# differential equations for $x^{x}$

In this entry, we will derive differential equations satisfied by the function $x^{x}$. 11In this entry, we restrict $x$, and hence $x^{x}$ to be strictly positive real numbers, hence it is justified to divide by these quantities. We begin by computing its derivative. To do this, we write $x^{x}=e^{x\log x}$ and apply the chain rule:

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

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

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

Applying the quotient rule and simplifying, this becomes

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

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, as above, $y=x^{x}$, we have $(d/dx)(y^{\prime}/y)=z$. Combining equations,

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

applying the quotient rule and simplifying,

 $y^{3}y^{\prime\prime\prime}-y^{2}(y^{\prime\prime})^{2}+2y(y^{\prime})^{2}y^{% \prime\prime}-3y^{2}y^{\prime}y^{\prime\prime}-(y^{\prime})^{4}+2y(y^{\prime})% ^{3}=0.$
Title differential equations for $x^{x}$ DifferentialEquationsForXx 2013-03-22 17:24:37 2013-03-22 17:24:37 rspuzio (6075) rspuzio (6075) 6 rspuzio (6075) Derivation msc 26A99