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

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

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

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Sort of: you can rewrite $x^x$ as $e^{x\ln x}$ and differentiate that. But the computations are essentially identical. –  Brian M. Scott Mar 24 '13 at 3:01
    
Thanks for the comment, I should have thought of that! I'll type up an answer. –  Gamma Function Mar 24 '13 at 3:03

2 Answers 2

up vote 7 down vote accepted

Thanks to Brian M. Scott for the comment that led to this solution:

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

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Logarithmic differentiation and this one is essentially the same. –  Moron plus plus Jun 13 '13 at 14:42

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 $y=f(x)^{g(x)}$.

If $g$ were constant, we'd get $f'(x)g(x)f(x)^{g(x)-1}$.

If $f$ were constant, we'd get $g'(x)f(x)^{g(x)}\log f(x)$.

Add these together to get the answer: $$y'=f'(x)g(x)f(x)^{g(x)-1}+g'(x)f(x)^{g(x)}\log f(x)$$

It's easy to see that in the original problem, where $f(x)=x$ and $g(x)=x$, this reduces to $x^x+x^x\log x$, as obtained by other methods.

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