ropowiec2gkc

2022-03-26

Derivative of $x}^{x$ without logarithmic differentiation

With logarithmic differentiation, it is quite simple to compute the derivative of$x}^{x$ :

$y={x}^{x}$

$\mathrm{ln}y=x\mathrm{ln}x$

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

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

Is there a method to compute the derivative of$x}^{x$ that does not rely on logarithmic differentiation?

With logarithmic differentiation, it is quite simple to compute the derivative of

Is there a method to compute the derivative of

Carter Lin

Beginner2022-03-27Added 13 answers

glikozyd3s68

Beginner2022-03-28Added 16 answers

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${g}^{\prime}\left(x\right){f\left(x\right)}^{g\left(x\right)}\mathrm{log}f\left(x\right)$

Add these together to get the answer:

${y}^{\prime}={f}^{\prime}\left(x\right)g\left(x\right){f\left(x\right)}^{g\left(x\right)-1}+{g}^{\prime}\left(x\right){f\left(x\right)}^{g\left(x\right)}\mathrm{log}f\left(x\right)$

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.

If g were constant, we'd get

If f were constant, we'd get

Add these together to get the answer:

It's easy to see that in the original problem, where

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