occur frequently in calculus problems involving differentiation, monotonicity, maxima and minima, limits and objective questions.

For rapid problem solving, the standard pattern

\[ \frac{d}{dx}\left[\log(f(x))\right] = \frac{f'(x)}{f(x)} \]
can be applied immediately.

Taking

\[ f(x)=\log x, \]
we get
\[ \frac{d}{dx}\left[\log(\log x)\right] = \frac{\frac{1}{x}}{\log x} = \frac{1}{x\log x}. \]

The domain condition \(x>1\) is also important in objective questions because it can be used to eliminate mathematically invalid answer choices.

🔑 Key Takeaways
Key Takeaways
›
Key Takeaways  ·  6 points

  1. When a logarithm contains another function, use the chain rule.

  2. For

    \[ y=\log u, \]
    we have
    \[ \frac{dy}{du}=\frac{1}{u}. \]

  3. For

    \[ u=\log x, \]
    we have
    \[ \frac{du}{dx}=\frac{1}{x}. \]

  4. Therefore,

    \[ \frac{d}{dx}\left[\log(\log x)\right] = \frac{1}{\log x}\cdot\frac{1}{x}. \]

  5. The final result is

    \[ \boxed{ \frac{dy}{dx} = \frac{1}{x\log x} }. \]

  6. The real-valued domain of the function is

    \[ \boxed{x>1}. \]
← Q7
8 / 10  ·  80%
Q9 →
Q9
NUMERIC3 marks
Differentiate the following with respect to \(x\): \[y=\frac{\cos x}{\log x}\]
📘 Concept & Theory
Concept/Theory
›

The given function is the quotient of two differentiable functions:

\[ \cos x \]
and
\[ \log x. \]
Therefore, the quotient rule of differentiation is the most direct method.

If

\[ y=\frac{u}{v}, \]
then
\[ \frac{dy}{dx} = \frac{ v\frac{du}{dx}-u\frac{dv}{dx} }{v^2}. \]

Here,

\[ u=\cos x \]
and
\[ v=\log x. \]

The required standard derivatives are

\[ \frac{d}{dx}(\cos x)=-\sin x \]
and
\[ \frac{d}{dx}(\log x)=\frac{1}{x}. \]

Thus, the quotient rule can be applied directly after finding the derivatives of the numerator and denominator.

🗺️ Solution Roadmap
Step-by-step Plan
›

  1. Identify the numerator \(u=\cos x\) and denominator \(v=\log x\).

  2. Find \(\dfrac{du}{dx}\).

  3. Find \(\dfrac{dv}{dx}\).

  4. Apply the quotient rule.

  5. Substitute the derivatives carefully, maintaining the order

    \[ v\frac{du}{dx}-u\frac{dv}{dx}. \]

  6. Simplify the numerator.

  7. Take a common denominator if necessary to obtain a compact final form.

✏️ Solution
Complete Solution
›
Step-by-step Solution  ·  16 steps
  1. Given —
    \[y=\frac{\cos x}{\log x}\]
  2. Let
    \[u=\cos x\]
    and
    \[v=\log x\]
  3. Differentiate the numerator:
    \[\frac{du}{dx}=\frac{d}{dx}(\cos x)=-\sin x\]
  4. Differentiate the denominator:
    \[\frac{dv}{dx}=\frac{d}{dx}(\log x)=\frac{1}{x}\]
  5. Using the quotient rule,
    \[\frac{dy}{dx}=\frac{v\frac{du}{dx}-u\frac{dv}{dx}}{v^2}\]
  6. Substituting
    \[u=\cos x,\quad v=\log x,\quad \frac{du}{dx}=-\sin x,\quad \frac{dv}{dx}=\frac{1}{x},\]
  7. we obtain
    \[\frac{dy}{dx}=\frac{(\log x)(-\sin x)-(\cos x)\left(\frac{1}{x}\right)}{(\log x)^2}\]
  8. Therefore,
    \[\frac{dy}{dx}=\frac{-(\log x)\sin x-\frac{\cos x}{x}}{(\log x)^2}\]
  9. To combine the two terms in the numerator, take \(x\) as the common denominator:
    \[-(\log x)\sin x-\frac{\cos x}{x}=-\frac{x(\log x)\sin x+\cos x}{x}\]
  10. Hence,
    \[\frac{dy}{dx}=\frac{-\dfrac{x(\log x)\sin x+\cos x}{x}}{(\log x)^2}\]
  11. Therefore,
    \[\frac{dy}{dx}=-\frac{x(\log x)\sin x+\cos x}{x(\log x)^2}\]
  12. Hence, the required derivative is
    \[\boxed{\frac{dy}{dx}=-\frac{x(\log x)\sin x+\cos x}{x(\log x)^2}}\]
  13. Equivalent Form of the Answer
  14. The result can also be left in the quotient-rule form
    \[ \boxed{ \frac{dy}{dx} = \frac{ -(\log x)\sin x-\frac{\cos x}{x} }{ (\log x)^2 } }. \]
  15. Multiplying the numerator and denominator appropriately gives
    \[ \boxed{ \frac{dy}{dx} = -\frac{x(\log x)\sin x+\cos x} {x(\log x)^2} }. \]
  16. Both forms are algebraically equivalent.
  17. Domain Consideration
  18. Since the denominator contains \(\log x\), the function is defined only when
    \[x>0\]
    and
    \[\log x\neq0\]
  19. Since
    \[\log x=0\]
    when
    \[x=1,\]
    the domain of the given function is
    \[ \boxed{x>0,\quad x\neq1}. \]
🎯 Exam Significance
Exam Significance
›

This problem is a direct application of the quotient rule combined with the standard derivatives of trigonometric and logarithmic functions. It is useful for developing accuracy in multi-step differentiation.

  • It reinforces the quotient-rule formula.
  • It tests the derivative of \(\cos x\), including its negative sign.
  • It tests the derivative of \(\log x\).
  • It develops skill in simplifying a numerator containing a fraction.
  • It demonstrates how to obtain a single compact fraction from the quotient-rule result.
  • The domain restriction is important because \(\log x\) occurs in the denominator.
Significance for Competitive Entrance Exam Aspirants

Quotients involving trigonometric and logarithmic functions can appear in objective questions on differentiation, monotonicity, maxima and minima, tangents and normals, and function analysis.

For rapid calculation, remember the quotient-rule pattern

\[ \left(\frac{u}{v}\right)' = \frac{vu'-uv'}{v^2}. \]

In this problem,

\[ u=\cos x,\qquad u'=-\sin x, \]
and
\[ v=\log x,\qquad v'=\frac{1}{x}. \]

Therefore,

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

The ability to retain this structure correctly is particularly useful when answer choices contain several similar-looking expressions with different signs or denominators.

🔑 Key Takeaways
Key Takeaways
›
Key Takeaways  ·  6 points

  1. For

    \[ y=\frac{u}{v}, \]
    use
    \[ \frac{dy}{dx} = \frac{vu'-uv'}{v^2}. \]

  2. Remember

    \[ \frac{d}{dx}(\cos x)=-\sin x. \]

  3. Remember

    \[ \frac{d}{dx}(\log x)=\frac{1}{x}. \]

  4. After applying the quotient rule,

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

  5. A compact equivalent form is

    \[ \boxed{ \frac{dy}{dx} = -\frac{x(\log x)\sin x+\cos x} {x(\log x)^2} }. \]

  6. The original function is defined for

    \[ \boxed{x>0,\quad x\neq1}. \]
← Q8
9 / 10  ·  90%
Q10 →
Q10
NUMERIC3 marks
Differentiate the following with respect to \(x\), \(x>0\): \[y=\cos\left(\log x+e^x\right)\]
📘 Concept & Theory
Concept/Theory
›

The given function is a composite function. The expression

\[ \log x+e^x \]
occurs inside the cosine function. Therefore, the chain rule is required.

If

\[ y=f(u),\qquad u=g(x), \]
then
\[ \frac{dy}{dx} = \frac{dy}{du}\cdot\frac{du}{dx}. \]

Here, the outer function is

\[ y=\cos u \]
and the inner function is
\[ u=\log x+e^x. \]

We use the standard derivatives

\[ \frac{d}{du}(\cos u)=-\sin u, \]
\[ \frac{d}{dx}(\log x)=\frac{1}{x}, \]
and
\[ \frac{d}{dx}(e^x)=e^x. \]

Since the inner function is a sum, its derivative is obtained using the sum rule:

\[ \frac{d}{dx}\left[f(x)+g(x)\right] = f'(x)+g'(x). \]

🗺️ Solution Roadmap
Step-by-step Plan
›

  1. Identify the inner function \(u=\log x+e^x\).

  2. Write the outer function as \(y=\cos u\).

  3. Differentiate \(y=\cos u\) with respect to \(u\).

  4. Differentiate \(u=\log x+e^x\) with respect to \(x\).

  5. Apply the chain rule.

  6. Substitute \(u=\log x+e^x\).

  7. Write the final answer in simplified form.

✏️ Solution
Complete Solution
›
Step-by-step Solution  ·  17 steps
  1. Given —
    \[y=\cos\left(\log x+e^x\right)\]
  2. Let
    \[u=\log x+e^x\]
  3. Therefore,
    \[y=\cos u\]
  4. Differentiate \(y=\cos u\) with respect to \(u\):
    \[\frac{dy}{du}=\frac{d}{du}(\cos u)=-\sin u\]
  5. Now differentiate
    \[u=\log x+e^x\]
    with respect to \(x\):
    \[\frac{du}{dx}=\frac{d}{dx}\left(\log x+e^x\right)\]
  6. Using the sum rule,
    \[\frac{du}{dx}=\frac{d}{dx}(\log x)+\frac{d}{dx}(e^x)\]
  7. Using the standard derivatives,
    \[\frac{du}{dx}=\frac{1}{x}+e^x\]
  8. Now apply the chain rule:
    \[\frac{dy}{dx}=\frac{dy}{du}\cdot\frac{du}{dx}\]
  9. Substituting the values obtained above,
    \[\frac{dy}{dx}=(-\sin u)\left(\frac{1}{x}+e^x\right)\]
  10. Therefore,
    \[\frac{dy}{dx}=-\sin u\left(\frac{1}{x}+e^x\right)\]
  11. Since
    \[u=\log x+e^x\]
  12. substituting back, we obtain
    \[\frac{dy}{dx}=-\sin\left(\log x+e^x\right)\left(\frac{1}{x}+e^x\right)\]
  13. Hence, the required derivative is
    \[\boxed{\frac{dy}{dx}=-\sin\left(\log x+e^x\right)\left(\frac{1}{x}+e^x\right)}\]
  14. Optional Algebraic Simplification
  15. The bracketed expression can be written as a single fraction:
    \[ \frac{1}{x}+e^x = \frac{1+xe^x}{x}. \]
  16. Therefore, an equivalent form is
    \[ \boxed{ \frac{dy}{dx} = -\frac{ (1+xe^x)\sin\left(\log x+e^x\right) }{x} } \]
  17. Both forms represent the same derivative. The first form more clearly displays the chain-rule structure.
  18. Domain Consideration
  19. The expression contains \(\log x\). For the real logarithm,
    \[ x>0. \]
  20. Therefore, the given function and its derivative are considered on the domain
    \[ \boxed{x>0}. \]
🎯 Exam Significance
Exam Significance
›

This problem is a standard application of the chain rule combined with the sum rule. It is particularly useful because the inner function contains two different types of functions, \(\log x\) and \(e^x\).

  • It reinforces the identification of an inner and outer function.
  • It tests the derivative of \(\cos u\).
  • It tests the standard derivatives of \(\log x\) and \(e^x\).
  • It demonstrates the sum rule inside the chain rule.
  • It helps students organise multi-step differentiation without skipping intermediate derivatives.
  • The condition \(x>0\) reinforces the importance of domain restrictions for logarithmic functions.
Significance for Competitive Entrance Exam Aspirants

Composite expressions of the form

\[ f(g(x)+h(x)) \]
are common in objective calculus problems. A useful general pattern is
\[ \frac{d}{dx}\left[f(g(x))\right] = f'(g(x))g'(x). \]

In this problem,

\[ f(u)=\cos u \]
and
\[ u=\log x+e^x. \]
Hence,
\[ f'(u)=-\sin u \]
and
\[ u'=\frac{1}{x}+e^x. \]

Recognising these two components quickly allows the derivative to be obtained efficiently while avoiding unnecessary expansion.

🔑 Key Takeaways
Key Takeaways
›
Key Takeaways  ·  7 points

  1. Identify the inner function before applying the chain rule.

  2. For

    \[ y=\cos u, \]
    \[ \frac{dy}{du}=-\sin u. \]

  3. For

    \[ u=\log x+e^x, \]
    \[ \frac{du}{dx}=\frac{1}{x}+e^x. \]

  4. Use the chain rule:

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

  5. The final result is

    \[ \boxed{ \frac{dy}{dx} = -\sin\left(\log x+e^x\right) \left( \frac{1}{x}+e^x \right) }. \]

  6. An equivalent single-fraction form is

    \[ \boxed{ \frac{dy}{dx} = -\frac{ (1+xe^x)\sin\left(\log x+e^x\right) }{x} }. \]

  7. The real domain is

    \[ \boxed{x>0}. \]
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NCERT Class 12 Maths Ex 5.4 Q10 Solution
NCERT Class 12 Maths Ex 5.4 Q10 Solution — Complete Notes & Solutions · academia-aeternum.com
NCERT Class 12 Mathematics Chapter 5 Continuity and Differentiability Exercise 5.4 Question 10 focuses on the application of the chain rule to differentiate a composite trigonometric function involving logarithmic and exponential terms. The problem asks students to differentiate \(\cos(\log x+e^x)\), where the expression \(\log x+e^x\) acts as the inner function. This question is important for understanding how multiple standard differentiation rules work together in a single expression. The…
🎓 Class 12 📐 Mathematics 📖 NCERT ✅ Free Access 🏆 CBSE · JEE
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    Frequently Asked Questions

    The derivative is -sin(log x + e^x)(1/x + e^x), obtained using the chain rule and sum rule.

    The chain rule is the main rule used because cos(log x + e^x) is a composite function. The sum rule is also used to differentiate log x + e^x.

    The inner function is u = log x + e^x, while the outer function is y = cos u.

    For x > 0, the derivative of log x is 1/x.

    The derivative of e^x with respect to x is e^x.

    First differentiate cos u with respect to u to obtain -sin u, then multiply it by du/dx, where du/dx = 1/x + e^x.

    The final answer is dy/dx = -sin(log x + e^x)(1/x + e^x).

    The condition x > 0 is required because log x is defined for positive real values of x.

    Yes. An equivalent form is dy/dx = -[(1 + xe^x)sin(log x + e^x)]/x.

    This question strengthens the application of the chain rule to composite functions and combines trigonometric, logarithmic, exponential, and sum-rule differentiation.

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