Question

If $y=x^x,\mathrm{find}\frac{dy}{dx}atx=e$.

Answer 1

$\mathrm{We}\mathrm{have},y=x^x...\left(i\right)$
Taking log on both sides,
$$\begin{gathered} \log y=\log x^x \\ \Rightarrow \log y=x\log x \end{gathered}$$

$$\begin{gathered} \Rightarrow \frac{1}{y}\frac{dy}{dx}=x\frac{d}{dx}\left(\log x\right)+\log x\frac{d}{dx}\left(x\right) \\ \Rightarrow \frac{1}{y}\frac{dy}{dx}=x\left(\frac{1}{x}\right)+\log x\left(1\right) \\ \Rightarrow \frac{1}{y}\frac{dy}{dx}=1+\log x \\ \Rightarrow \frac{dy}{dx}=y\left(1+\log x\right) \\ \Rightarrow \frac{dy}{dx}=x^x\left(1+\log x\right)\left[\mathrm{using}\mathrm{equation}\left(\mathrm{i}\right)\right] \\ \mathrm{Puting}x=e,\mathrm{we}\mathrm{get}, \\ \frac{dy}{dx}=e^e\left(1+{\log }_{e}e\right) \\ \Rightarrow \frac{dy}{dx}=e^e\left(1+1\right)\left[\because {\log }_{e}e=1\right] \\ \Rightarrow \frac{dy}{dx}=2e^e \end{gathered}$$