The derivative of x to the x to the x

To find the derivative of $x^{x^x}$, use logarithmic differentiation. In order to apply this technique, let the expression being differentiated equal $y$:

$$y=x^{x^x}.$$

Now take the natural logarithm of both sides:

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

Use the properties of the logarithm to get

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

$x^x$ is still problematic, so take the natural logarithm of both sides again:

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

Using the properties of the logarithm once more, arrive at

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

Take the derivative of both sides with respect to $x$. Use the chain rule and the product rule. Simplify the equation, then solve it for $\frac{\textrm{d}y}{\textrm{d}x}$:

$$\begin{aligned} \frac{\textrm{d}}{\textrm{d}x}\ln\left(\ln{y}\right)&=\frac{\textrm{d}}{\textrm{d}x}\left(x\ln x+\ln\left(\ln x\right)\right)\newline \frac{1}{\ln y}\cdot\frac{1}{y}\cdot\frac{\textrm{d}y}{\textrm{d}x}&=\ln x+x\cdot\frac{1}{x}+\frac{1}{\ln x}\cdot\frac{1}{x}\newline \frac{1}{\ln y}\cdot\frac{1}{y}\cdot\frac{\textrm{d}y}{\textrm{d}x}&=\ln x+1+\frac{1}{x\ln x}\newline \frac{\textrm{d}y}{\textrm{d}x}&=y\ln y\left(\ln x+1+\frac{1}{x\ln x}\right). \end{aligned}$$

We can make our result a bit more useful by substituting $x^{x^x}$ for $y$. This produces

$$\begin{aligned} \frac{\textrm{d}y}{\textrm{d}x}&=x^{x^x}\ln x^{x^x}\left(\ln x+1+\frac{1}{x\ln x}\right)\newline &=x^{x^x}x^x\ln x\left(\ln x+1+\frac{1}{x\ln x}\right)\newline &=x^{x^x+x}\left(\ln^2 x+\ln x+\frac{1}{x}\right). \end{aligned}$$