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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.

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up vote 6 down vote accepted

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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Oh okay so I had the rule memorized incorrectly. – user138246 Oct 13 '11 at 22:29
You just forgot to multiply by the function in the end. The correct formula is: $$f'(x)=f(x)\frac{d}{dx}(\ln(f(x))$$. – Joe Johnson 126 Oct 13 '11 at 23:06
I actually didn't even do the chain rule I was just kind of throwing out what I thought was a rule. – user138246 Oct 13 '11 at 23:07
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. – Mathemagician1234 Oct 14 '11 at 3:59

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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What happened to the 1/y? – user138246 Oct 14 '11 at 0:39
I multiplied both sides by $y$. – Gerry Myerson Oct 14 '11 at 1:23
That seems wrong though, the answer shouldn't have a y in it. – user138246 Oct 14 '11 at 1:28
The answer doesn't have a $y$ in it; the term with the $y$ in it is just a stage on the way to the last thing I wrote down, $x^x(\log x+1)$, which is the answer. – Gerry Myerson Oct 14 '11 at 3:22
Oh okay I sort of see it now, I am not familiar with that method. – user138246 Oct 14 '11 at 3:30

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