# What is the derivative of ln(x)^x?

Apr 21, 2015

We're going to use implicit differentiation to solve this problem.

We'll also be using the product rule and the chain rule .

Know that:

If $q = u \cdot v$ and $u = g \left(x\right)$ and $v = h \left(x\right)$,

$\frac{\mathrm{dq}}{\mathrm{dx}} = u \cdot \frac{\mathrm{dv}}{\mathrm{dx}} + v \cdot \frac{\mathrm{du}}{\mathrm{dx}}$

So, say that:

$q = x \cdot \ln \left(\ln \left(x\right)\right) = u \cdot v$

Therefore:

$u = x$, $\frac{\mathrm{du}}{\mathrm{dx}} = 1$

Say that:

$v = \ln \left(\ln \left(x\right)\right) = \ln p$

Therefore:

$\frac{\mathrm{dv}}{\mathrm{dp}} = \frac{1}{p} = \frac{1}{\ln} \left(x\right)$

$p = \ln \left(x\right)$, $\frac{\mathrm{dp}}{\mathrm{dx}} = \frac{1}{x}$

This means that:

$\frac{\mathrm{dv}}{\mathrm{dx}} = \frac{1}{x \cdot \ln \left(x\right)}$

And as a result...

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

$= \frac{1}{\ln \left(x\right)} + \ln \left(\ln \left(x\right)\right)$

Now let's differentiate the function you were talking about using implicit differentiation...

$y = \ln {\left(x\right)}^{x}$

$\ln y = \ln \left(\ln {\left(x\right)}^{x}\right)$

$\ln y = x \ln \left(\ln \left(x\right)\right)$

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

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

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

And finally...

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