# How do you use logarithmic differentiation to find the derivative of the function y=x^x?

Mar 28, 2015

$y = {x}^{x}$

$\ln y = \ln {x}^{x}$

$\ln y = x \ln x$

Now differentiate implicitly:

$\frac{1}{y} \frac{\mathrm{dy}}{\mathrm{dx}} = \left(1\right) \left(\ln x\right) + \left(x\right) \left(\frac{1}{x}\right)$

$\frac{1}{y} \frac{\mathrm{dy}}{\mathrm{dx}} = 1 + \ln x$

Solve for $\frac{\mathrm{dy}}{\mathrm{dx}}$

$\frac{\mathrm{dy}}{\mathrm{dx}} = y \left(1 + \ln x\right)$

Recall that $y = {x}^{x}$, so

$\frac{\mathrm{dy}}{\mathrm{dx}} = {x}^{x} \left(1 + \ln x\right)$

Mar 28, 2015

It's easy if you remember the logarithmic rule:

[f(x)]^g(x)=e^ln([f(x)]^g(x))=e^(g(x)*lnf(x),

so:

$y = {x}^{x} = {e}^{x \ln x} \Rightarrow$

$y ' = {e}^{x \ln x} \left(1 \cdot \ln x + x \cdot \frac{1}{x}\right) = {e}^{x \ln x} \left(\ln x + 1\right)$,

(or, if you want: $y ' = {x}^{x} \left(\ln x + 1\right)$).