Class 11: Derivatives – Exercise 30.3 – ICSE / ISC / CBSE Mathematics Portal for K12 Students

Differentiate the following functions with respect to $x$ :

$\displaystyle \text{Question 1: } x^4 - 2 \sin x + 3 \cos x$

Answer:

$\displaystyle \frac{d}{dx} ( x^4 - 2 \sin x + 3 \cos x )$

$\displaystyle = \frac{d}{dx} (x^4) -2 \frac{d}{dx} ( \sin x ) + 3 \frac{d}{dx} (\cos x)$

$\displaystyle = 4 x^3 - 2 \cos x - 3 \sin x$

$\\$

$\displaystyle \text{Question 2: } 3^x + x^3 + 3^3$

Answer:

$\displaystyle \frac{d}{dx} (3^x + x^3 + 3^3)$

$\displaystyle = \frac{d}{dx} (3^x) + \frac{d}{dx} (x^3) + \frac{d}{dx} (3^3)$

$\displaystyle = 3^x \log 3 + 3 x^2$

$\\$

$\displaystyle \text{Question 3: } \frac{x^3}{3} - 2 \sqrt{x} + \frac{5}{x^2}$

Answer:

$\displaystyle \frac{d}{dx} (\frac{x^3}{3} - 2 \sqrt{x} + \frac{5}{x^2})$

$\displaystyle = \frac{d}{dx} (\frac{x^3}{3}) -2 \frac{d}{dx} (\sqrt{x}) + \frac{d}{dx} (\frac{5}{x^2})$

$\displaystyle = \frac{1}{3} ( 3 x^2) - 2 \cdot \frac{1}{2} x^{\frac{-1}{2}} + 5 ( -2) x^{-3}$

$\displaystyle = x^2 - x^{\frac{-1}{2}} - 10 x^{-3}$

$\\$

$\displaystyle \text{Question 4: } e^{x \log a} + e^{a \log x} + e^{ a \log a}$

Answer:

$\displaystyle \frac{d}{dx} (e^{x \log a} + e^{a \log x} + e^{ a \log a})$

$\displaystyle = \frac{d}{dx} (e^{x \log a}) + \frac{d}{dx} (e^{a \log x}) + \frac{d}{dx} (e^{ a \log a})$

$\displaystyle = \frac{d}{dx} (a^x) + \frac{d}{dx} (x^a) + \frac{d}{dx} (a^a)$

$\displaystyle = a^x \log a + a x^{a-1}$

$\\$

$\displaystyle \text{Question 5: } (2x^2+1)(3x+2)$

Answer:

$\displaystyle \frac{d}{dx} ( (2x^2+1)(3x+2))$

$\displaystyle \frac{d}{dx} ( 6x^3+4x^2+3x+2)$

$\displaystyle = \frac{d}{dx} (6x^3) + \frac{d}{dx} (4x^2) + \frac{d}{dx} (3x) + \frac{d}{dx} (2)$

$\displaystyle = 18x^2+8x+3$

$\\$

$\displaystyle \text{Question 6: } \log_3 x + 3 \log_e x+ 2 \tan x$

Answer:

$\displaystyle \frac{d}{dx} (\log_3 x + 3 \log_e x+ 2 \tan x)$

$\displaystyle = \frac{d}{dx} (\log_3 x) + 3 \frac{d}{dx} (\log_e x) + 2 \frac{d}{dx} ( \tan x)$

$\displaystyle = \frac{d}{dx} (\frac{\log x}{\log 3}) + 3 \frac{d}{dx} (\log_e x) + 2 \frac{d}{dx} ( \tan x)$

$\displaystyle = \frac{1}{\log 3} \cdot \frac{1}{x} + 3 \frac{1}{x} + 2 \sec^2 x$

$\displaystyle = \frac{1}{x \log 3} + \frac{3}{x} + 2 \sec^2 x$

$\\$

$\displaystyle \text{Question 7: } \Big(x + \frac{1}{x} \Big) \Big( \sqrt{x}+ \frac{1}{\sqrt{x}} \Big)$

Answer:

$\displaystyle \frac{d}{dx} \Big(x + \frac{1}{x} \Big) \Big( \sqrt{x}+ \frac{1}{\sqrt{x}} \Big)$

$\displaystyle = \frac{d}{dx} (x+x^{-1}) ( x^{\frac{1}{2}} + x^{\frac{-1}{2} })$

$\displaystyle = \frac{d}{dx} (x^{\frac{3}{2}} + x^{\frac{1}{2}} + x^{\frac{-1}{2}} + x^{\frac{-3}{2}} )$

$\displaystyle = \frac{d}{dx} (x^{\frac{3}{2}}) + \frac{d}{dx} (x^{\frac{1}{2}}) + \frac{d}{dx} (x^{\frac{-1}{2}}) + \frac{d}{dx} (x^{\frac{-3}{2}})$

$\displaystyle = \frac{3}{2} x^{\frac{1}{2}} + \frac{1}{2} x^{\frac{-1}{2}} - \frac{1}{2} x^{\frac{-3}{2}} - \frac{3}{2} x^{\frac{-5}{2}}$

$\\$

$\displaystyle \text{Question 8: } \Big(\sqrt{x}+ \frac{1}{\sqrt{x}} \Big)^3$

Answer:

$\displaystyle \frac{d}{dx} (\Big(\sqrt{x}+ \frac{1}{\sqrt{x}} \Big)^3)$

$\displaystyle =\frac{d}{dx} \Big( (\sqrt{x})^3+ 3 ( \sqrt{x})^2 \Big( \frac{1}{\sqrt{x} } \Big) + 3 (\sqrt{x}) \Big(\frac{1}{\sqrt{x}}\Big)^2 + \Big( \frac{1}{\sqrt{x}} \Big)^3 \Big)$

$\displaystyle = \frac{d}{dx} \Big(x^{\frac{3}{2} } \Big) + 3 \frac{d}{dx} \Big(x^{\frac{1}{2}}\Big) + 3\frac{d}{dx} \Big(x^{\frac{-1}{2}} \Big) + \frac{d}{dx} \Big( x^{\frac{-3}{2}} \Big)$

$\displaystyle = \frac{3}{2} x^{\frac{1}{2}} + 3 \cdot \frac{1}{2} x^{\frac{-1}{2}} + 3 \cdot \frac{-1}{2} x^{ \frac{-3}{2}} - \frac{3}{2} x^{\frac{-5}{2}}$

$\displaystyle = \frac{3}{2} x^{\frac{1}{2}} + \frac{3}{2} x^{\frac{-1}{2}} - \frac{3}{2} x^{ \frac{-3}{2}} - \frac{3}{2} x^{\frac{-5}{2}}$

$\\$

$\displaystyle \text{Question 9: } \frac{2x^2+ 3x+ 4}{x}$

Answer:

$\displaystyle \frac{d}{dx} (\frac{2x^2+ 3x+ 4}{x})$

$\displaystyle = \frac{d}{dx} (\frac{2x^2}{x}) + \frac{d}{dx} (\frac{3x}{x}) + \frac{d}{dx} (\frac{4}{x})$

$\displaystyle = 2\frac{d}{dx} (x) + 3\frac{d}{dx} (1) + 4\frac{d}{dx} (x^{-1})$

$\displaystyle = 2 + 3(0) + 4 ( -1) x^{-2}$

$\displaystyle = 2 - \frac{4}{x^2}$

$\\$

$\displaystyle \text{Question 10: } \frac{(x^3+1)(x-2)}{x^2}$

Answer:

$\displaystyle \frac{d}{dx} (\frac{(x^3+1)(x-2)}{x^2})$

$\displaystyle = \frac{d}{dx} (\frac{x^4 - 2 x^3 + x - 2}{x^2})$

$\displaystyle = \frac{d}{dx} (x^2) -2 \frac{d}{dx} (x) + \frac{d}{dx} (x^{-1}) -2 \frac{d}{dx} (x^{-2})$

$\displaystyle = 2x - 2 - x^{-2} -2 ( -2) x^{-3}$

$\displaystyle = 2x - 2 - \frac{1}{x^2} + \frac{4}{x^3}$

$\\$

$\displaystyle \text{Question 11: } \frac{a \cos x + b \sin x + c }{\sin x}$

Answer:

$\displaystyle \frac{d}{dx} (\frac{a \cos x + b \sin x + c }{\sin x})$

$\displaystyle = \frac{d}{dx} \Big(\frac{a \cos x}{\sin x} \Big) + \frac{d}{dx} \Big(\frac{b \sin x}{\sin x} \Big) + \frac{d}{dx} \Big(\frac{c}{\sin x} \Big)$

$\displaystyle = a \frac{d}{dx} cot x + \frac{d}{dx} \Big(\frac{b \sin x}{\sin x} \Big) + \frac{d}{dx} \Big(\frac{c}{\sin x} \Big)$

$\displaystyle = a \frac{d}{dx} (\cot x) + \frac{d}{dx} (b) + c\frac{d}{dx} (\mathrm{cosec} x)$

$\displaystyle = - a \mathrm{cosec}^2 x + 0 - c \mathrm{cosec} x \cdot \cot x$

$\displaystyle = - a \mathrm{cosec}^2 x - c \mathrm{cosec} x \cdot \cot x$

$\\$

$\displaystyle \text{Question 12: } 2 \sec x + 3 \cot x- 4 \tan x$

Answer:

$\displaystyle \frac{d}{dx} (2 \sec x + 3 \cot x- 4 \tan x)$

$\displaystyle = 2\frac{d}{dx} (\sec x) + 3\frac{d}{dx} (\cot x) -4 \frac{d}{dx} (\tan x)$

$\displaystyle = 2 \sec x \tan x - 3 \mathrm{cosec}^2 x - 4 \sec^2 x$

$\\$

$\displaystyle \text{Question 13: } a_0 x^n + a_1 x^{n-1} + a_2 x^{n-2} + \cdots + a_{n-1}x + a_n$

Answer:

$\displaystyle \frac{d}{dx} (a_0 x^n + a_1 x^{n-1} + a_2 x^{n-2} + \cdots + a_{n-1}x + a_n)$

$\displaystyle = a_0 \frac{d}{dx} (x^n) + a_1\frac{d}{dx} (x^{n-1}) +a_2 \frac{d}{dx} (x^{n-2}) + \cdots + a_{n-1} \frac{d}{dx} (x) + a_n\frac{d}{dx} (1)$

$\displaystyle = na_0 x^{n-1} +(n-1) a_1 x^{n-2} + ( n-2) a_2 x^{n-3} + \cdots + a_{n-1}$

$\\$

$\displaystyle \text{Question 14: } \frac{1}{\sin x} + 2^{x+3} + \frac{4}{\log_x 3}$

Answer:

$\displaystyle \frac{d}{dx} \Big(\frac{1}{\sin x} +2^{x+3} + \frac{4}{\log_x 3} \Big)$

$\displaystyle = \frac{d}{dx} \Big( \mathrm{cosec} x + 2^3 \cdot 2^x+ \frac{4}{\frac{\log 3}{\log x}} \Big)$

$\displaystyle = \frac{d}{dx} ( \mathrm{cosec} x) + 2^3\frac{d}{dx} (2^x) + \frac{4}{\log 3} \frac{d}{dx} (\log x)$

$\displaystyle = - \mathrm{cosec} x \cot x + 2^3 \cdot 2^x \cdot \log 2 + \frac{4}{\log 3} \cdot \frac{1}{x}$

$\displaystyle = - \mathrm{cosec} x \cot x + 2^{x+3} \cdot \log 2 + \frac{4}{x\log 3}$

$\\$

$\displaystyle \text{Question 15: } \frac{(x+5)(2x^2-1)}{x}$

Answer:

$\displaystyle \frac{d}{dx} \Big(\frac{(x+5)(2x^2-1)}{x} \Big)$

$\displaystyle = \frac{d}{dx} \Big(\frac{2x^3+10x^2-x-5}{x} \Big)$

$\displaystyle = \frac{d}{dx} (2x^2) + \frac{d}{dx} (10x) - \frac{d}{dx} (1) - \frac{d}{dx} (5x^{-1})$

$\displaystyle = 4x + 10 + 5x^{-2}$

$\displaystyle = 4x + 10 + \frac{5}{x^2}$

$\\$

$\displaystyle \text{Question 16: } \log \Bigg( \frac{1}{\sqrt{x}} \Bigg) + 5x^a - 3a^x + \sqrt[3]{x^2}+ 6 \sqrt[4]{x^{-3}}$

Answer:

$\displaystyle \frac{d}{dx} \Bigg[ \log \Bigg( \frac{1}{\sqrt{x}} \Bigg) + 5x^a - 3a^x + \sqrt[3]{x^2}+ 6 \sqrt[4]{x^{-3}} \Bigg]$

$\displaystyle = \frac{d}{dx} \Big( \log ( x^{\frac{-1}{2}} \Big) + 5\frac{d}{dx} (x^a) -3 \frac{d}{dx} (a^x) + \frac{d}{dx} (x^{\frac{2}{3}}) + 6\frac{d}{dx} (x^{\frac{-3}{4}})$

$\displaystyle = \frac{d}{dx} \Big( \frac{-1}{2} \log x \Big) + 5\frac{d}{dx} (x^a) -3 \frac{d}{dx} (a^x) + \frac{d}{dx} (x^{\frac{2}{3}}) + 6\frac{d}{dx} (x^{\frac{-3}{4}})$

$\displaystyle = \frac{-1}{2} \cdot \frac{1}{x} + 5a x^{a-1} - 3a^x \log a + \frac{2}{3} x^{\frac{-1}{3}} + 6 \Big( \frac{-3}{4} \Big) x^{\frac{-7}{4}}$

$\displaystyle = \frac{-1}{2x} + 5a x^{a-1} - 3a^x \log a + \frac{2}{3} x^{\frac{-1}{3}} - \frac{9}{2} x^{\frac{-7}{4}}$

$\\$

$\displaystyle \text{Question 17: } \cos ( x+a)$

Answer:

$\displaystyle \frac{d}{dx} (\cos ( x+a) )$

$\displaystyle = \frac{d}{dx} (\cos x \cos a - \sin x \sin a )$

$\displaystyle = \cos a\frac{d}{dx} (\cos x) - \sin a \frac{d}{dx} (\sin x)$

$\displaystyle = - \cos a \sin x - \sin a \cos x$

$\displaystyle = - ( \sin x \cos a + \cos x \sin a )$

$\displaystyle = - \sin ( x+ a)$

$\\$

$\displaystyle \text{Question 18: } \frac{\cos ( x-2)}{\sin x}$

Answer:

$\displaystyle \frac{d}{dx} \Big( \frac{\cos ( x-2)}{\sin x} \Big)$

$\displaystyle = \frac{d}{dx} \Big( \frac{\cos x \cos 2 + \sin x \sin 2}{\sin x} \Big)$

$\displaystyle = \cos 2 \frac{d}{dx} (\cot x) + \sin 2\frac{d}{dx} (1)$

$\displaystyle = - \cos 2 \ \mathrm{cosec}^2 x + 0$

$\displaystyle = - \cos 2 \ \mathrm{cosec}^2 x$

$\\$

$\displaystyle \text{Question 19: } \text{If } y = \Bigg( \sin \frac{x}{2} + \cos \frac{x}{2} \Bigg)^2, \text{ find } \frac{dy}{dx} \text{ at } x = \frac{\pi}{4}$

Answer:

$\displaystyle \frac{d}{dx} \Bigg( \sin \frac{x}{2} + \cos \frac{x}{2} \Bigg)^2$

$\displaystyle = \frac{d}{dx} \Bigg( \sin^2 \frac{x}{2} + \cos^2 \frac{x}{2} + 2 \sin \frac{x}{2} \cos \frac{x}{2} \Bigg)$

$\displaystyle = \frac{d}{dx} ( 1 + \sin x)$

$\displaystyle = 0 + \cos x$

$\displaystyle = \cos x$

$\displaystyle \frac{dy}{dx} \text{ at } x = \frac{\pi}{6} = \cos \frac{\pi}{6} = \frac{\sqrt{3}}{2}$

$\\$

$\displaystyle \text{Question 20: } \text{If } y = \Bigg( \frac{2 - 3 \cos x}{\sin x} \Bigg) , \text{ find } \frac{dy}{dx} \text{ at } x = \frac{\pi}{4}$

Answer:

$\displaystyle \frac{d}{dx} (\Bigg( \frac{2 - 3 \cos x}{\sin x} \Bigg))$

$\displaystyle = \frac{d}{dx} ( 2 \mathrm{cosec} x - 3 \cot x)$

$\displaystyle = 2\frac{d}{dx} ( \mathrm{cosec} x) -3 \frac{d}{dx} (\cot x)$

$\displaystyle = - 2 \mathrm{cosec} x \cdot \cot x + 3 \mathrm{cosec}^2 x$

$\displaystyle \frac{dy}{dx} \text{ at } x = \frac{\pi}{4} = - 2 \mathrm{cosec} \frac{\pi}{4} \cdot \cot \frac{\pi}{4} + 3 \mathrm{cosec}^2 \frac{\pi}{4} \\ \\ { \hspace{2.5cm} = -2 \sqrt{2} \cdot 1 + 3 \cdot 2 = - 2\sqrt{2} + 6 = 6 - 2\sqrt{2}}$

$\\$

$\displaystyle \text{Question 21: } \text{Find the slope of the tangent to the curve } \\ \\ f(x) = 2x^6 + x^4 - 1 \text{ at } x = 1$

Answer:

$\displaystyle \frac{d}{dx} (2x^6 + x^4 - 1)$

$\displaystyle = \frac{d}{dx} (2x^6) + \frac{d}{dx} (x^4) - \frac{d}{dx} (1)$

$\displaystyle = 12x^5 + 4 x^3$

$\displaystyle \frac{dy}{dx} \text{ at } x =1 = 12(1)^5 + 4(1)^3 = 12 + 4 = 16$

$\\$

$\displaystyle \text{Question 22: } \text{If } y = \sqrt{\frac{x}{a}} + \sqrt{\frac{a}{x}}, \text{ prove that } 2xy \frac{dy}{dx} = \Bigg( \frac{x}{a} - \frac{a}{x} \Bigg)$

Answer:

$\displaystyle \frac{dy}{dx} = \frac{d}{dx} \Big( \sqrt{\frac{x}{a}} + \sqrt{\frac{a}{x}} \Big)$

$\displaystyle \frac{dy}{dx} = \frac{1}{\sqrt{a}}\frac{d}{dx} (\sqrt{x}) + \sqrt{a}\frac{d}{dx} (\frac{1}{\sqrt{x}})$

$\displaystyle \frac{dy}{dx} = \frac{1}{\sqrt{a}} \frac{1}{2\sqrt{x}} + \sqrt{a} \cdot \frac{-1}{2} \cdot \frac{1}{x\sqrt{x}}$

$\displaystyle \frac{dy}{dx} = \frac{1}{2x} \Bigg( \sqrt{ \frac{x}{a} } - \sqrt{ \frac{a}{x} } \Bigg)$

$\displaystyle 2x\frac{dy}{dx} = \Bigg( \sqrt{ \frac{x}{a} } - \sqrt{ \frac{a}{x} } \Bigg)$

Multiplying both sides by y we get

$\displaystyle 2xy\frac{dy}{dx} = \Bigg( \sqrt{ \frac{x}{a} } - \sqrt{ \frac{a}{x} } \Bigg) \Bigg( \sqrt{ \frac{x}{a} } + \sqrt{ \frac{a}{x} } \Bigg)$

$\displaystyle 2xy\frac{dy}{dx} = \Bigg( \frac{x}{a} - \frac{a}{x} \Bigg)$

$\\$

$\displaystyle \text{Question 23: } \text{Find the rate at which the function } \\ \\ f(x) = x^4 - 2x^3 + 3x^2 + x + 5 \text{ changes with respect to } x$

Answer:

$\displaystyle \text{Rate } = f'(x)$

$\displaystyle = \frac{d}{dx} (x^4 - 2x^3 + 3x^2 + x + 5)$

$\displaystyle = \frac{d}{dx} (x^4) -2 \frac{d}{dx} (x^3) +3 \frac{d}{dx} (x^2) + \frac{d}{dx} (5)$

$\displaystyle = 4x^3- 6x^2 + 6x + 1$

$\\$

$\displaystyle \text{Question 24: } \text{If } y = \frac{2x^9}{3} - \frac{5}{7} x^7 + 6x^3 - x , \text{ find } \frac{dy}{dx} \text{ at } x = 1$

Answer:

$\displaystyle \frac{dy}{dx} = \frac{d}{dx} \Big( \frac{2x^9}{3} - \frac{5}{7} x^7 + 6x^3 - x \Big)$

$\displaystyle = \frac{2}{3}\frac{d}{dx} (x^9) -\frac{5}{7} \frac{d}{dx} (x^7) + 6\frac{d}{dx} (x^3) - \frac{d}{dx} (x)$

$\displaystyle = \frac{2}{3} ( 9x^8) -\frac{5}{7} (7x^6) + 6 (3x^2) - 1$

$\displaystyle = 6x^8 - 5x^6 + 18x^2 - 1$

$\displaystyle \frac{dy}{dx} \text{ at } x =1 = 6(1)^8 - 5(1)^6 + 18(1)^2 - 1 = 6-5+18-1 = 18$

$\\$

$\displaystyle \text{Question 25: } \text{If for } f(x) = \lambda x^2 + \mu x + 12, f'(4) = 15 \text{ and } f'(2) = 11 , \text{ then find} \lambda \text{ and } \mu .$

Answer:

$\displaystyle f'(x) = \frac{d}{dx} (\lambda x^2 + \mu x + 12)$

$\displaystyle = \lambda \frac{d}{dx} (x^2) + \mu \frac{d}{dx} (x) + \frac{d}{dx} (12)$

$\displaystyle = 2 \lambda x + \mu$

Given $\displaystyle f'(4) = 15$

$\displaystyle \Rightarrow 2 \lambda ( 4) + \mu = 15$

$\displaystyle \Rightarrow 8 \lambda + u = 15 \text{ ... ... ... ... ... ... (i) }$

Also $\displaystyle f'(2) = 11$

$\displaystyle \Rightarrow 2 \lambda ( 2) + \mu = 11$

$\displaystyle \Rightarrow 4 \lambda + u = 11 \text{ ... ... ... ... ... ... (ii) }$

Solving (i) and (ii) we get $\displaystyle \lambda = 1$ and $\displaystyle \mu = 7$

$\\$

$\displaystyle \text{Question 26: } \text{For the function } f(x) = \frac{x^{100}}{100} + \frac{x^{99}}{99} + \cdots + \frac{x^2}{2}+ x + 1. \\ \\ \text{Prove that } f'(1) = 100 f'(0) .$