\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 

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\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

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\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

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\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

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\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

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\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

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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