ropowiec2gkc

2022-03-26

Derivative of ${x}^{x}$ without logarithmic differentiation
With logarithmic differentiation, it is quite simple to compute the derivative of ${x}^{x}$:
$y={x}^{x}$
$\mathrm{ln}y=x\mathrm{ln}x$
$\frac{1}{y}\frac{dy}{dx}=\mathrm{ln}\left\{x\right\}+1$
$\frac{dy}{dx}=\left(\mathrm{ln}\left(x\right)+1\right){x}^{x}.$
Is there a method to compute the derivative of ${x}^{x}$ that does not rely on logarithmic differentiation?

Carter Lin

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

glikozyd3s68

There's another way that looks like a gross blunder but actually is perfectly correct. I will illustrate it on the more general question, differentiate $y={f\left(x\right)}^{g\left(x\right)}$
If g were constant, we'd get ${f}^{\prime }\left(x\right)g\left(x\right){f\left(x\right)}^{g\left(x\right)-1}$
If f were constant, we'd get ${g}^{\prime }\left(x\right){f\left(x\right)}^{g\left(x\right)}\mathrm{log}f\left(x\right)$
${y}^{\prime }={f}^{\prime }\left(x\right)g\left(x\right){f\left(x\right)}^{g\left(x\right)-1}+{g}^{\prime }\left(x\right){f\left(x\right)}^{g\left(x\right)}\mathrm{log}f\left(x\right)$
It's easy to see that in the original problem, where $f\left(x\right)=x$ and $g\left(x\right)=x$, this reduces to ${x}^{x}+{x}^{x}\mathrm{log}x$ , as obtained by other methods.