Question

If $y=\sin \left(x^x\right)$, prove that $\frac{dy}{dx}=\cos \left(x^x\right)·x^x\left(1+\log x\right)$

Answer 1

$$\begin{gathered} \mathrm{Let}y=\sin \left(x^x\right)...\left(i\right) \\ \mathrm{Also},\mathrm{Let}u=x^x...\left(ii\right) \\ \mathrm{Taking}\log \mathrm{on}\mathrm{both}\mathrm{sides}, \\ \Rightarrow \log u=\log x^x \\ \Rightarrow \log u=x\log x \end{gathered}$$
Differentiating both sides with respect to x,
$$\begin{gathered} \frac{1}{u}\frac{du}{dx}=\frac{d}{dx}\left(x\log x\right) \\ \Rightarrow \frac{1}{u}\frac{du}{dx}=x\frac{d}{dx}\left(\log x\right)+\log x\frac{d}{dx}x \\ \Rightarrow \frac{1}{u}\frac{du}{dx}=x\left(\frac{1}{x}\right)+\log x\left(1\right) \\ \Rightarrow \frac{1}{u}\frac{du}{dx}=1+\log x \\ \Rightarrow \frac{du}{dx}=u\left(1+\log x\right) \\ \Rightarrow \frac{du}{dx}=x^x\left(1+\log x\right)...\left(iii\right)\left[\mathrm{using}\mathrm{equation}\left(\mathrm{ii}\right)\right] \\ \mathrm{Now},\mathrm{using}\mathrm{equation}\left(\mathrm{ii}\right)\mathrm{in}\mathrm{equation}\left(\mathrm{i}\right), \\ y=\sin u \\ \mathrm{Differentiating}\mathrm{with}\mathrm{respect}\mathrm{to}x, \\ \frac{dy}{dx}=\frac{d}{dx}\left(\sin u\right) \\ \Rightarrow \frac{dy}{dx}=\cos u\frac{du}{dx} \\ \mathrm{Using}\mathrm{equation}\left(\mathrm{ii}\right)\mathrm{and}\left(\mathrm{iii}\right), \\ \frac{dy}{dx}=\cos \left(x^x\right)\times x^x\left(1+\log x\right) \end{gathered}$$