Class 11, Derivatives, Exercise 30.1

Class 11: Derivatives – Exercise 30.1

Date: January 28, 2022Author: ICSE CBSE ISC Board Mathematics Portal for Students 0 Comments

$\displaystyle \text{Question 1: Find the derivative of } f(x) = 3x \text{ at } x = 2$

Answer:

We have:

$\displaystyle f'(2) = \lim \limits_{h \to 0 } \frac{f(2+h) - f(2) }{h}$

$\displaystyle = \lim \limits_{h \to 0 } \frac{3(2+h) - 3(2)}{h}$

$\displaystyle = \lim \limits_{h \to 0 } \frac{6+3h-6}{h}$

$\displaystyle = \lim \limits_{h \to 0 } \frac{3h}{h}$

$\displaystyle = 3$

$\\$

$\displaystyle \text{Question 2: Find the derivative of } f(x) = x^2 - 2 \text{ at } x = 10$

Answer:

We have:

$\displaystyle f'(x) = \lim \limits_{h \to 0 } \frac{f(10+h) - f(10) }{h}$

$\displaystyle = \lim \limits_{h \to 0 } \frac{(10+h)^2 - 2 - (10^2-2)}{h}$

$\displaystyle = \lim \limits_{h \to 0 } \frac{100+h^2 + 20 h - 2 - 100 + 2}{h}$

$\displaystyle = \lim \limits_{h \to 0 } \frac{h^2 + 20 h}{h}$

$\displaystyle = \lim \limits_{h \to 0 } \frac{h( h + 20)}{h}$

$\displaystyle = \lim \limits_{h \to 0 } h+20$

$\displaystyle = 20$

$\\$

$\displaystyle \text{Question 3: Find the derivative of } f(x) = 99x \text{ at } x = 100$

Answer:

We have:

$\displaystyle f'(100) = \lim \limits_{h \to 0 } \frac{f(100+h) - f(100) }{h}$

$\displaystyle = \lim \limits_{h \to 0 } \frac{99(100+h) - 99(100)}{h}$

$\displaystyle = \lim \limits_{h \to 0 } \frac{9900+99h-9900}{h}$

$\displaystyle = \lim \limits_{h \to 0 } \frac{99h}{h}$

$\displaystyle = \lim \limits_{h \to 0 } 99$

$\displaystyle = 99$

$\\$

$\displaystyle \text{Question 4: Find the derivative of } f(x) = x \text{ at } x = 1$

Answer:

We have:

$\displaystyle f'(x) = \lim \limits_{h \to 0 } \frac{f(1+h) - f(1) }{h}$

$\displaystyle = \lim \limits_{h \to 0 } \frac{1+h-1}{h}$

$\displaystyle = \lim \limits_{h \to 0 } 1$

$\displaystyle = 1$

$\\$

$\displaystyle \text{Question 5: Find the derivative of } f(x) = \cos x \text{ at } x = 0$

Answer:

We have:

$\displaystyle f'(x) = \lim \limits_{h \to 0 } \frac{f(0+h) - f(0) }{h}$

$\displaystyle = \lim \limits_{h \to 0 } \frac{\cos h - \cos 0}{h}$

$\displaystyle = \lim \limits_{h \to 0 } \frac{\cos h - 1}{h}$

$\displaystyle = \lim \limits_{h \to 0 } \frac{-2 \sin^2 \frac{h}{2}}{h}$

$\displaystyle = \lim \limits_{h \to 0 } \frac{-2 \sin^2 \frac{h}{2}}{\frac{h^2}{4}} \times \frac{h}{4}$

$\displaystyle = \lim \limits_{h \to 0 } -1 \times \frac{h}{2}$

$\displaystyle = 0$

$\\$

$\displaystyle \text{Question 6: Find the derivative of } f(x) = \tan x \text{ at } x = 0$

Answer:

We have:

$\displaystyle f'(x) = \lim \limits_{h \to 0 } \frac{f(0+h) - f(0) }{h}$

$\displaystyle = \lim \limits_{h \to 0 } \frac{\tan h - \tan 0}{h}$

$\displaystyle = \lim \limits_{h \to 0 } \frac{\tan h }{h}$

$\displaystyle = 1$

$\\$

Question 7: Find the derivation of the following functions at the indicated points:

$\displaystyle \text{(i) } \sin x \text{ at } x = \frac{\pi}{2}$

$\displaystyle \text{(ii) } x \text{ at } x = 1$

$\displaystyle \text{(iii) } 2 \cos x \text{ at } x = \frac{\pi}{2}$

$\displaystyle \text{(vi) } \sin 2x \text{ at } x \frac{\pi}{2}$

Answer:

(i) We have:

$\displaystyle f'(\frac{\pi}{2}) = \lim \limits_{h \to 0 } \frac{f(\frac{\pi}{2}+h) - f(\frac{\pi}{2}) }{h}$

$\displaystyle = \lim \limits_{h \to 0 } \frac{\sin (\frac{\pi}{2}+h) - \sin (\frac{\pi}{2})}{h}$

$\displaystyle = \lim \limits_{h \to 0 } \frac{\cos h - 1}{h}$

$\displaystyle = \lim \limits_{h \to 0 } \frac{-2 \sin^2 \frac{h}{2}}{h}$

$\displaystyle = \lim \limits_{h \to 0 } \frac{-2 \sin^2 \frac{h}{2}}{\frac{h^2}{4}} \times \frac{h}{4}$

$\displaystyle = \lim \limits_{h \to 0 } -1 \times \frac{h}{2}$

$\displaystyle = 0$

(ii) We have:

$\displaystyle f'(x) = \lim \limits_{h \to 0 } \frac{f(1+h) - f(1) }{h}$

$\displaystyle = \lim \limits_{h \to 0 } \frac{1+h-1}{h}$

$\displaystyle = \lim \limits_{h \to 0 } 1$

$\displaystyle = 1$

(iii) We have:

$\displaystyle f'(\frac{\pi}{2}) = \lim \limits_{h \to 0 } \frac{f(\frac{\pi}{2}+h) - f(\frac{\pi}{2}) }{h}$

$\displaystyle = \lim \limits_{h \to 0 } \frac{2\cos (\frac{\pi}{2}+h) - \cos (\frac{\pi}{2})}{h}$

$\displaystyle = \lim \limits_{h \to 0 } \frac{-2 \sin h - 0}{h}$

$\displaystyle = -2 \lim \limits_{h \to 0 } \frac{\sin h}{h}$

$\displaystyle = -2$

(vi) We have:

$\displaystyle f'(\frac{\pi}{2}) = \lim \limits_{h \to 0 } \frac{f(\frac{\pi}{2}+h) - f(\frac{\pi}{2}) }{h}$

$\displaystyle = \lim \limits_{h \to 0 } \frac{\sin 2(\frac{\pi}{2}+h) - \sin 2(\frac{\pi}{2})}{h}$

$\displaystyle = \lim \limits_{h \to 0 } \frac{\sin ( \pi + 2h) - 0 }{h}$

$\displaystyle = \lim \limits_{h \to 0 } \frac{-\sin 2h}{h} \times \frac{2}{2}$

$\displaystyle = \lim \limits_{h \to 0 } \frac{-\sin 2h}{2h} \times 2$

$\displaystyle = -2$

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