Question

Find $\frac{dy}{dx}$

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

Answer 1

$$\begin{gathered} \mathrm{Let}y=x^{\sin x}+{\left(\sin x\right)}^x \\ \mathrm{Also},\mathrm{let}u=x^{\sin x}\mathrm{and}v={\left(\sin x\right)}^x \\ \therefore y=u+v \\ \Rightarrow \frac{dy}{dx}=\frac{du}{dx}+\frac{dv}{dx}...\left(i\right) \\ \mathrm{Now},u=x^{\sin x} \\ \mathrm{Taking}\log \mathrm{on}\mathrm{both}\mathrm{sides}, \\ \Rightarrow \log u=\log \left(x^{\sin x}\right) \\ \Rightarrow \log u=\sin x\log x \\ \mathrm{Differentiating}\mathrm{both}\mathrm{sides}\mathrm{with}\mathrm{respect}\mathrm{to}x, \\ \frac{1}{u}\frac{du}{dx}=\log x\frac{d}{dx}\left(\sin x\right)+\sin x\frac{d}{dx}\left(\log x\right) \\ \Rightarrow \frac{du}{dx}=u\left[\cos x\log x+\sin x\frac{1}{x}\right] \\ \Rightarrow \frac{du}{dx}=x^{\sin x}\left[\cos x\log x+\frac{\sin x}{x}\right]...\left(ii\right) \\ \mathrm{Again},v={\left(\sin x\right)}^x \\ \mathrm{Taking}\log \mathrm{on}\mathrm{both}\mathrm{sides}, \\ \Rightarrow \log v=\log {\left(\sin x\right)}^x \\ \Rightarrow \log v=x\log \left(\sin x\right) \\ \mathrm{Differentiating}\mathrm{both}\mathrm{sides}\mathrm{with}\mathrm{respect}\mathrm{to}x, \\ \frac{1}{v}\frac{dv}{dx}=\log \left(\sin x\right)\frac{d}{dx}\left(x\right)+x\frac{d}{dx}\left[\log \left(\sin x\right)\right] \\ \Rightarrow \frac{dv}{dx}=v\left[\log \left(\sin x\right)+x\frac{1}{\sin x}\frac{d}{dx}\left(\sin x\right)\right] \\ \Rightarrow \frac{dv}{dx}={\left(\sin x\right)}^x\left[\log \sin x+\frac{x}{\sin x}\cos x\right] \\ \Rightarrow \frac{dv}{dx}={\left(\sin x\right)}^x\left[\log \sin x+x\cot x\right]..\left(iii\right) \\ \mathrm{From}\left(\mathrm{i}\right),\left(\mathrm{ii}\right)\mathrm{and}\left(\mathrm{iii}\right),\mathrm{we}\mathrm{obtain} \\ \frac{dy}{dx}=x^{\sin x}\left(\cos x\log x+\frac{\sin x}{x}\right)+{\left(\sin x\right)}^x\left[\log \sin x+x\cot x\right] \end{gathered}$$