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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6$\begingroup$ What do you mean "it doen't works"? $\endgroup$– Alex BeckerCommented Aug 1, 2012 at 21:05
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$\begingroup$ And "by definition", using the limit of the incremental ratio? $\endgroup$– Vincenzo TibulloCommented Aug 1, 2012 at 21:06
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1$\begingroup$ In fact, that "trick" does work. $\endgroup$– Michael HardyCommented Aug 1, 2012 at 21:07
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$\begingroup$ Yes @enzotib, using th limit of the incremental ratio. $\endgroup$– jon jonesCommented Aug 1, 2012 at 21:14
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$\begingroup$ Oh, I see: "by definition" in the title means you want to find the derivative by using the definition. $\endgroup$– Michael HardyCommented Aug 1, 2012 at 21:20
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2 Answers
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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 $t=(1+y)\log(1+y).$
answered Aug 1, 2012 at 21:24
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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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$\begingroup$ 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. $\endgroup$ Commented Aug 1, 2012 at 21:19
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1$\begingroup$ 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? $\endgroup$ Commented Aug 1, 2012 at 21:33
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$\begingroup$ @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. $\endgroup$– JavierCommented Aug 1, 2012 at 22:20