Question

if y=x^cosx find dy/dx
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Answer 1

Step-by-step explanation:

$\cos( {x}^{2} )^{ \cos(x - 1) }$

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Answer 2

Given,

$\sf{y=x^{cosx}}$

To find :-

$\sf{\dfrac{dy}{dx}}$

Solution :-

$\displaystyle{\sf{\dfrac{d(x^{cosx})}{dx} }}\\\\\displaystyle{\sf{\dashrightarrow x^{cosx}.\dfrac{d}{dx}[ln(x)cos(x)] }}\\\\\displaystyle{\sf{\dashrightarrow x^{cosx}\bigg(\dfrac{d}{dx}cos(x).ln(x) +cos(x).\dfrac{d}{dx}[ln(x)] \bigg)}}\\\\\displaystyle{\sf{\dashrightarrow x^{cosx}\bigg( [-sin(x)]ln(x)+cos(x).\dfrac{1}{x} \bigg) }}\\\\\displaystyle{\sf{\dashrightarrow x^{cosx}\Bigg( \dfrac{cos(x)}{x}-ln(x)sin(x) \Bigg) }}$