Question

What is the derivative of $x^x$?

Answer 1

Find the derivative of $x^x$.

Let, $y=x^x$

Take $\log$ on both sides.

$$\begin{gathered} \log y=\log x^x \\ \Rightarrow \log y=x\log x...\left(1\right)\left[\because \log m^n=n\log m\right] \end{gathered}$$

Differentiate $\left(1\right)$ on both sides, w.r.t $x$

$$\begin{gathered} \frac{d}{dx}\log y=\frac{d}{dx}x\log x \\ \Rightarrow \frac{1}{y}\times \frac{dy}{dx}=x\frac{d}{dx}\log x+\log x\frac{d}{dx}x\left[\begin{array}{c}\because \frac{d}{dx}\log x=\frac{1}{x},\\ \frac{d}{dx}uv=u\frac{d}{dx}v+v\frac{d}{dx}u\end{array}\right] \\ \Rightarrow \frac{1}{y}\times \frac{dy}{dx}=x\times \frac{1}{x}+\log x\times 1 \\ \Rightarrow \frac{1}{y}\times \frac{dy}{dx}=1+\log x \\ \Rightarrow \frac{dy}{dx}=y\left(1+\log x\right) \\ \Rightarrow \frac{dy}{dx}=x^x\left(1+\log x\right)\left[\because y=x^x\right] \end{gathered}$$

Therefore, the derivative of $x^x$ is $x^x\left(1+\log x\right)$