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Q9 Differentiate the functions w.r.t. x  x ^ { \sin x } + ( \sin x )^ \cos x

 

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Given function is
y = x ^ { \sin x } + ( \sin x )^ \cos x
Now, take t = x^{\sin x}
Now, take log on both sides
\log t = \sin x \log x
Now, differentiate it w.r.t. x 
we get,
\frac{1}{t}\frac{dt}{dx} = \cos x \log x+\frac{1}{x}.\sin x\\ \frac{dt}{dx}=t\left ( \cos x \log x+\frac{1}{x}.\sin x \right )\\ \frac{dt}{dx}= x^{\sin x}\left ( \cos x \log x+\frac{1}{x}.\sin x \right )
Similarly, take k = (\sin x)^{\cos x}
Now, take log on both the sides
\log k = \cos x \log (\sin x)
Now, differentiate it w.r.t. x
we get,
\frac{1}{k}\frac{dk}{dt} = (-\sin x)(\log (\sin x)) + \cos x.\frac{1}{\sin x}.\cos x=-\sin x\log(\sin x)+\cot x.\cos x\\ \frac{dk}{dt}= k\left ( -\sin x\log(\sin x)+\cot x.\cos x \right )\\ \frac{dk}{dt}=(\sin x)^{\cos x}\left ( -\sin x\log(\sin x)+\cot x.\cos x \right )
Now,
\frac{dy}{dx} = x^{\sin x}\left ( \cos x \log x+\frac{1}{x}.\sin x \right )+ (\sin x)^{\cos x}\left ( -\sin x\log(\sin x)+\cot x.\cos x \right )
Therefore, the answer is  x^{\sin x}\left ( \cos x \log x+\frac{1}{x}.\sin x \right )+ (\sin x)^{\cos x}\left ( -\sin x\log(\sin x)+\cot x.\cos x \right )

Posted by

Gautam harsolia

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