Differentiate x^{\sin x}+\left ( \sin x \right )^{\cos x} with respect to x.

 

 

 

 
 
 
 
 

Answers (1)

Let  y= x^{\sin x}+\left ( \sin x \right )^{\cos x}
Let  x^{\sin x}= u---\left ( i \right )
and \left ( \sin x \right )^{\cos x}= v---\left ( ii \right )
Now, y= u+v
\frac{dy}{dx}= \frac{du}{dx}+\frac{dv}{dx}|
Now,u= x^{\sin x}
\log u= \sin x\log x
On differentiating wrt x,
\frac{1}{u}\frac{du}{dx}= \sin x\times \frac{1}{x}+\log x\cos x
\frac{du}{dx}= u\left [ \frac{\sin x}{x}+\log x\cos x \right ]
\frac{du}{dx}= x^{\sin x}\left [ \frac{\sin x}{x}+\log x\cos x \right ]
Now, v= \left ( \sin x \right )^{\cos x}
\log v= \cos x\log \sin x
On Differentiating w.r.t  x
\frac{1}{v}\frac{dv}{dx}= \cos x\frac{1}{\sin x}\: \: \cos x+\log \: \sin x\left ( -\sin x \right )
\frac{1}{v}\frac{dv}{dx}= \cos t\, x \: x\cos x+\sin x\: \log \left ( \sin x \right )
\frac{dv}{dx}= v\left ( \cos t\, x\: \cos x-\sin x\: \log\: \sin x \right )
       = \left ( \sin x \right )^{\cos x}\left ( \cot x-\cos x-\sin x\: \log \sin x \right )---\left ( iv \right )
\frac{dy}{dx}= x^{\\sin x}\left ( \frac{\sin x}{x} +\log x\, \cos x\right )+\left ( \sin x \right )^{\cos x}\left ( \cot x\, \cos x-\sin x\, \log \sin x \right )

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