Find dy dx y=x^ x +( )^x - Vidyadip

Question

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

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Answer

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

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