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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 ) $\large\frac{x}{1+\tan x}$

1 Answer

Let $f(x) = \large\frac{x}{1+\tan x}$
$f'(x) =\large\frac{(1+\tan x)\large\frac{d}{dx}(x)-x\large\frac{d}{dx}(1+tan x)}{(1+\tan x)^2}$
$f'(x) =\large\frac{(1+\tan x)-x.\large\frac{d}{dx}(1+tan x)}{(1+\tan x)^2}$----------(i)
Let $g(x)=1+ tan x$. Accordingly, $g(x+h),=1+\tan (x+h)$
By first principle,
$g'(x) = \lim\limits_{ h \to 0} = \large\frac{g(x+h)-g(x)}{h}$
$= \lim\limits_{h \to 0}\bigg[ \large\frac{1+\tan (x+h)-1-\tan x}{h}\bigg]$
$= \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}$$= \sec^2x$
$ \Rightarrow \large\frac{d}{dx}$$(1+ \tan x)=\sec^2x$-----------(ii)
From (i) and (ii) , we obtain,
$ f'(x) =\large\frac{1+\tan x- x \sec^2x}{(1+\tan x)^2}$
answered Apr 16, 2014 by thanvigandhi_1

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