Find dy dx y=( )^x+ ^ -1 x

Question

Find $\frac{\text{dy}}{\text{dx}}$
$\text{y}=(\sin\text{x})^\text{x}+\sin^{-1}\sqrt{\text{x}}$

✓

Answer

Here,
$\text{y}=(\sin\text{x})^{\text{x}}+\sin^{-1}\sqrt{\text{x}}$
$=\text{e}^{\log(\sin\text{x})^\text{x}}+\sin^{-1}\sqrt{\text{x}}$
$\text{y}=\text{e}^{\text{x}\log\sin\text{x}}+\sin^{-1}\sqrt{\text{x}}$
$\big[\text{Since},\log_\text{e}^\text{e}=1,\log\text{a}^\text{b}=\text{b}\log\text{a}\big]$
Differentitating it with respect to x using chain rule and product rule,
$\frac{\text{dy}}{\text{dx}}=\frac{\text{d}}{\text{dx}}\big(\text{e}^{\text{x}\log\sin\text{x}}\big)+\frac{\text{d}}{\text{dx}}\sin^{-1}\big(\sqrt{\text{x}}\big)$
$=\text{e}^{\text{x}\log\sin\text{x}}\frac{\text{d}}{\text{dx}}(\text{x}\log\sin\text{x})+\frac{1}{\sqrt{1-\big(\sqrt{\text{x}}\big)^2}}\frac{\text{d}}{\text{dx}}\big(\sqrt{\text{x}}\big)$
$=\text{e}^{\log(\sin\text{x})^\text{x}}\Big[\text{x}\frac{\text{d}}{\text{dx}}\log\sin\text{x}+\log\sin\text{x}\frac{\text{d}}{\text{dx}}(\text{x})+\frac{1}{\sqrt{1-\text{x}}}\times\frac{1}{2\sqrt{\text{x}}}\Big]$
$=(\sin\text{x})^\text{x}\Big[\text{x}\times\frac{1}{\sin\text{x}}\frac{\text{d}}{\text{dx}}(\sin\text{x})+\log\sin\text{x}(1)\Big]+\frac{1}{2\sqrt{\text{x}-\text{x}^2}}$
$=(\sin\text{x})^\text{x}\Big[\frac{\text{x}}{\sin\text{x}}(\cos\text{x})+\log\sin\text{x}\Big]+\frac{1}{2\sqrt{\text{x}-\text{x}^2}}$
$\frac{\text{dy}}{\text{dx}}=(\sin\text{x})^\text{x}\Big[\text{x}\cot\text{x}+\log\sin\text{x}\Big]+\frac{1}{2\sqrt{\text{x}-\text{x}^2}}$

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