Question

derivative of y=(cosx)^cotx

Answer 1

Answer:

rative

Step-by-step explanation:

Answer 2

$\large\underline{\sf{Given- }}$

$\rm :\longmapsto\:y = {(cosx)}^{cotx}$

$\large\underline{\sf{To\:Find - }}$

$\rm :\longmapsto\:\dfrac{dy}{dx}$

$\begin{gathered}\large{\bold{{\underline{Formula \: Used - }}}}  \end{gathered}$

$\boxed{ \rm \: log( {x}^{y} ) = y \: logx}$

$\boxed{ \rm \: \dfrac{d}{dx}cosx = - \: sinx}$

$\boxed{ \rm \: \dfrac{d}{dx}cotx = - {cosec}^{2}x}$

$\boxed{ \rm \: \dfrac{d}{dx}logy = \frac{1}{y}}$

$\large\underline{\sf{Solution-}}$

Given that,

$\rm :\longmapsto\:y = {(cosx)}^{cotx}$

Taking log on both sides, we get

$\rm :\longmapsto\:logy =log{(cosx)}^{cotx}$

$\rm :\longmapsto\:logy \: = \: cotx \: log{(cosx)}$

On differentiating both sides, w. r. t. x, we get

$\rm :\longmapsto\:\dfrac{d}{dx}logy \: = \dfrac{d}{dx}\: cotx \: log{(cosx)}$

$\rm :\longmapsto\:\dfrac{1}{y}\dfrac{dy}{dx} = cotx\dfrac{d}{dx}logcosx \: + \: logcosx\dfrac{d}{dx}cotx$

$\rm :\longmapsto\:\dfrac{1}{y}\dfrac{dy}{dx} = cotx\dfrac{1}{cosx} \dfrac{d}{dx}cosx \: + \: logcosx( { - cosec}^{2}x)$

$\rm :\longmapsto\:\dfrac{1}{y}\dfrac{dy}{dx} = \dfrac{cosx}{sinx} \dfrac{1}{cosx}( - sinx) \: - \: logcosx( {cosec}^{2}x)$

$\rm :\longmapsto\:\dfrac{1}{y}\dfrac{dy}{dx} = - 1 \: - \: logcosx( {cosec}^{2}x)$

$\rm :\longmapsto\:\dfrac{1}{y}\dfrac{dy}{dx} = - \bigg (1 \: + \: logcosx( {cosec}^{2}x) \bigg)$

$\rm :\longmapsto\:\dfrac{dy}{dx} = -y \bigg (1 \: + \: logcosx( {cosec}^{2}x) \bigg)$

$\rm :\longmapsto\:\dfrac{dy}{dx} = - {(cosx)}^{cotx} \bigg (1 \: + \: logcosx( {cosec}^{2}x) \bigg)$

Additional Information :-

$\boxed{ \rm \: \dfrac{d}{dx}tanx = {sec}^{2}x}$

$\boxed{ \rm \: \dfrac{d}{dx} {e}^{x} = {e}^{x}}$

$\boxed{ \rm \: \dfrac{d}{dx} {a}^{x} = {a}^{x}loga}$

$\boxed{ \rm \: \dfrac{d}{dx} \sqrt{x} = \frac{1}{2 \sqrt{x} }}$

$\boxed{ \rm \: \dfrac{d}{dx}secx = secx \: tanx}$

$\boxed{ \rm \: \dfrac{d}{dx}cosecx = - cosecx \: cotx}$