Find the derivative of the following functions

Let $f(x)=(x+\cos x)(x-\tan x)$

By product rule,

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

$=(x+\cos x)\left[\frac{d}{d x}(x)-\frac{d}{d x}(\tan x)\right]+(x-\tan x)(1-\sin x)$

$=(x+\cos x)\left[1-\frac{d}{d x} \tan x\right]+(x-\tan x)(1-\sin x)$ (i)

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

By first principle,

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

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

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

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

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

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

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

$=\frac{1}{\cos x} \cdot 1 \cdot \frac{1}{\cos (x+0)}$

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

$=\sec ^{2} x$ (ii)

Therefore, from (i) and (ii), we obtain

$f^{\prime}(x)=(x+\cos x)\left(1-\sec ^{2} x\right)+(x-\tan x)(1-\sin x)$

$=(x+\cos x)\left(-\tan ^{2} x\right)+(x-\tan x)(1-\sin x)$

$=-\tan ^{2} x(x+\cos x)+(x-\tan x)(1-\sin x)$