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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.

jon jones
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2 Answers2

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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 $t=(1+y)\log(1+y).$

Vincenzo Tibullo
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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.

user758556
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  • Thanks @user758556 . I know to do in this way. What I wanted to do is find this limit using the limit of the incremental ratio. – jon jones Aug 01 '12 at 21:19
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    Not sure what you mean by limit of the incremental ratio (I'm really a physicist, so apologies for sloppy nonrigorous math). But just using the definition of derivative:

    $\displaystyle \lim_{h \rightarrow 0} \frac{(x+h)^{(x+h)} - x^x}{h} = \displaystyle \lim_{h \rightarrow 0} x^x \frac{[(x+h)^h - 1]}{h}$. Since you want this for $x=1$, you have 0/0, so evaluate the limit using L'Hospital's rule at $x = 1$. Apologies if using L'Hospital's rule is assuming too much for whatever you're trying to do?

     – user758556 Aug 01 '12 at 21:33
  • @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. – Javier Aug 01 '12 at 22:20