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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 ) $ cosec \: x \cot x$

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Let $f(x) = cosec \: x \cot x$
By Leibnitz product rule,
$f'(x) = cosec\: x( \cot x)'+ (\cot x) cosec \: x'$---------(1)
Let $f_1(x) = \cot x$. Accordingly, $f_1(x+h)=\cot(x+h)$
By first principle,
$ f'_1(x) = \lim\limits_{h \to 0} \Large\frac{f_1(x+h)-f_1(x)}{h}$
$ = \lim\limits_{h \to 0} \Large\frac{\cot(x+h)-\cot x}{h}$
$ = \lim\limits_{ h \to 0} \large\frac{1}{h}$$ \bigg[ \Large\frac{\cos (x+h)}{\sin (x+h)}$$ - \Large\frac{\cos x}{\sin x} \bigg]$
$ = \lim\limits_{ h \to 0} \large\frac{1}{h}$$ \bigg[ \Large\frac{\sin x \cos (x+h)- \cos x \sin (x+h)}{\sin x \sin (x+h)} \bigg]$
$ = \lim\limits_{ h \to 0} \large\frac{1}{h}$$ \bigg[ \Large\frac{\sin (x-x-h)}{\sin x \sin (x+h)}\bigg]$
$ =\large\frac{1}{\sin x}$$ \lim\limits_{ h \to 0} \large\frac{1}{h}$$ \bigg[ \Large\frac{\sin (-h)}{\sin (x+h)} \bigg]$
$ =\large\frac{-1}{\sin x}$$ \bigg(\lim\limits_{ h \to 0} \large\frac{\sin h}{h}\bigg)$$\bigg(\lim\limits_{ h \to 0} \large\frac{1}{\sin (x+h)}\bigg)$
$ =\large\frac{-1}{\sin x}$$.1 \bigg(\large\frac{1}{\sin (x+0)}\bigg)$
$ =\large\frac{-1}{\sin^2 x}$
$ = - cosec^2x$
$ \therefore (\cot x)' = -cosec^2x'$-----------(2)
Now, let $f_2(x) = cosec\:x$ Accordingly,$f_2(x+h) = cosec(x+h)$
By first principle,
$ f'_2(x) = \lim\limits_{h \to 0} \Large\frac{f_2(x+h)-f_2(x)}{h}$
$ = \lim\limits_{h \to 0}\large\frac{1}{h}$$ [cosec (x+h)- cosec x]
$ = \lim\limits_{h \to 0}\large\frac{1}{h}$$ \bigg[ \large\frac{1}{\sin (x+h)}$$ - \large\frac{1}{\sin x}\bigg]$
$ = \lim\limits_{h \to 0}\large\frac{1}{h}$$ \bigg[ \large\frac{\sin x-\sin (x+h)}{\sin x \sin(x+h)} \bigg]$
$ =\large\frac{1}{\sin x}$$. \lim\limits_{h \to 0}\large\frac{1}{h}$$ \bigg[ \large\frac{2\cos \bigg( \Large\frac{x+x+h}{2} \bigg) \sin \bigg(\Large\frac{x-x-h}{2} \bigg)}{\sin (x+h)} \bigg]$
$ =\large\frac{1}{\sin x}$$. \lim\limits_{h \to 0}\large\frac{1}{h}$$ \bigg[ \large\frac{2\cos \bigg( \Large\frac{2x+h}{2} \bigg) \sin \bigg(\Large\frac{-h}{2} \bigg)}{\sin (x+h)} \bigg]$
$ =\large\frac{1}{\sin x}$$. \lim\limits_{h \to 0} \bigg[ \large\frac{-\sin \bigg( \Large\frac{h}{2} \bigg)}{\bigg(\Large\frac{h}{2} \bigg)}. \Large\frac{\cos \bigg( \Large\frac{2x+h}{2} \bigg)}{\sin (x+h)} \bigg]$
$ =\large\frac{-1}{\sin x}$$. \lim\limits_{h \to 0}\large\frac{\sin \bigg( \Large\frac{h}{2} \bigg)}{\bigg(\Large\frac{h}{2} \bigg)}$$\lim\limits_{ h \to 0} \Large\frac{ \cos \bigg( \Large\frac{2x+h}{2} \bigg)}{\sin (x+h)} $
$ =\large\frac{-1}{\sin x}$$.1. \Large\frac{ \cos \bigg( \Large\frac{2x+0}{2} \bigg)}{\sin (x+0)} $
$ =\large\frac{-1}{\sin x}$$. \Large\frac{\cos x}{\sin x}$
$ = -cosec x. \cot x$
$ \therefore ( cosec\: x)' = -cosec x \cot x$--------(3)
From (1),(2) and (3) we obtain,
$f'(x) = cosec \: x (- cosec^2 x)+ \cot x ( - cosecx \cot x)$
$ = -cosec^3x-\cot^2x cosecx$
answered Apr 14, 2014 by thanvigandhi_1

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