Find the derivative of the following functions

Solution:

Let $f(x)=\frac{x}{1+\tan x}$

$f^{\prime}(x)=\frac{(1+\tan x) \frac{d}{d x}(x)-x \frac{d}{d x}(1+\tan x)}{(1+\tan x)^{2}}$

$f^{\prime}(x)=\frac{(1+\tan x)-x \cdot \frac{d}{d x}(1+\tan x)}{(1+\tan x)^{2}}$ (i)

Let $g(x)=1+\tan x$. Accordingly, $g(x+h)=1+\tan (x+h)$.

By first principle,

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

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

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

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

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

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

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

$=1 \times \frac{1}{\cos ^{2} x}=\sec ^{2} x$

$\Rightarrow \frac{d}{d x}(1+\tan x)=\sec ^{2} x$ (ii)

From (i) and (ii), we obtain

$f^{\prime}(x)=\frac{1+\tan x-x \sec ^{2} x}{(1+\tan x)^{2}}$