Find the derivative of the following functions from first principle:

(i) Let $f(x)=-x .$ Accordingly, $f(x+h)=-(x+h)$

By first principle,

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

$=\lim _{h \rightarrow 0} \frac{-(x+h)-(-x)}{h}$

$=\lim _{h \rightarrow 0} \frac{-x-h+x}{h}$

$=\lim _{h \rightarrow 0} \frac{-h}{h}$

$=\lim _{h \rightarrow 0}(-1)=-1$

(ii) Let $f(x)=(-x)^{-1}=\frac{1}{-x}=\frac{-1}{x} \cdot$ Accordingly, $f(x+h)=\frac{-1}{(x+h)}$

By first principle,

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

$=\lim _{h \rightarrow 0} \frac{1}{h}\left[\frac{-1}{x+h}-\left(\frac{-1}{x}\right)\right]$

$=\lim _{h \rightarrow 0} \frac{1}{h}\left[\frac{-1}{x+h}+\frac{1}{x}\right]$

$=\lim _{h \rightarrow 0} \frac{1}{h}\left[\frac{-x+(x+h)}{x(x+h)}\right]$

$=\lim _{h \rightarrow 0} \frac{1}{h}\left[\frac{-x+x+h}{x(x+h)}\right]$

$=\lim _{h \rightarrow 0} \frac{1}{h}\left[\frac{h}{x(x+h)}\right]$

$=\lim _{h \rightarrow 0} \frac{1}{x(x+h)}$

$=\frac{1}{x \cdot x}=\frac{1}{x^{2}}$

(iii) Let $f(x)=\sin (x+1)$. Accordingly, $f(x+h)=\sin (x+h+1)$

By first principle,

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

$=\lim _{h \rightarrow 0} \frac{1}{h}[\sin (x+h+1)-\sin (x+1)]$

$=\lim _{h \rightarrow 0} \frac{1}{h}\left[2 \cos \left(\frac{x+h+1+x+1}{2}\right) \sin \left(\frac{x+h+1-x-1}{2}\right)\right]$

$=\lim _{\mathrm{h} \rightarrow 0} \frac{1}{\mathrm{~h}}\left[2 \cos \left(\frac{2 \mathrm{x}+\mathrm{h}+2}{2}\right) \sin \left(\frac{\mathrm{h}}{2}\right)\right]$

$=\lim _{h \rightarrow 0} \frac{1}{h}\left[2 \cos \left(\frac{2 x+h+2}{2}\right) \sin \left(\frac{h}{2}\right)\right]$

$=\lim _{\mathrm{h} \rightarrow 0}\left[\cos \left(\frac{2 \mathrm{x}+\mathrm{h}+2}{2}\right) \cdot \frac{\sin \left(\frac{\mathrm{h}}{2}\right)}{\left(\frac{\mathrm{h}}{2}\right)}\right]$

$=\lim _{h \rightarrow 0} \cos \left(\frac{2 x+h+2}{2}\right) \cdot \lim _{h \rightarrow 0} \frac{\sin \left(\frac{h}{2}\right)}{\left(\frac{h}{2}\right)} \quad\left[\right.$ As $\left.h \rightarrow 0 \Rightarrow \frac{h}{2} \rightarrow 0\right]$

$=\cos \left(\frac{2 x+0+2}{2}\right) \cdot 1 \quad\left[\lim _{x \rightarrow 0} \frac{\sin x}{x}=1\right]$

$=\cos (x+1)$

(iv) Letf $(x)=\cos \left(x-\frac{\pi}{8}\right) .$ Accordingly, $f(x+h)=\cos \left(x+h-\frac{\pi}{8}\right)$

By first principle,

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

$=\lim _{h \rightarrow 0} \frac{1}{h}\left[\cos \left(x+h-\frac{\pi}{8}\right)-\cos \left(x-\frac{\pi}{8}\right)\right]$

$=\lim _{h \rightarrow 0} \frac{1}{h}\left[-2 \sin \frac{\left(x+h-\frac{\pi}{8}+x-\frac{\pi}{8}\right)}{2} \sin \left(\frac{x+h-\frac{\pi}{8}-x+\frac{\pi}{8}}{2}\right]\right]$

$=\lim _{h \rightarrow 0} \frac{1}{h}\left[-2 \sin \left(\frac{2 x+h-\frac{\pi}{4}}{2}\right) \sin \frac{h}{2}\right]$

$=\lim _{h \rightarrow 0}\left[-\sin \left(\frac{2 x+h-\frac{\pi}{4}}{2}\right) \frac{\sin \left(\frac{h}{2}\right)}{\left(\frac{h}{2}\right)}\right]$

$=\lim _{\mathrm{h} \rightarrow 0}\left[-\sin \left(\frac{2 \mathrm{x}+\mathrm{h}-\frac{\pi}{4}}{2}\right]\right] \lim _{\frac{\mathrm{h}}{2} \rightarrow 0} \frac{\sin \left(\frac{\mathrm{h}}{2}\right)}{\left(\frac{\mathrm{h}}{2}\right)} \quad\left[\right.$ As $\left.\mathrm{h} \rightarrow 0 \Rightarrow \frac{\mathrm{h}}{2} \rightarrow 0\right]$

$=-\sin \left(\frac{2 x+0-\frac{\pi}{4}}{2}\right) .1$

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