# Solve this

Question:

Find $\frac{\mathrm{dy}}{\mathrm{dx}}$, when

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

Solution:

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

$\Rightarrow y=a+b$

where, $a=(\sin x)^{x} ; b=\sin ^{-1} \sqrt{x}$

$\frac{\mathrm{dy}}{\mathrm{dx}}=\frac{\mathrm{da}}{\mathrm{dx}}+\frac{\mathrm{db}}{\mathrm{dx}}$

$\left\{\right.$ Using chain rule, $\frac{\mathrm{d}(\mathrm{u}+\mathrm{a})}{\mathrm{dx}}=\frac{\mathrm{du}}{\mathrm{dx}}+\frac{\mathrm{da}}{\mathrm{dx}}$ where $\mathrm{a}$ and $\mathrm{u}$ are any variables $\}$

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

Taking log both the sides:

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

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

$\left\{\log x^{a}=\operatorname{alog} x\right\}$

Differentiating with respect to $x$ :

$\Rightarrow \frac{\mathrm{d}(\log \mathrm{a})}{\mathrm{dx}}=\frac{\mathrm{d}(\mathrm{x} \log (\sin \mathrm{x}))}{\mathrm{dx}}$

$\Rightarrow \frac{\mathrm{d}(\log \mathrm{a})}{\mathrm{dx}}=\mathrm{x} \times \frac{\mathrm{d}(\log (\sin \mathrm{x}))}{\mathrm{dx}}+\log (\sin \mathrm{x}) \times \frac{\mathrm{dx}}{\mathrm{dx}}$

$\left\{\right.$ Using product rule, $\left.\frac{d(u v)}{d x}=u \frac{d v}{d x}+v \frac{d u}{d x}\right\}$

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

$\left\{\frac{d(\sin x)}{d x}=\cos x\right\}$

$\Rightarrow \frac{1}{a} \frac{d a}{d x}=\frac{x}{\sin x}(\cos x)+\log (\sin x)$

$\Rightarrow \frac{1}{a} \frac{d a}{d x}=\frac{x \cos x}{\sin x}+\log (\sin x)$

$\Rightarrow \frac{1}{a} \frac{d a}{d x}=x \cot x+\log (\sin x)$

$\Rightarrow \frac{d a}{d x}=a\{x \cot x+\log (\sin x)\}$Put the value of $a=(\sin x)^{x}$ :

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

$\mathrm{b}=\sin ^{-1} \sqrt{\mathrm{x}}$

$\Rightarrow \mathrm{b}=\sin ^{-1}(\mathrm{x})^{\frac{1}{2}}$

Differentiating with respect to $x$ :

$\frac{\mathrm{db}}{\mathrm{dx}}=\frac{\mathrm{d}\left(\sin ^{-1}(\mathrm{x})^{\frac{1}{2}}\right)}{\mathrm{dx}}$

$\Rightarrow \frac{\mathrm{db}}{\mathrm{dx}}=\frac{1}{\sqrt{1-\left(\mathrm{x}^{\frac{1}{2}}\right)^{2}}} \frac{\mathrm{d}(\mathrm{x})^{\frac{1}{2}}}{\mathrm{dx}}$

$\left\{\frac{\mathrm{d}\left(\sin ^{-1} \mathrm{u}\right)}{\mathrm{dx}}=\frac{1}{\sqrt{1-\mathrm{u}^{2}}} \frac{\mathrm{du}}{\mathrm{dx}}\right\}$

$\Rightarrow \frac{\mathrm{db}}{\mathrm{dx}}=\frac{1}{\sqrt{1-\mathrm{x}}}\left(\frac{1}{2} \mathrm{x}\left(\frac{1}{2}-1\right)\right)$

$\left\{\frac{d\left(u^{n}\right)}{d x}=n u^{n-1} \frac{d u}{d x}\right\}$

$\Rightarrow \frac{\mathrm{db}}{\mathrm{dx}}=\frac{1}{2 \sqrt{1-\mathrm{x}}}\left(\mathrm{x}\left(-\frac{1}{2}\right)\right)$

$\Rightarrow \frac{\mathrm{db}}{\mathrm{dx}}=\frac{1}{2 \sqrt{1-\mathrm{x}}}\left(\frac{1}{\sqrt{\mathrm{x}}}\right)$

$\Rightarrow \frac{\mathrm{db}}{\mathrm{dx}}=\frac{1}{2 \sqrt{\mathrm{x}} \sqrt{1-\mathrm{x}}}$

$\Rightarrow \frac{\mathrm{db}}{\mathrm{dx}}=\frac{1}{2 \sqrt{\mathrm{x}(1-\mathrm{x})}}$

$\frac{\mathrm{dy}}{\mathrm{dx}}=\frac{\mathrm{da}}{\mathrm{dx}}+\frac{\mathrm{db}}{\mathrm{dx}}$

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