# Differentiate the function with respect to x.

Question:

Differentiate the function with respect to x.

$(\sin x)^{x}+\sin ^{-1} \sqrt{x}$

Solution:

Let $y=(\sin x)^{x}+\sin ^{-1} \sqrt{x}$

Also, let $u=(\sin x)^{x}$ and $v=\sin ^{-1} \sqrt{x}$

$\therefore y=u+v$

$\Rightarrow \frac{d y}{d x}=\frac{d u}{d x}+\frac{d v}{d x}$   ...(1)

$u=(\sin x)^{x}$

$\Rightarrow \log u=\log (\sin x)^{x}$

$\Rightarrow \log u=x \log (\sin x)$

Differentiating both sides with respect to $x$, we obtain

$\Rightarrow \frac{1}{u} \frac{d u}{d x}=\frac{d}{d x}(x) \times \log (\sin x)+x \times \frac{d}{d x}[\log (\sin x)]$

$\Rightarrow \frac{d u}{d x}=u\left[1 \cdot \log (\sin x)+x \cdot \frac{1}{\sin x} \cdot \frac{d}{d x}(\sin x)\right]$

$\Rightarrow \frac{d u}{d x}=(\sin x)^{x}\left[\log (\sin x)+\frac{x}{\sin x} \cdot \cos x\right]$

$\Rightarrow \frac{d u}{d x}=(\sin x)^{x}(x \cot x+\log \sin x)$   ...(2)

$v=\sin ^{-1} \sqrt{x}$

Differentiating both sides with respect to x, we obtain

$\frac{d v}{d x}=\frac{1}{\sqrt{1-(\sqrt{x})^{2}}} \cdot \frac{d}{d x}(\sqrt{x})$

$\Rightarrow \frac{d v}{d x}=\frac{1}{\sqrt{1-x}} \cdot \frac{1}{2 \sqrt{x}}$

$\Rightarrow \frac{d v}{d x}=\frac{1}{2 \sqrt{x-x^{2}}}$

Therefore, from (1), (2), and (3), we obtain]

$\frac{d y}{d x}=(\sin x)^{x}(x \cot x+\log \sin x)+\frac{1}{2 \sqrt{x-x^{2}}}$