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Home  >>  CBSE XI  >>  Math  >>  Limits and Derivatives
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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 ) $(x + \sec x)(x-\tan x)$

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Let $f(x) = (x + \sec x)(x-\tan x)$
By product rule,
$f'(x) = (x \sec x)\large\frac{d}{dx}$$(x-\tan x)+(x-\tan x)\large\frac{d}{dx}$$(x+ \sec x)$
$= ( x + \sec x) \bigg[ \large\frac{d}{dx}$$(x) - \large\frac{d}{dx}$$ \tan x \bigg]+ (x - \tan x) \bigg[\large\frac{d}{dx}$$(x) + \large\frac{d}{dx}$$ \sec x \bigg]$
$ = (x+ \sec x) \bigg[1- \large\frac{d}{dx}$$\tan x \bigg]+ (x - \tan x)\bigg[1+ \large\frac{d}{dx}$$\sec x \bigg]$-----------(i)
Let $f_1(x)= \tan x, f_2(x) = \sec x$
Accordingly, $f_1 (x+h)= \tan (x+h) \: and \: f_2(x+h) =\sec (x+h)$
$f'_1(x) = \lim\limits_{h \to 0} \bigg( \large\frac{f_1(x+h)-f_1(x)}{h}\bigg)$
$ = \lim\limits_{ h \to 0}\bigg( \large\frac{\tan (x+h)-\tan (x)}{h}\bigg)$
$ = \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}$$ \bigg[ \large\frac{\sin (x+h) \cos x- \sin x \cos(x+h)}{\cos (x+h) \cos x}\bigg]$
$= \lim\limits_{h to 0} \large\frac{1}{h}$$ \bigg[ \large\frac{\sin (x+h-x)}{\cos (x+h) \cos x} \bigg]$
$= \lim\limits_{h to 0} \large\frac{1}{h}$$ \bigg[ \large\frac{\sin h}{\cos (x+h) \cos x} \bigg]$
$ = \bigg( \lim\limits_{ h \to 0} \large\frac{\sin h}{h} \bigg)$$.\bigg( \lim\limits_{ h \to 0} \large\frac{1}{\cos (x+h)\cos x} \bigg)$
$ = 1 \times \large\frac{1}{\cos^2x}$$= \sec^2x$
$ \Rightarrow \large\frac{d}{dx}$$\tan x=\sec^2x$---------(ii)
$f'_2(x) = \lim\limits_{h \to 0} \bigg( \large\frac{f_2(x+h)-f_2(x)}{h}\bigg)$
$ = \lim\limits_{ h \to 0}\bigg( \large\frac{\sec (x+h)-\sec (x)}{h}\bigg)$
$= \lim\limits_{h to 0} \large\frac{1}{h}$$ \bigg[ \large\frac{1}{\cos (x+h)}$$ - \large\frac{1 }{\cos x} \bigg]$
$= \lim\limits_{h to 0} \large\frac{1}{h}$$ \bigg[ \large\frac{\cos x - \cos(x+h)}{\cos (x+h) \cos x}\bigg]$
$ =\large\frac{1}{\cos x}$$. \lim\limits_{h \to 0}\large\frac{1}{h}$$ \bigg[ \large\frac{-2\sin \bigg( \Large\frac{x+x+h}{2} \bigg). \sin \bigg(\Large\frac{x-x-h}{2} \bigg)}{\cos (x+h)} \bigg]$
$ =\large\frac{1}{\cos x}$$. \lim\limits_{h \to 0}\large\frac{1}{h}$$ \bigg[ \large\frac{-2\sin \bigg( \Large\frac{2x+h}{2} \bigg) \sin \bigg(\Large\frac{-h}{2} \bigg)}{\cos (x+h)} \bigg]$
$ =\large\frac{1}{\cos x}$$. \lim\limits_{h \to 0} \bigg[\Large\frac{\sin \bigg( \Large\frac{2x+h}{2} \bigg) \bigg\{\Large\frac{\sin \bigg( \Large\frac{h}{2} \bigg)}{\Large\frac{h}{2} }\bigg\}}{\cos (x+h)} \bigg]$
$ = \sec x \large\frac{ \bigg\{ \lim\limits_{h \to 0} \sin \bigg( \large\frac{2x+h}{2} \bigg) \bigg\} \bigg\{ \lim\limits_{ \Large\frac{h}{2} \to 0}\Large\frac{\sin \bigg( \Large\frac{h}{2} \bigg)}{\large\frac{h}{2} }\bigg\}}{ \lim\limits_{h \to 0} \cos(x+h)}$
$ = \sec x . \large\frac{\sin x.1}{\cos x}$
$ \Rightarrow \large\frac{d}{dx}$$\sec x = \sec x \tan x$---------(iii)
From (i), (ii) and (iii) , we obtain,
$f'(x) =(x+ \sec x)(1-\sec^2x)+ (x-\tan x)( 1 + \sec x \tan x)$
answered Apr 16, 2014 by thanvigandhi_1
edited Apr 16, 2014 by thanvigandhi_1
 

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