# What is the derivative of x to the x? d/dx (x^x)

## If you answer, thank you! I always see this problem but I don't know how to answer it.

May 26, 2018

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

#### Explanation:

$y = {x}^{x}$

$L n y = x \ln x$

Apply implicit differentiation, standard differential and the product rule.

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

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

Substitute $y = {x}^{x}$

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

May 26, 2018

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

#### Explanation:

$\frac{\mathrm{dy}}{\mathrm{dx}} \left[{x}^{x}\right] = \frac{\mathrm{dy}}{\mathrm{dx}} \left[{e}^{x \ln \left(x\right)}\right]$
Let $u = x \ln \left(x\right)$ and thus, ${x}^{x} = {e}^{u}$

Apply chain rule:
$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{\mathrm{dy}}{\mathrm{du}} \cdot \frac{\mathrm{du}}{\mathrm{dx}}$
$= \frac{d}{\mathrm{du}} \left[{e}^{u}\right] \cdot \frac{d}{\mathrm{dx}} \left[x \ln \left(x\right)\right]$

Derivative of ${e}^{u}$ is itself, Derivative of $\ln \left(x\right)$ is $\setminus \frac{1}{x}$ and also apply product rule $\frac{d}{\mathrm{dx}} \left[f \left(x\right) g \left(x\right)\right] = f ' \left(x\right) g \left(x\right) + g ' \left(x\right) f \left(x\right)$

$= \left({e}^{u}\right) \left[\left(x\right) \left(\frac{1}{x}\right) + \left(1\right) \left(\ln \left(x\right)\right)\right]$
$= \left({x}^{x}\right) \left[\left(x\right) \left(\frac{1}{x}\right) + \left(1\right) \left(\ln \left(x\right)\right)\right]$
$= \left({x}^{x}\right) \left[1 + \ln \left(x\right)\right]$