\displaystyle \textbf{Question 1: } \text{Find the derivative of }f(x)=3x\text{ at }x=2.
\displaystyle \text{Answer:}
\displaystyle \text{We have,}
\displaystyle f'(2)=\lim\limits_{h\to0}\frac{f(2+h)-f(2)}{h}
\displaystyle =\lim\limits_{h\to0}\frac{3(2+h)-3(2)}{h}
\displaystyle =\lim\limits_{h\to0}\frac{6+3h-6}{h}
\displaystyle =\lim\limits_{h\to0}\frac{3h}{h}
\displaystyle =3
\displaystyle \\

\displaystyle \textbf{Question 2: } \text{Find the derivative of }f(x)=x^2-2\text{ at }x=10.
\displaystyle \text{Answer:}
\displaystyle \text{We have,}
\displaystyle f'(10)=\lim\limits_{h\to0}\frac{f(10+h)-f(10)}{h}
\displaystyle =\lim\limits_{h\to0}\frac{(10+h)^2-2-(10^2-2)}{h}
\displaystyle =\lim\limits_{h\to0}\frac{100+20h+h^2-2-100+2}{h}
\displaystyle =\lim\limits_{h\to0}\frac{h^2+20h}{h}
\displaystyle =\lim\limits_{h\to0}(h+20)
\displaystyle =20
\displaystyle \\

\displaystyle \textbf{Question 3: } \text{Find the derivative of }f(x)=99x\text{ at }x=100.
\displaystyle \text{Answer:}
\displaystyle \text{We have,}
\displaystyle f'(100)=\lim\limits_{h\to0}\frac{f(100+h)-f(100)}{h}
\displaystyle =\lim\limits_{h\to0}\frac{99(100+h)-99(100)}{h}
\displaystyle =\lim\limits_{h\to0}\frac{9900+99h-9900}{h}
\displaystyle =\lim\limits_{h\to0}\frac{99h}{h}
\displaystyle =99
\displaystyle \\

\displaystyle \textbf{Question 4: } \text{Find the derivative of }f(x)=x\text{ at }x=1.
\displaystyle \text{Answer:}
\displaystyle \text{We have,}
\displaystyle f'(1)=\lim\limits_{h\to0}\frac{f(1+h)-f(1)}{h}
\displaystyle =\lim\limits_{h\to0}\frac{1+h-1}{h}
\displaystyle =\lim\limits_{h\to0}1
\displaystyle =1
\displaystyle \\

\displaystyle \textbf{Question 5: } \text{Find the derivative of }f(x)=\cos x\text{ at }x=0.
\displaystyle \text{Answer:}
\displaystyle \text{We have,}
\displaystyle f'(0)=\lim\limits_{h\to0}\frac{f(0+h)-f(0)}{h}
\displaystyle =\lim\limits_{h\to0}\frac{\cos h-\cos0}{h}
\displaystyle =\lim\limits_{h\to0}\frac{\cos h-1}{h}
\displaystyle =\lim\limits_{h\to0}\frac{-2\sin^2\frac{h}{2}}{h}
\displaystyle =\lim\limits_{h\to0}\left(-\frac{\sin\frac{h}{2}}{\frac{h}{2}}\cdot\sin\frac{h}{2}\right)
\displaystyle =-1\times0=0
\displaystyle \\

\displaystyle \textbf{Question 6: } \text{Find the derivative of }f(x)=\tan x\text{ at }x=0.
\displaystyle \text{Answer:}
\displaystyle \text{We have,}
\displaystyle f'(0)=\lim\limits_{h\to0}\frac{f(0+h)-f(0)}{h}
\displaystyle =\lim\limits_{h\to0}\frac{\tan h-\tan0}{h}
\displaystyle =\lim\limits_{h\to0}\frac{\tan h}{h}
\displaystyle =1
\displaystyle \\

\displaystyle \textbf{Question 7: } \text{Find the derivative 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{(iv) }\sin2x\text{ at }x=\frac{\pi}{2}
\displaystyle \text{Answer:}

\displaystyle \text{(i) We have,}
\displaystyle f'\!\left(\frac{\pi}{2}\right)=\lim\limits_{h\to0}\frac{f\!\left(\frac{\pi}{2}+h\right)-f\!\left(\frac{\pi}{2}\right)}{h}
\displaystyle =\lim\limits_{h\to0}\frac{\sin\left(\frac{\pi}{2}+h\right)-\sin\frac{\pi}{2}}{h}
\displaystyle =\lim\limits_{h\to0}\frac{\cos h-1}{h}
\displaystyle =\lim\limits_{h\to0}\frac{-2\sin^2\frac{h}{2}}{h}
\displaystyle =\lim\limits_{h\to0}\left(-\frac{\sin\frac{h}{2}}{\frac{h}{2}}\cdot\sin\frac{h}{2}\right)
\displaystyle =-1\times0=0

\displaystyle \text{(ii) We have,}
\displaystyle f'(1)=\lim\limits_{h\to0}\frac{f(1+h)-f(1)}{h}
\displaystyle =\lim\limits_{h\to0}\frac{1+h-1}{h}
\displaystyle =\lim\limits_{h\to0}1
\displaystyle =1

\displaystyle \text{(iii) We have,}
\displaystyle f'\!\left(\frac{\pi}{2}\right)=\lim\limits_{h\to0}\frac{f\!\left(\frac{\pi}{2}+h\right)-f\!\left(\frac{\pi}{2}\right)}{h}
\displaystyle =\lim\limits_{h\to0}\frac{2\cos\left(\frac{\pi}{2}+h\right)-2\cos\frac{\pi}{2}}{h}
\displaystyle =\lim\limits_{h\to0}\frac{-2\sin h}{h}
\displaystyle =-2\lim\limits_{h\to0}\frac{\sin h}{h}
\displaystyle =-2

\displaystyle \text{(iv) We have,}
\displaystyle f'\!\left(\frac{\pi}{2}\right)=\lim\limits_{h\to0}\frac{f\!\left(\frac{\pi}{2}+h\right)-f\!\left(\frac{\pi}{2}\right)}{h}
\displaystyle =\lim\limits_{h\to0}\frac{\sin2\left(\frac{\pi}{2}+h\right)-\sin2\left(\frac{\pi}{2}\right)}{h}
\displaystyle =\lim\limits_{h\to0}\frac{\sin(\pi+2h)-0}{h}
\displaystyle =-\lim\limits_{h\to0}\frac{\sin2h}{2h}\times2
\displaystyle =-1\times2=-2
\displaystyle \\

 

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