Q1. Find the derivative of \(x^2 – 2\) at \(x = 10\).

Solution

Let the given function be \( f(x) = x^2 - 2 \). To find the derivative at a specific point, we first differentiate the function with respect to \(x\).

The derivative of \(x^2\) is obtained using the power rule \( \dfrac{d}{dx}(x^n) = nx^{n-1} \), and the derivative of a constant is zero.

\[ \begin{aligned} f'(x) &= \frac{d}{dx}(x^2 - 2) \\ &= 2x - 0 \\ &= 2x \end{aligned} \]

Now substitute \(x = 10\) in the derivative to find the required value.

\[ \begin{aligned} f'(10) &= 2 \times 10 \\ &= 20 \end{aligned} \]

Hence, the derivative of \(x^2 - 2\) at \(x = 10\) is \(20\).

Q2. Find the derivative of \(x\) at \(x\) = 1.

Solution

Let the given function be defined by \( f(x)=x \). This is a polynomial function of degree one, so its derivative can be found directly using the standard power rule.

\[ \begin{aligned} f'(x) &= 1 \cdot x^{1-1} \\ &= x^{0} \\ &= 1 \end{aligned} \]

The derivative of the function is constant and does not depend on the value of \(x\). Hence, evaluating it at \(x=1\) gives \( f'(1)=1 \). Therefore, the derivative of \(x\) at \(x=1\) is equal to 1.

Q3. Find the derivative of \(99x\) at \(x = l00\).

Solution

Let the given function be defined by \( f(x)=99x \). This is a linear polynomial function, so its derivative can be obtained by applying the power rule.

\[ \begin{aligned} f'(x) &= 99 \cdot 1 \cdot x^{1-1} \\ &= 99 \cdot x^{0} \\ &= 99 \end{aligned} \]

The derivative of the function is a constant and is independent of the value of \(x\). Therefore, at \(x=100\), the value of the derivative remains the same. Hence, the derivative of \(99x\) at \(x=100\) is equal to 99.

Q4. Find the derivative of the following functions from first principle.
(i) \(x^3-27\)
(ii) \((x-1)(x-2)\)
(iii) \(\dfrac{1}{x^2}\)
(iv) \(\dfrac{x+1}{x-1}\)

Solution

(i) Let the function be defined as \( f(x)=x^{3}-27 \). The derivative is obtained from the first principle, using the definition of the derivative as a limit. Since the derivative of a constant is zero, only the term \(x^{3}\) contributes to the derivative.

\[ \begin{aligned} f'(x) &= \lim_{h\to 0}\dfrac{f(x+h)-f(x)}{h} \\ &= \lim_{h\to 0}\dfrac{(x+h)^{3}-27-(x^{3}-27)}{h} \\ &= \lim_{h\to 0}\dfrac{(x+h)^{3}-x^{3}}{h} \\ &= \lim_{h\to 0}\dfrac{x^{3}+3x^{2}h+3xh^{2}+h^{3}-x^{3}}{h} \\ &= \lim_{h\to 0}\left(3x^{2}+3xh+h^{2}\right) \\ &= 3x^{2} \end{aligned} \]

Hence, the derivative of \(x^{3}-27\) is \(3x^{2}\).

(ii) Let the function be \( f(x)=(x-1)(x-2) \). Using the definition of the derivative from first principle, the increment form of the function is substituted and simplified.

\[ \begin{aligned} f'(x) &= \lim_{h\to 0}\dfrac{(x+h-1)(x+h-2)-(x-1)(x-2)}{h} \\ &= \lim_{h\to 0}\dfrac{x^{2}-3x+2+2hx-3h+h^{2}-x^{2}+3x-2}{h} \\ &= \lim_{h\to 0}\dfrac{2hx-3h+h^{2}}{h} \\ &= \lim_{h\to 0}\left(2x-3+h\right) \\ &= 2x-3 \end{aligned} \]

Therefore, the derivative of \((x-1)(x-2)\) is \(2x-3\).

(iii) Let the function be \( f(x)=\dfrac{1}{x^{2}} \). Applying the first principle definition of derivative and simplifying step by step gives the required result.

\[ \begin{aligned} f'(x) &= \lim_{h\to 0}\dfrac{\dfrac{1}{(x+h)^{2}}-\dfrac{1}{x^{2}}}{h} \\ &= \lim_{h\to 0}\dfrac{x^{2}-(x+h)^{2}}{h\,x^{2}(x+h)^{2}} \\ &= \lim_{h\to 0}\dfrac{x^{2}-x^{2}-2xh-h^{2}}{h\,x^{2}(x+h)^{2}} \\ &= \lim_{h\to 0}\dfrac{-2x-h}{x^{2}(x+h)^{2}} \\ &= -\dfrac{2}{x^{3}} \end{aligned} \]

Hence, the derivative of \(\dfrac{1}{x^{2}}\) is \(-\dfrac{2}{x^{3}}\).

(iv) Let the function be \( f(x)=\dfrac{x+1}{x-1} \). The derivative is found directly from the definition of the derivative by forming the difference quotient and simplifying.

\[ \begin{aligned} f'(x) &= \lim_{h\to 0}\dfrac{\dfrac{x+h+1}{x+h-1}-\dfrac{x+1}{x-1}}{h} \\ &= \lim_{h\to 0}\dfrac{(x-1)(x+h+1)-(x+1)(x+h-1)}{h(x+h-1)(x-1)} \\ &= \lim_{h\to 0}\dfrac{x^{2}+hx-x-h-1-x^{2}-hx+x-h+1}{h(x+h-1)(x-1)} \\ &= \lim_{h\to 0}\dfrac{-2h}{h(x+h-1)(x-1)} \\ &= \lim_{h\to 0}\dfrac{-2}{(x+h-1)(x-1)} \\ &= -\dfrac{2}{(x-1)^{2}} \end{aligned} \]

Thus, the derivative of \(\dfrac{x+1}{x-1}\) is \(-\dfrac{2}{(x-1)^{2}}\).

Q5. For the function

\(f(x)=\dfrac{x^{100}}{100}+\dfrac{x^{99}}{99}+\ldots+\dfrac{x^2}{2}+x+1\)

Prove that \(f^\prime(1)=100f^\prime(0)\)

Solution

Given the function \( f(x)=\dfrac{x^{100}}{100}+\dfrac{x^{99}}{99}+\cdots+\dfrac{x^{2}}{2}+x+1 \), we first differentiate it term by term. Each term is a power of \(x\), so the derivative is obtained using the standard power rule.

\[ \begin{aligned} f'(x) &= \dfrac{100x^{99}}{100}+\dfrac{99x^{98}}{99}+\cdots+\dfrac{2x}{2}+1 \\ &= x^{99}+x^{98}+\cdots+x+1 \end{aligned} \]

To evaluate the derivative at \(x=0\), we substitute \(x=0\) in the obtained expression. All terms containing positive powers of \(x\) become zero, leaving only the constant term.

\[ \begin{aligned} f'(0) &= 1 \end{aligned} \]

Next, we evaluate the derivative at \(x=1\). Substituting \(x=1\) makes each term equal to 1, and since there are 100 such terms in the sum, their total is 100.

\[ \begin{aligned} f'(1) &= 1+1+\cdots+1 \\ &= 100 \end{aligned} \]

Since \( f'(0)=1 \), we have \( f'(1)=100\times 1 \). Hence, it follows that \( f'(1)=100f'(0) \), which proves the required result.

Q6. Find the derivative of \(x^n+ax^{n-1}+a^2x^{n-2}+\ldots+a^{n-1}+a^n\) for some fixed real number \(a\)

Solution

Let the given function be \( f(x)=x^{n}+ax^{\,n-1}+a^{2}x^{\,n-2}+\cdots+a^{n} \), where \(a\) is a fixed real number. Since the function is a sum of power functions of \(x\) with constant coefficients, its derivative can be found by differentiating each term separately.

\[ \begin{aligned} f'(x) &= nx^{\,n-1}+a(n-1)x^{\,n-2}+a^{2}(n-2)x^{\,n-3}+\cdots+0 \end{aligned} \]

The last term \(a^{n}\) is a constant and hence its derivative is zero. Therefore, the derivative of the given function is \( nx^{\,n-1}+a(n-1)x^{\,n-2}+a^{2}(n-2)x^{\,n-3}+\cdots \).

Q7. For some constants a and b, find the derivative of
(i) \((x-a)(x-b)\)
(ii) \(\left(ax^2+b\right)^2\)
(iii) \(\dfrac{x-a}{x-b}\)

Solution

(i) Let the function be defined as \( f(x)=(x-a)(x-b) \), where \(a\) and \(b\) are constants. Since the function is a product of two linear functions of \(x\), its derivative is obtained by applying the product rule.

\[ \begin{aligned} f'(x) &= (x-a)(x-b)' + (x-a)'(x-b) \\ &= (x-a)\cdot 1 + 1\cdot (x-b) \\ &= (x-a)+(x-b) \\ &= 2x-(a+b) \end{aligned} \]

Thus, the derivative of \((x-a)(x-b)\) is \(2x-(a+b)\).

(ii) Let the function be \( f(x)=(ax^{2}+b)^{2} \). This is a composite function, so its derivative is obtained using the chain rule.

\[ \begin{aligned} f'(x) &= 2(ax^{2}+b)\cdot \dfrac{d}{dx}(ax^{2}+b) \\ &= 2(ax^{2}+b)\cdot 2ax \\ &= 4ax(ax^{2}+b) \end{aligned} \]

Hence, the derivative of \((ax^{2}+b)^{2}\) is \(4ax(ax^{2}+b)\).

(iii) Let the function be \( f(x)=\dfrac{x-a}{x-b} \). Since this is a quotient of two functions of \(x\), its derivative is obtained by applying the quotient rule.

\[ \begin{aligned} f'(x) &= \dfrac{(x-b)\dfrac{d}{dx}(x-a)-(x-a)\dfrac{d}{dx}(x-b)}{(x-b)^{2}} \\ &= \dfrac{(x-b)-(x-a)}{(x-b)^{2}} \\ &= \dfrac{x-b-x+a}{(x-b)^{2}} \\ &= \dfrac{a-b}{(x-b)^{2}} \end{aligned} \]

Therefore, the derivative of \(\dfrac{x-a}{x-b}\) is \(\dfrac{a-b}{(x-b)^{2}}\).

Q8. Find the derivative of \(\left(\dfrac{x^n-a^n}{x-a}\right)\) for some constant a.

Solution

Let the given function be defined as \( f(x)=\dfrac{x^{n}-a^{n}}{x-a} \), where \(a\) is a constant. Since the function is expressed as a quotient of two functions of \(x\), its derivative is obtained by applying the quotient rule.

\[ \begin{aligned} f'(x) &= \dfrac{(x-a)\dfrac{d}{dx}(x^{n}-a^{n})-(x^{n}-a^{n})\dfrac{d}{dx}(x-a)}{(x-a)^{2}} \\ &= \dfrac{(x-a)\cdot nx^{n-1}-(x^{n}-a^{n})\cdot 1}{(x-a)^{2}} \\ &= \dfrac{nx^{n}-anx^{n-1}-x^{n}+a^{n}}{(x-a)^{2}} \end{aligned} \]

This expression gives the derivative of \( \dfrac{x^{n}-a^{n}}{x-a} \) for a constant \(a\). The result follows directly from the quotient rule together with the fact that the derivative of the constant term \(a^{n}\) is zero.

Q9. Find the derivative of

1. \(2x-\frac{3}{4}\)


2. \(\left(5x^3+3x-1\right)\left(x-1\right)\)


3. \(x^{-3}\left(5x+3\right)\)


4. \(x^5\left(3-6x^{-9}\right)\)


5. \(x^{-4}\left(3-4x^{-5}\right)\)


6. \(\dfrac{2}{x+1}-\dfrac{x^2}{3x-1}\)

Solution

(i) Let the function be defined as \( f(x)=2x-\dfrac{3}{4} \). The derivative of a linear term is its coefficient, while the derivative of a constant is zero.

\[ \begin{aligned} f'(x) &= \dfrac{d}{dx}(2x)-\dfrac{d}{dx}\left(\dfrac{3}{4}\right) \\ &= 2-0 \\ &= 2 \end{aligned} \]

(ii) Let \( f(x)=\left(5x^{3}+3x-1\right)(x-1) \). Since the function is a product of two functions of \(x\), the product rule is applied.

\[ \begin{aligned} f'(x) &= (x-1)\dfrac{d}{dx}(5x^{3}+3x-1)+(5x^{3}+3x-1)\dfrac{d}{dx}(x-1) \\ &= (x-1)(15x^{2}+3)+(5x^{3}+3x-1) \\ &= 15x^{3}+3x-15x^{2}-3+5x^{3}+3x-1 \\ &= 20x^{3}-15x^{2}+6x-4 \end{aligned} \]

(iii) Let \( f(x)=x^{-3}(5x+3) \). Using the product rule and simplifying the result gives the derivative.

\[ \begin{aligned} f'(x) &= (5x+3)\dfrac{d}{dx}(x^{-3})+x^{-3}\dfrac{d}{dx}(5x+3) \\ &= (5x+3)(-3x^{-4})+x^{-3}\cdot 5 \\ &= \dfrac{-15x-9}{x^{4}}+\dfrac{5x}{x^{4}} \\ &= \dfrac{-10x-9}{x^{4}} \end{aligned} \]

(iv) Let \( f(x)=x^{5}(3-6x^{-9}) \). First simplify the expression and then differentiate term by term.

\[ \begin{aligned} f(x) &= 3x^{5}-6x^{-4} \\ f'(x) &= \dfrac{d}{dx}(3x^{5})-\dfrac{d}{dx}(6x^{-4}) \\ &= 15x^{4}+24x^{-5} \\ &= 15x^{4}+\dfrac{24}{x^{5}} \end{aligned} \]

(v) Let \( f(x)=x^{-4}(3-4x^{-5}) \). Simplifying the function first makes differentiation straightforward.

\[ \begin{aligned} f(x) &= 3x^{-4}-4x^{-9} \\ f'(x) &= \dfrac{d}{dx}(3x^{-4})-\dfrac{d}{dx}(4x^{-9}) \\ &= -12x^{-5}+36x^{-10} \\ &= -\dfrac{12}{x^{5}}+\dfrac{36}{x^{10}} \end{aligned} \]

(vi) Let \( f(x)=\dfrac{2}{x+1}-\dfrac{x^{2}}{3x-1} \). The derivative is obtained by differentiating each term separately, using the chain rule for the first term and the quotient rule for the second.

\[ \begin{aligned} \dfrac{d}{dx}\left(\dfrac{2}{x+1}\right) &= -\dfrac{2}{(x+1)^{2}} \\ \dfrac{d}{dx}\left(\dfrac{x^{2}}{3x-1}\right) &= \dfrac{(3x-1)\cdot 2x-x^{2}\cdot 3}{(3x-1)^{2}} \\ &= \dfrac{3x^{2}-2x}{(3x-1)^{2}} \\ f'(x) &= -\dfrac{2}{(x+1)^{2}}-\dfrac{3x^{2}-2x}{(3x-1)^{2}} \end{aligned} \]

Q10. Find the derivative of cos x from first principle.

Solution

Let the given function be defined as \( f(x)=\cos x \). To find its derivative from first principle, we use the definition of the derivative as a limit.

\[ \begin{aligned} f'(x) &= \lim_{h\to 0}\dfrac{f(x+h)-f(x)}{h} \\ &= \lim_{h\to 0}\dfrac{\cos(x+h)-\cos x}{h} \end{aligned} \]

Using the trigonometric identity \( \cos A-\cos B=-2\sin\left(\dfrac{A+B}{2}\right)\sin\left(\dfrac{A-B}{2}\right) \), the numerator is simplified.

\[ \begin{aligned} f'(x) &= \lim_{h\to 0}\dfrac{-2\sin\left(\dfrac{2x+h}{2}\right)\sin\left(\dfrac{h}{2}\right)}{h} \\ &= \lim_{h\to 0}\left[-\sin\left(x+\dfrac{h}{2}\right)\cdot \dfrac{\sin\left(\dfrac{h}{2}\right)}{\dfrac{h}{2}}\right] \end{aligned} \]

As \( h\to 0 \), we have \( \sin\left(\dfrac{h}{2}\right)/\left(\dfrac{h}{2}\right)\to 1 \) and \( \sin\left(x+\dfrac{h}{2}\right)\to \sin x \). Hence, the limit can be evaluated directly.

\[ \begin{aligned} f'(x) &= -\sin x \end{aligned} \]

Therefore, the derivative of \( \cos x \) from first principle is \( -\sin x \).

Q11. Find the derivative of the following functions:

1. \(\sin x \cos x\)


2. \(\sec x\)


3. \(5\sec x + 4\cos x\)


4. \(\text{coec }x\)


5. \(3\cot x + 5\text{cosec }x\)


6. \(5\sin x-6\cos x\,+7\)


7. \(2\tan x-7\sec x\)

Solution

(i) Let the function be defined as \( f(x)=\sin x\cos x \). Since the function is a product of two trigonometric functions, the product rule is applied.

\[ \begin{aligned} f'(x) &= \cos x\dfrac{d}{dx}(\sin x)+\sin x\dfrac{d}{dx}(\cos x) \\ &= \cos x\cdot \cos x+\sin x\cdot(-\sin x) \\ &= \cos^{2}x-\sin^{2}x \\ &= \cos 2x \end{aligned} \]

(ii) Let the function be defined as \( f(x)=\sec x \). Writing it in terms of cosine allows the use of the quotient rule.

\[ \begin{aligned} f(x) &= \dfrac{1}{\cos x} \\ f'(x) &= \dfrac{\cos x\dfrac{d}{dx}(1)-1\dfrac{d}{dx}(\cos x)}{\cos^{2}x} \\ &= \dfrac{\sin x}{\cos^{2}x} \\ &= \sec x\tan x \end{aligned} \]

(iii) Let the function be defined as \( f(x)=5\sec x+4\cos x \). The derivative of a sum is the sum of the derivatives.

\[ \begin{aligned} f'(x) &= 5\dfrac{d}{dx}(\sec x)+4\dfrac{d}{dx}(\cos x) \\ &= 5(\sec x\tan x)+4(-\sin x) \\ &= 5\sec x\tan x-4\sin x \end{aligned} \]

(iv) Let the function be defined as \( f(x)=\csc x \). Expressing it in terms of sine helps in applying the quotient rule.

\[ \begin{aligned} f(x) &= \dfrac{1}{\sin x} \\ f'(x) &= \dfrac{\sin x\dfrac{d}{dx}(1)-1\dfrac{d}{dx}(\sin x)}{\sin^{2}x} \\ &= \dfrac{-\cos x}{\sin^{2}x} \\ &= -\csc x\cot x \end{aligned} \]

(v) Let the function be defined as \( f(x)=3\cot x+5\csc x \). Each term is differentiated separately.

\[ \begin{aligned} \dfrac{d}{dx}(\cot x) &= -\csc^{2}x \\ \dfrac{d}{dx}(\csc x) &= -\csc x\cot x \\ f'(x) &= 3(-\csc^{2}x)+5(-\csc x\cot x) \\ &= -3\csc^{2}x-5\csc x\cot x \end{aligned} \]

(vi) Let the function be defined as \( f(x)=5\sin x-6\cos x+7 \). The derivative of a constant is zero.

\[ \begin{aligned} f'(x) &= 5\dfrac{d}{dx}(\sin x)-6\dfrac{d}{dx}(\cos x)+\dfrac{d}{dx}(7) \\ &= 5\cos x-6(-\sin x) \\ &= 5\cos x+6\sin x \end{aligned} \]

(vii) Let the function be defined as \( f(x)=2\tan x-7\sec x \). Using standard derivatives of trigonometric functions gives the result.

\[ \begin{aligned} \dfrac{d}{dx}(\tan x) &= \sec^{2}x \\ \dfrac{d}{dx}(\sec x) &= \sec x\tan x \\ f'(x) &= 2\sec^{2}x-7\sec x\tan x \end{aligned} \]