How to differentiate $x^{\frac1x}$?

Question

How should $x^{\frac{1}{x}}$ be differentiated? I know the answer is $$\frac{1-\ln(x)}{x^{2-\frac{1}{x}}}$$

but I do not understand how to get there.

Attempt at solution.

I believe the following is true:

$$ \begin{aligned}\frac{d}{dx}x^u&=ux^{u-1}\cdot u^\prime\\ \frac{d}{dx}a^x&=a^x\cdot\ln(a) \end{aligned}$$ but I don't know what to do when both the base and the exponent are functions of $x$.

Answer 1

Hint: Let $y=x^{1/x}$. Now take the natural log of both sides. $$\ln(y)=\ln(x^{1/x})=\frac{1}{x}\ln(x).$$ Now you can differentiate both sides and solve to find $y'$. I'll even do the left side for you: $$\frac{d}{dx}\ln(y)=\frac{y'}{y}=\frac{y'}{x^{1/x}}.$$

Answer 2

Use the general definition of a real exponent: $$x^{\tfrac1x}\overset{\text{def}}{=}\mathrm e^{\tfrac{\ln x}{x}}.$$ Hence $$\Bigl(x^{\tfrac1x}\Bigr)'=\mathrm e^{\tfrac{\ln x}{x}}\cdot\frac{\cfrac1x\cdot x-\ln x}{x^2}=x^{\tfrac1x}\cdot\frac{1-\ln x}{x^2}.$$

Answer 3

$x^{\frac{1}{x}} = e^{\frac{1}{x} ln(x)}$ therefore $\left(x^{\frac{1}{x}}\right)' = \left(e^{\frac{1}{x} ln(x)}\right)' = e^{\frac{1}{x} ln(x)}\left(\frac{1}{x} ln(x)\right)' = e^{\frac{1}{x} ln(x)} \left(\frac{-1}{x^2}ln(x) + \frac{1}{x}\frac{1}{x}\right) = e^{\frac{1}{x} ln(x)}\left(\frac{1-ln(x)}{x^2}\right) = x^{\frac{1}{x}}x^{-2}\left(1-ln(x)\right) = \frac{\left(1-ln(x)\right)}{x^{2 - \frac{1}{x}}}$