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Home  >>  CBSE XI  >>  Math  >>  Limits and Derivatives
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Find the derivative of the given function $ 3 \cot x + 5 cosec\: x$

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Let $f(x) = 3 \cot x + 5 cosec\: x $. Accordingly, from the first principle,
$ f'(x) = \lim\limits_{h \to 0}= \large\frac{f(x+h)-f(x)}{h}$
$= \lim\limits_{h \to 0} \large\frac{3 \cot (x+h)+5 cosec (x+h)-3 \cot x- 5 cosec\: x}{h}$
$= 3 \lim\limits_{h \to 0} \large\frac{1}{h}$$[\cot(x+h)-\cot x]= 5 \lim\limits_{h \to 0} \large\frac{1}{h}$$[cosec(x+h)- cosec\: x]$-------(1)
Now, $= \lim\limits_{h \to 0} \large\frac{1}{h}$$[\cot(x+h)-\cot x]$
$= \lim\limits_{h \to 0} \large\frac{1}{h}$$\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}$$\bigg[\large\frac{\cos (x+h)\sin x - \cos x sin (x+h)}{\sin x \sin(x+h)} \bigg]$
$= \lim\limits_{h \to 0} \large\frac{1}{h}$$\bigg[\large\frac{\sin(x-x-h)}{\sin x \sin(x+h)} \bigg]$
$= \lim\limits_{h \to 0} \large\frac{1}{h}$$\bigg[\large\frac{\sin(-h)}{\sin x \sin(x+h)} \bigg]$
$ =- \bigg( \lim\limits_{h \to 0}\large\frac{\sin h}{h} \bigg)$$ \bigg( \lim\limits_{h \to 0}\large\frac{1}{\sin x.\sin(x+h)} \bigg)$
$ = -1. \large\frac{1}{\sin x.\sin(x+0)}$$= \large\frac{-1}{\sin^2x}$$ = -cosec^2x$--------------(2)
$= \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}$$ \bigg[ \large\frac{\sin x- \sin(x+h)}{\sin (x+h) \sin x} \bigg]$
$= \lim\limits_{h \to 0} \large\frac{1}{h}$$ \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)\sin x} \bigg]$
$= \lim\limits_{h \to 0} \large\frac{1}{h}$$ \bigg[ \large\frac{2 \cos \bigg( \Large\frac{2x+h}{2} \bigg)\sin\bigg( -\Large\frac{ h}{2}\bigg)}{\sin (x+h)\sin x} \bigg]$
$ = \lim\limits_{h \to 0}= \large\frac{ -\cos \bigg( \Large\frac{2x+h}{2} \bigg). \Large\frac{\sin\bigg( \Large\frac{h}{2} \bigg)}{ \Large\frac{h}{2} } }{\sin (x+h)\sin x}$
$= \lim\limits_{h \to 0}\bigg[\large\frac{-\cos \bigg(\large\frac{ 2x+h}{2} \bigg)}{\sin (x+h)\sin x}\bigg]$$. \lim\limits_{\large\frac{h}{2} \to 0} \Large\frac{sin\bigg( \Large\frac{h}{2} \bigg)}{\bigg( \Large\frac{h}{2} \bigg)} $
$ = \bigg( \large\frac{- \cos x}{\sin x\sin x} \bigg).1$
$ = cosec x \cot x$-----------(3)
From (1), (2) and (3), we obtain
$ f'(x) =-3 cosec^2x- 5 cosec \: x \cot x$
answered Apr 10, 2014 by thanvigandhi_1
 

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