Patrick Michalik

The derivative of x to the x to the x

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

y=xxx.y=x^{x^x}.

Now take the natural logarithm of both sides:

lny=lnxxx.\ln{y}=\ln x^{x^x}.

Use the properties of the logarithm to get

lny=xxlnx.\ln{y}=x^x\ln x.

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

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

Using the properties of the logarithm once more, arrive at

ln(lny)=xlnx+ln(lnx).\ln\left(\ln{y}\right)=x\ln x+\ln\left(\ln x\right).

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

ddxln(lny)=ddx(xlnx+ln(lnx))1lny1ydydx=lnx+x1x+1lnx1x1lny1ydydx=lnx+1+1xlnxdydx=ylny(lnx+1+1xlnx).\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 xxxx^{x^x} for yy. This produces

dydx=xxxlnxxx(lnx+1+1xlnx)=xxxxxlnx(lnx+1+1xlnx)=xxx+x(ln2x+lnx+1x).\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}
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