# Derivatives

## The derivative of ${\displaystyle {\frac {d}{dx}}x^{x}}$

### Question

What is ${\displaystyle {\frac {d}{dx}}x^{x}}$?

### Solution 1

• Let ${\displaystyle y=x^{x}}$.
• Take ${\displaystyle \ln }$ of both sides: ${\displaystyle \ln y=x\ln x}$.
• Differentiate both sides: ${\displaystyle {\frac {d}{dx}}\ln y={\frac {d}{dx}}x\ln x}$.
• Apply the chain rule on the left-hand side: ${\displaystyle {\frac {d}{dx}}\ln y={\frac {1}{y}}\cdot {\frac {dy}{dx}}}$.
• Apply the product rule on the right-hand side: ${\displaystyle {\frac {d}{dx}}x\ln x=1\cdot \ln x+x\cdot {\frac {1}{x}}=\ln x+1}$.
• Putting it together, we have ${\displaystyle {\frac {1}{y}}\cdot {\frac {dy}{dx}}=\ln x+1}$.
• Hence ${\displaystyle {\frac {dy}{dx}}=y(\ln x+1)=x^{x}(\ln x+1)}$.

### Solution 2

• Note that ${\displaystyle x=e^{\ln x}}$, so ${\displaystyle x^{x}=(e^{\ln x})^{x}=e^{x\ln x}}$.
• Applying the chain rule, ${\displaystyle {\frac {d}{dx}}x^{x}={\frac {d}{dx}}e^{x\ln x}=e^{x\ln x}{\frac {d}{dx}}x\ln x}$.
• Applying the product rule, ${\displaystyle {\frac {d}{dx}}x\ln x=1\cdot \ln x+x\cdot {\frac {1}{x}}=\ln x+1}$.
• Therefore ${\displaystyle {\frac {d}{dx}}x^{x}=e^{x\ln x}(\ln x+1)=x^{x}(\ln x+1)}$.