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Find the derivative of $x^x$ at $x=1$ by definition (i.e. using the limit of the incremental ratio).

The only trick I know is $x^x = e^{x \ln x}$ but it doesn't work.

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    Oh, I see: "by definition" in the title means you want to find the derivative by using the definition.2012-08-01

2 Answers 2

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Using the definition: $ \begin{align} f'(1)&=\lim_{x\rightarrow1}\frac{x^x-1}{x-1}\\ &=\lim_{x\rightarrow1}\frac{e^{x\log{x}}-1}{x-1}\\ &=\lim_{y\rightarrow0}\frac{e^{(1+y)\log(1+y)}-1}{y}\\ &=\lim_{y\rightarrow0}\frac{e^{(1+y)\log(1+y)}-1}{(1+y)\log(1+y)}\frac{(1+y)\log(1+y)}{y}\\ &=\lim_{t\rightarrow0}\frac{e^{t}-1}{t}\lim_{y\rightarrow0}(1+y)\frac{\log(1+y)}{y}=1 \end{align} $ where $y=x-1$, and $t=(1+y)\log(1+y).$

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    @Neat answer. I like it (+1).2012-09-17
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The trick you mentioned $\frac{d}{dx}[x^{x}] = \frac{d}{dx} e^{x \ln{x}}$ still works. :)

Apply the chain rule: $e^{x \ln{x}}\frac{d}{dx}[x \ln{x}]$

And then the product rule: $e^{x \ln{x}}(\ln{x}+x\frac{1}{x})$

Simplify: $x^x(1+\ln{x})$

Edit: You wanted the value of the derivative evaluated at $x = 1$, so just substitute in and you get 1.

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    @user758556: I guess to use L'Hôpital's you would have to know $\frac{\mathrm{d}}{\mathrm{d}h} (x+h)^h$, which sorts of defeats the purpose of using the limit definition in the first place.2012-08-01