Math, asked by saisahishnu, 1 month ago

find (dy)/(dx) if y=e^(x^(2)tan x)

Answers

Answered by mathdude500
1

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

\rm :\longmapsto\:y =  {e}^{ {x}^{2}tanx }

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

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

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

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

\boxed{ \bf \: \dfrac{d}{dx} {x}^{n} =  {nx}^{n - 1}}

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

\boxed{ \bf \: \dfrac{d}{dx}u.v \:  =  \: v\dfrac{d}{dx}u \:  +  \: u\dfrac{d}{dx}v}

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

Given that,

\rm :\longmapsto\:y =  {e}^{ {x}^{2}tanx }

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

\rm :\longmapsto\:\dfrac{d}{dx}y =\dfrac{d}{dx}{e}^{ {x}^{2}tanx }

\rm :\longmapsto\:\dfrac{dy}{dx} =  {e}^{ {x}^{2}tanx}\dfrac{d}{dx} {x}^{2}tanx

\rm :\longmapsto\:\dfrac{dy}{dx} =  {e}^{ {x}^{2}tanx}\bigg( {x}^{2}\dfrac{d}{dx}tanx + tanx\dfrac{d}{dx} {x}^{2}\bigg)

\rm :\longmapsto\:\dfrac{dy}{dx} =  {e}^{ {x}^{2}tanx}\bigg( {x}^{2} {sec}^{2}x + tanx.2x   \bigg)

\rm :\longmapsto\:\dfrac{dy}{dx} =x{e}^{ {x}^{2}tanx}\bigg(x {sec}^{2}x + 2tanx  \bigg)

Additional Information :-

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

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

\boxed{ \bf \: \dfrac{d}{dx}sinx =   cosx}

\boxed{ \bf \: \dfrac{d}{dx}x =1}

\boxed{ \bf \: \dfrac{d}{dx}k =0}

\boxed{ \bf \: \dfrac{d}{dx}secx =secx \: tanx}

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

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