English, asked by anjalinetam2909, 3 months ago

find dy/dx=? y=esinx/sinxn.​

Answers

Answered by aryan073
3

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Question :

(1) \huge \green{ \rm{y =  \frac{esinx}{sinxn} }}

To Find :

   \pink \bigstar\large \rm{ \:  \frac{dy}{dx}  = }

Formula :

 \large  \red\bigstar \rm \:  \frac{dy}{dx}  =  \:  \frac{x \:  \frac{dy}{dx} \:( y) \:  -  y \:  \frac{dy}{dx} (x)}{ {(x)}^{2} }  \:  \: ......quoitent \: rule

Solutions :

 \large\qquad \implies \red{ \bf \: y =  \frac{esinx}{sinxn} }

 \:  \qquad \large  \red \star  \pink{{\underline{ \bf{differentiating \: equation \: with \: respect \: to \: x}}}}

  \\ \qquad \implies \large \sf \:  \frac{dy}{dx} =  \frac{sinxn \frac{dy}{dx}(esinx) -  \frac{dy}{dx}(sinxn)esinx  }{ {(sinxn)}^{2} }

  \\ \qquad \large \implies \sf \:  \frac{dy}{dx}  =  \frac{sinxn(e sinx \:  \times  \: cosx) - \: cosn \times esinx }{ {(sinxn)}^{2} }

  \\  \qquad \large \implies \sf \:  \frac{dy}{dx}  =  \frac{e {sin}^{2}xn.cosx  - \: esinx.cosn }{ {(sinxn)}^{2} } is the answer

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