# How do you find the first and second derivative of (lnx)^(x)?

Aug 11, 2016

Let:

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

The easiest way to differentiate this is to take the natural logarithm of both sides.

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

Simplify using the logarithm rule:

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

Differentiate both sides. The left-hand side will need chain rule, similar to implicit differentiation, and the right-hand side will need chain rule and product rule.

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

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

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

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

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

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

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

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

The steps to find the second derivative are far too lengthy, and the work is rather pointless, but if you wish try to find it, the second derivative is:

$\frac{{d}^{2} y}{{\mathrm{dx}}^{2}} = \frac{{\left(\ln x\right)}^{x - 2} \left(\ln x \left(x \ln \left(\ln x\right) \left(\ln x \cdot \ln \left(\ln x\right) + 2\right) + 1\right) + x - 1\right)}{x}$