Question

Find $\frac{dy}{dx}$

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

Answer 1

$$\begin{gathered} \mathrm{We}\mathrm{have},y={\left(\sin x\right)}^x+{\sin }^{-1}\sqrt{x} \\ \Rightarrow y=e^{\log {\left(\sin x\right)}^x}+{\sin }^{-1}\sqrt{x} \\ \Rightarrow y=e^{x\log \sin x}+{\sin }^{-1}\sqrt{x} \end{gathered}$$
Differentiating with respect to x using chain rule,
$$\begin{gathered} \frac{dy}{dx}=\frac{d}{dx}\left(e^{x\log \sin x}\right)+\frac{d}{dx}\left({\sin }^{-1}\sqrt{x}\right) \\ =e^{x\log \sin x}\frac{d}{dx}\left(x\log \sin x\right)+\frac{1}{\sqrt{1-{\left(\sqrt{x}\right)}^2}}\frac{d}{dx}\left(\sqrt{x}\right) \\ =e^{\log {\left(\sin x\right)}^x}\left[x\frac{d}{dx}\left(\log \sin x\right)+\log \sin x\frac{d}{dx}\left(x\right)+\frac{1}{\sqrt{1-x}}\times \frac{1}{2\sqrt{x}}\right] \\ ={\left(\sin x\right)}^x\left[x\frac{1}{\sin x}\frac{d}{dx}\left(\sin x\right)+\log \sin x\right]+\frac{1}{2\sqrt{x-x^2}} \\ ={\left(\sin x\right)}^x\left[\frac{x}{\sin x}\left(\cos x\right)+\log \sin x\right]+\frac{1}{2\sqrt{x-x^2}} \\ ={\left(\sin x\right)}^x\left[x\cot x+\log \sin x\right]+\frac{1}{2\sqrt{x-x^2}} \end{gathered}$$