Problem: Differentiate \(x^x\)

Hint: Rewrite \(x^x\) as \(e^{\ln(x^x)}\)

Solution:

\[\frac{d}{dx} (x^x)= \frac{d}{dx} \left( e^{\ln(x^x)}\right)= \frac{d}{dx} \left( e^{x \ln x}\right)\]

By the chain rule: \[=\left( e^{x \ln x}\right) \frac{d}{dx} ( x \ln x)\]

\[=\left( x^x \right) \frac{d}{dx} (x \ln x)\]

Now using the product rule:

\[=\left( x^x \right) \left[ 1 \times \ln x + x \times \frac{1}{x} \right]\]

\[=\left( x^x \right) \left[ \ln x + 1 \right]\]

\[\therefore \frac{d}{dx} (x^x)=\left( x^x \right) ( \ln x + 1 )\]