Mathematics/Calculus/Corner cases

Derivatives

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

Question

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

Solution 1

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

Solution 2

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