Differentiate the function with respect to x: | vedclass

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Let $y=x^{\sin x}+(\sin x)^{\cos x}$.Let $u=x^{\sin x}$ and $v=(\sin x)^{\cos x}$.Then $y=u+v$,so $\frac{dy}{dx}=\frac{du}{dx}+\frac{dv}{dx}$  $(1)$.F

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Let $y=x^{\sin x}+(\sin x)^{\cos x}$.Let $u=x^{\sin x}$ and $v=(\sin x)^{\cos x}$.Then $y=u+v$,so $\frac{dy}{dx}=\frac{du}{dx}+\frac{dv}{dx}$  $(1)$.For $u=x^{\sin x}$,taking log on both sides: $\log u = \sin x \log x$.Differentiating with respect to $x$: $\frac{1}{u} \frac{du}{dx} = \cos x \log x + \sin x \cdot \frac{1}{x}$.So,$\frac{du}{dx} = x^{\sin x} \left( \cos x \log x + \frac{\sin x}{x} \right)$  $(2)$.For $v=(\sin x)^{\cos x}$,taking log on both sides: $\log v = \cos x \log(\sin x)$.Differentiating with respect to $x$: $\frac{1}{v} \frac{dv}{dx} = -\sin x \log(\sin x) + \cos x \cdot \frac{1}{\sin x} \cdot \cos x$.So,$\frac{dv}{dx} = (\sin x)^{\cos x} [\cot x \cos x - \sin x \log(\sin x)]$  $(3)$.Substituting $(2)$ and $(3)$ into $(1)$,we get:$\frac{dy}{dx} = x^{\sin x} \left( \cos x \log x + \frac{\sin x}{x} \right) + (\sin x)^{\cos x} [\cot x \cos x - \sin x \log(\sin x)]$.

Differentiate the function with respect to $x$: $x^{\sin x}+(\sin x)^{\cos x}$

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