differential equations for $x^x$

In this entry, we will derive differential equations satisfied by the function $x^x$.We begin by computing its derivative. To do this, we write $x^x=e^{x\log x}$ andapply the chain rule:

$$\frac{d}{dx}{x}^x=\frac{d}{dx}{e}^{x\log x}=e^{x\log x}(1+\log x)=x^x(1+\log x)$$

Set $y=x^x$. Then we have $y^{\prime }/y=1+\log x$. Takinganother derivative, we have

$$\frac{d}{dx}\left(\frac{y^{\prime }}{y}\right)=\frac{1}{x}\cdot$$

Applying the quotient rule and simplifying, this becomes

$$y{y}^{\prime \prime }-{(y^{\prime })}^2-y^2/x=0.$$

It is also possible to derive an equation in which $x$ does not appear.We start by noting that, if $z=1/x$, then $z^{\prime }+z^2=0$. If, asabove, $y=x^x$, we have $(d/dx)(y^{\prime }/y)=z$. Combining equations,

$$\frac{d^2}{d{x}^2}\left(\frac{y^{\prime }}{y}\right)+{\left(\frac{d}{dx}\left(\frac{y^{\prime }}{y}\right)\right)}^2=0;$$

applying the quotient rule and simplifying,

$$y^3{y}^{\prime \prime \prime }-y^2{(y^{\prime \prime })}^2+2y{(y^{\prime })}^2{y}^{\prime \prime }-3y^2{y}^{\prime }{y}^{\prime \prime }-{(y^{\prime })}^4+2y{(y^{\prime })}^3=0.$$