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# Find the derivative of the following functions ( it is to be understood that a, b, c, d, p, q, r and s are fixed non-zero constants and m and n are integers ) $(x+\cos x)(x-\tan x)$

Let $f(x) = (x+\cos x)(x-\tan x)$
By product rule,
$f'(x) =(x+\cos x) \large\frac{d}{dx}$$(x-\tan x)+(x-\tan x)\large\frac{d}{dx}$$(x+\cos x)$
$x+ \cos x \bigg[ \large\frac{d}{dx}(x) - \large\frac{d}{dx}$$(\tan x) \bigg]$$+( x - \tan x)(1- \sin x)$
$(x+ \cos x) \bigg[ 1 - \large\frac{d}{dx}$$\tan x \bigg]$$+(x- \tan x)(1-\sin x)$----------(i)
Let $g(x) = \tan x$. Accordingly, $g(x+h)= \tan (x+h)$
By first principle,
$g'(x) = \lim\limits_{h \to 0} \large\frac{g(x+h)-g(x)}{h}$
$= \lim\limits_{h \to 0} \bigg( \large\frac{\tan (x+h)-\tan x}{h} \bigg)$
$= \lim\limits_{ h \to 0} \large\frac{1}{h}$$\bigg[ \large\frac{\sin (x+h)}{\cos (x+h)}$$- \large\frac{\sin x}{\cos x} \bigg]$
$= \lim\limits_{ h \to 0} \large\frac{1}{h}$$\bigg[ \large\frac{\sin (x+h) \cos x- \sin x \cos(x+h)}{\cos (x+h) \cos x}\bigg] =\large\frac{1}{\cos x}$$. \lim\limits_{ h \to 0} \large\frac{1}{h}$$\bigg[ \large\frac{\sin (x+h-x)}{\cos (x+h)}\bigg] =\large\frac{1}{\cos x}$$. \lim\limits_{ h \to 0} \large\frac{1}{h}$$\bigg[ \large\frac{\sin h}{\cos (x+h)}\bigg] = \large\frac{1}{\cos x}$$. \bigg( \lim\limits_{h \to 0} \large\frac{\sin h}{h} \bigg)$$\bigg( \lim\limits_{h \to 0} \large\frac{1}{\cos (x+h)} \bigg) = \large\frac{1}{\cos x}$$.1. \large\frac{1}{\cos (x+0)}$
$= \large\frac{1}{\cos^2x}$
$= \sec^2 x$------------(ii)
Therefore from (i) and (ii), we obtain,
$f'(x) =(x+ \cos x)(1- \sec^2x)+(x-\tan x)(1- \sin x)$
$= ( x + \cos x) ( - \tan^2x)+(x-\tan x)(1- \sin x)$
$= -\tan^2x(x+ \cos x) +(x-\tan x)(1- \sin x)$