Originally Posted by

**DavidM** $\displaystyle y=x^\frac{1}{x}$

$\displaystyle \frac{dy}{dx} = x^ \frac{1}{x}(\frac{1}{x^2} - \frac{log(x)}{x^2})$

When you graph x^(1/x), the maximum point (gradient zero) should be e (2.71828...) for x. But it doesn't.

$\displaystyle \frac{dy}{dx} = x^\frac{1}{x} (\frac{1}{x^2} - \frac{In(x)}{x^2})$

By replacing the log with in, I get the correct answer. When dy/dx = 0, x = e.

Can anyone explain the difference? And which one should I use the next time I use the chain rule to differentiate powers, In or Log?