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I can't get any of these on my own and I am attempting to do $x^x$ I had it explained to me for about an hour and I still can't do it on my own. I thought I was supposed to make it into $x(\ln x)$ which should be equivalent to the original term. Then the differentiation should be easy for most people from here.

$1(\ln x)+x(\frac1x) $ or $\ln x+1$

This is of course wrong but I do not know why.

2 Answers 2

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Use the fact that $y=e^{\ln(y)}$ and that $\ln(a^b)=b\ln(a)$ to get

$ x^x=e^{\ln(x^x}=e^{x\ln(x)}. $

Then the derivative you want will be the derivative of this:

$ \frac{d}{dx}(x^x)=\frac{d}{dx}(e^{x\ln(x)}). $

By chain rule this derivative is then:

$ \frac{d}{dx}(e^{x\ln(x)})=e^{x\ln(x)}\left(\ln(x)+x\cdot\frac{1}{x}\right). $

Thus the final answer is $x^x(\ln(x)+1)$.

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    It never ceases to amaze me how in calculus/analysis, the simplest looking problems can sometimes require some clever problem solving. I always thought of this as a nice example and I'll be assigning it regularly as a homework problem when I teach calculus.2011-10-14
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Start with $y=x^x$ Take logarithms on both sides: $\log y=\log(x^x)$ Simplify on the right side: $\log y=x\log x$ Differentiate with respect to $x$, remembering to use the chain rule on the left side: $(1/y)(dy/dx)=\log x+1$ So, $dy/dx=y\times(\log x+1)=x^x(\log x+1)$

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    Oh okay I sort of see it now, I am not familiar with that method.2011-10-14