Question

Differentiate each of the following from first principles:
(i) -x
(ii) $(-x{)}^{-1}$
(iii) sin (x+1)
(iv) $cos(x-\frac{\pi }{8})$

Answer 1

(i) Let f(x) =-x

We have,

$f^{\prime }\left(x\right)={lim}_{h\to 0}\frac{f(x+h)-f\left(x\right)}{h}$

$={lim}_{h\to 0}\frac{-(x+h)-(-x)}{h}$ (By first principle)

$={lim}_{h\to 0}\frac{-x-h+x}{h}$ $[\because f(x)=-x]$

$\Rightarrow f^{\prime }\left(x\right)=limh\to 0\frac{-h}{h}=-1$

(ii) Let $f\left(x\right)=(-x{)}^{-1}$

$\Rightarrow f\left(x\right)=-\frac{1}{x}$

We have, $f^{\prime }\left(x\right)={lim}_{h\to 0}\frac{f(x+h)-f\left(x\right)}{h}$ (By first principle)

$\Rightarrow f^{\prime }\left(x\right)={lim}_{h\to 0}\frac{-\frac{1}{x+h}+\frac{1}{x}}{h}$ $(\because f(x)=\frac{1}{x})$

$\Rightarrow f^{\prime }\left(x\right){lim}_{h\to 0}\frac{\frac{1}{x}-\frac{1}{x+h}}{h}={lim}_{h\to 0}\frac{x+h-x}{x(x+h)h}$

$={lim}_{h\to 0}\frac{h}{x(x+h)h}=\frac{1}{x(x+0)}=\frac{1}{x^2}$

(iii) f(x)= sin(x+1)
We have, $f^{\prime }\left(x\right)={lim}_{h\to 0}\frac{f(x+h)-f\left(x\right)}{h}$ (By first principle)

$\Rightarrow f^{\prime }\left(x\right)={lim}_{h\to 0}\frac{sin(x+h+1)-sin(x+1)}{h}$ $[\because f(x)=sin(x+1\left)\right]$

$\Rightarrow f^{\prime }\left(x\right)={lim}_{h\to 0}\frac{20cos\frac{x+h+1+x+1}{2}sin\frac{x+h+1-x-1}{2}}{h}$ $[\because sinC-sinD=2cos\frac{C+D}{2}sin\frac{C-D}{2}]$

${lim}_{h\to 0}\frac{2cos\frac{2x+h+2}{2}sin\frac{h}{2}}{2\times \frac{h}{2}}=cos\frac{2x+0+2}{2}\times 1$ $(\because {lim}_{h\to 0}\frac{sin\frac{h}{2}}{\frac{h}{2}}=1)$

$\Rightarrow f^{\prime }\left(x\right)=cos(x+1)$

(iv) Let $f\left(x\right)=cos(x-\frac{\pi }{8})$

We have, $f^{\prime }\left(x\right)={lim}_{h\to 0}\frac{f(x+h)-f\left(x\right)}{h}$ (By first principle)

$\Rightarrow f^{\prime }\left(x\right)={lim}_{h\to 0}\frac{cos(x+h-\frac{\pi }{8})-cos(x-\frac{\pi }{8})}{h}$ $(\because f(x)=cos(x-\frac{\pi }{8}))$

$\Rightarrow {lim}_{h\to 0}\frac{-2sin\frac{x+h-\frac{\pi }{8}+x-\frac{\pi }{8}}{2}sin\frac{x+h-\frac{\pi }{8}-x+\frac{\pi }{8}}{2}}{h}$ $[\because cosC-cosD=-2sin\frac{C+D}{2}sin\frac{C-D}{2}]$

$={lim}_{h\to 0}\frac{-2sin\frac{2x-2(\frac{\pi }{8}+h)sin\frac{h}{2}}{2}}{2\times \frac{h}{2}}$ = $-sin\frac{2x-2\left(\frac{\pi }{8}\right)+0}{2}\times 1$ $(\because {lim}_{h\to 0}\frac{sin\frac{h}{2}}{\frac{h}{2}}=1)$

$=-sin\frac{2(x-\frac{\pi }{8})}{2}$

$\Rightarrow f^{\prime }\left(x\right)=-sin(x-\frac{\pi }{8})$