# 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 ${\displaystyle {\frac {d}{dx}}x^{x}=e^{x\ln x}(\ln x+1)=x^{x}(\ln x+1)}$.

# Integrals

## The integral ${\displaystyle \int x^{x}\,dx}$

### Question

What is ${\displaystyle \int x^{x}\,dx}$?

### Solution

• We can write ${\displaystyle x^{x}}$ as ${\displaystyle (e^{\ln x})^{x}=e^{x\ln x}}$.
• Consider the series expansion of ${\displaystyle e^{x\ln x}}$:
   ${\displaystyle e^{x\ln x}=1+(x\ln x)+{\frac {(x\ln x)^{2}}{2!}}+{\frac {(x\ln x)^{3}}{3!}}+\ldots +{\frac {(x\ln x)^{i}}{i!}}+\ldots =\sum _{i=0}^{\infty }{\frac {(x\ln x)^{i}}{i!}}}$.

• We can interchange the integration and summation (we can recognize this as a special case of the Fubini/Tonelli theorems) and write
   ${\displaystyle \int x^{x}\,dx=\int \left(\sum _{i=0}^{\infty }{\frac {(x\ln x)^{i}}{i!}}\right)\,dx=\sum _{i=0}^{\infty }\left(\int {\frac {(x\ln x)^{i}}{i!}}\,dx\right)=\sum _{i=0}^{\infty }\left({\frac {1}{i!}}\int x^{i}(\ln x)^{i}\,dx\right).}$