# Proof of why $\sqrt[x]{x}$ is greatest when $x=e$

Stated above question. If the mathjax I used was wrong, it should be: Why does the xth root of x reach the greatest y at x=e

• $\log\left(\sqrt[x]x\right)=\frac{\log(x)}x$. – David C. Ullrich Jul 4 '15 at 0:06

## 1 Answer

So, you're asking to maximize $x^{1/x}$. We'll restrict out attention to positive $x$'s because otherwise we need to deal with complex numbers.

Taking the derivative of this expression is sometimes tricky. There are more fun ways to do this, but you could write this as $$x^{1/x}=e^{(\ln x)/x}.$$ The derivative of this is $$e^{(\ln x)/x}\left(\frac{1}{x^2}-\frac{\ln x}{x^2}\right).$$ The maximum occurs when the derivative vanishes, i.e., $\ln(x)=1$. This happens when $x=e$.

Observe that at $x=e$ the sign of the derivative changes from positive to negative, so this is a local maximum and since it's the only zero of the derivative, it is the global maximum.

A bonus: a fun way to take the derivative of $x^{1/x}$. One way you could do this is as above. On the other hand, we can do the following:

First, treat the base $x$ as a variable and the exponent $1/x$ as a constant to get $\frac{1}{x}x^{\frac{1}{x}-1}=x^{\frac{1}{x}}\frac{1}{x^2}$.

Second, treat the base as a constant and the exponent $1/x$ as a variable to get $x^{1/x}\ln(x)\frac{-1}{x^2}$.

Add those two together and you get the derivative of $x^{\frac{1}{x}}$.