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  • If \left ( \sin x \right )^{y}= x+y,\: find\:\: \frac{dy}{dx

If  \left ( \sin x \right )^{y}= x+y,\: find\:\: \frac{dy}{dx}

 

 

 

 
 
 
 
 

Answers (1)

given:  \left ( \sin x \right )^{y}= x+y
\log \left ( \sin x \right )^{y}= \log \left ( x+y \right )
\Rightarrow y \ \log \left ( \sin x \right )-\log \left ( x+y \right )---\left ( i \right )
\log \left ( \sin x \right )\frac{dy}{dx}+y\frac{d}{dx}\left [ \log \left ( \sin x \right ) \right ]= \frac{d}{dx}\left [ \log \left ( x+y \right ) \right ]
\Rightarrow \log \left ( \sin x \right )\frac{dy}{dx}+y\: \frac{\cos x}{\sin x}= \frac{1}{\left ( x+y \right )}\left ( 1+\frac{dy}{dx} \right )
\Rightarrow \frac{dy}{dx}\left [ \log \left ( \sin x \right ) -\frac{1}{\left ( x+y \right )}\right ]= \frac{1}{\left ( x+y \right )}-y\cdot \cot x
\Rightarrow \frac{dy}{dx}= \frac{1-\left ( xy+y^{2} \right )\cdot \cot x}{\left ( x+y \right )\log \left ( \sin x \right )-1}
 

Posted by

Ravindra Pindel

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