Find dy dx y=x^ +( )^

Question

Find $\frac{\text{dy}}{\text{dx}}$
$\text{y}=\text{x}^{\cos\text{x}}+(\sin\text{x})^{\tan\text{x}}$

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Answer

We have, $\text{y}=\text{x}^{\cos\text{x}}+(\sin\text{x})^{\tan\text{x}}$
$\text{y}=\text{e}^{\log\text{x}^{\cos\text{x}}}+\text{e}^{\log(\sin\text{x})^{\tan\text{x}}}$
$\text{y}=\text{e}^{\cos\text{x}\log\text{x}}+\text{e}^{\tan\text{x}\log\sin\text{x}}$
Differentiating with respect to x using chain rule,
$\frac{\text{dy}}{\text{dx}}=\frac{\text{d}}{\text{dx}}\big(\text{e}^{\cos\text{x}\log\text{x}}\big)+\frac{\text{d}}{\text{dx}}\big(\text{e}^{\tan\text{x}\log\sin\text{x}}\big)$
$=\text{e}^{\cos\text{x}\log\text{x}}\frac{\text{d}}{\text{dx}}(\cos\text{x}\log\text{x}) \\ +\text{e}^{\tan\text{x}\log\sin\text{x}}\frac{\text{d}}{\text{dx}}(\tan\text{x}\log\sin\text{x})$
$=\text{e}^{\log\text{x}^{\cos\text{x}}}\Big[\cos\text{x}\frac{\text{d}}{\text{dx}}(\log\text{x})+\log\text{x}\frac{\text{d}}{\text{dx}}(\cos\text{x})\Big] \\ +\text{e}^{\log(\sin\text{x})^{\tan\text{x}}}\Big[\tan\text{x}\frac{\text{d}}{\text{dx}}\log\sin\text{x}+\log\sin\text{x}\frac{\text{d}}{\text{dx}}(\tan\text{x})\Big]$
$=\text{x}^{\cos\text{x}}\Big[\cos\text{x}\Big(\frac{1}{\text{x}}\Big)+\log\text{x}(-\sin\text{x})\Big] \\ +(\sin\text{x})^{\tan\text{x}}\Big[\tan\text{x}\Big(\frac{1}{\sin\text{x}}\Big)\frac{\text{d}}{\text{dx}}(\sin)\text{x}+\log\sin\text{x}(\sec^2\text{x})\Big]$
$=\text{x}^{\cos\text{x}}\Big[\frac{\cos\text{x}}{\text{x}}-\sin\text{x}\log\text{x}\Big] \\ +(\sin\text{x})^{\tan\text{x}}\Big[\tan\text{x}\Big(\frac{1}{\sin\text{x}}\Big)(\cos\text{x})+\sec^2\text{x}\log\sin\text{x}\Big]$
$=\text{x}^{\cos\text{x}}\Big[\frac{\cos\text{x}}{\text{x}}-\sin\text{x}\log\text{x}\Big] \\ +(\sin\text{x})^{\tan\text{x}}\big[1+\sec^2\text{x}\log\sin\text{x}\big]$