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I want to prove that $\displaystyle\lim_{h\to 0}\frac{(x+h)^{x+h}-x^x}{h}=x^x(\ln(x)+1).$

If I write $x^x$ as $e^{x\ln(x)}$ I get: $\displaystyle\lim_{h\to0}\frac{e^{(x+h)\ln(x+h)}-e^{x\ln(x)}}{h}$ but then I'm stuck. What are the next steps? Thanks in advance.

Cristopher
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    If you insist on evaluating this limit from the definition, have a look at how you prove the chain rule from first principles. Use that as a model for your limit. – Simon S Jun 14 '15 at 23:00
  • @SimonS Okay, I'm looking at the first proof here: https://en.wikipedia.org/wiki/Chain_rule#Proofs Am I supposed to define $f(x)=x^x$ and $g(x)=x+h$? Could you maybe show me the first steps of how to use this as a model so I can see if I am able to work it out from there? Thanks – Cristopher Jun 15 '15 at 00:42
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    @Christopher I think he meant define $f(x) = e^x$ and $g(x) = x \log x$. This is how this derivative is generally obtained when not restricting oneself to using the definition directly. However, you will in essence be using the chain rule anyway, it is just you will show the limits explicitly. If you are looking to truly avoid the chain and l' Hopital's rules, AnalysisStudent0414's answer seems to be the way to go. – John Colanduoni Jun 15 '15 at 03:50
  • [Related Question](https://math.stackexchange.com/questions/433135/derivative-of-xx-using-first-principle) – Axion004 Jul 29 '20 at 02:23

2 Answers2

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Here is a simple evaluation of the derivative. \begin{align} f'(x) &= \lim_{h \to 0}\frac{f(x + h) - f(x)}{h}\notag\\ &= \lim_{h \to 0}\frac{(x + h)^{x + h} - x^{x}}{h}\notag\\ &= \lim_{h \to 0}\frac{\exp\{(x + h)\log(x + h)\} - \exp(x\log x)}{h}\notag\\ &= \exp(x\log x)\lim_{h \to 0}\frac{\exp\{(x + h)\log(x + h) - x\log x\} - 1}{h}\notag\\ &= x^{x}\lim_{h \to 0}\frac{\exp\{(x + h)\log(x + h) - x\log x\} - 1}{(x + h)\log(x + h) - x\log x}\cdot\frac{(x + h)\log(x + h) - x\log x}{h}\notag\\ &= x^{x}\lim_{t \to 0}\frac{e^{t} - 1}{t}\cdot\lim_{h \to 0}\frac{(x + h)\log(x + h) - x\log x}{h}\notag\\ &= x^{x}\cdot 1\cdot\lim_{h \to 0}\frac{(x + h)\log(x + h) - x\log(x + h) + x\log(x + h) - x\log x}{h}\notag\\ &= x^{x}\lim_{h \to 0}\dfrac{h\log(x + h) + x\log\left(1 + \dfrac{h}{x}\right)}{h}\notag\\ &= x^{x}\lim_{h \to 0}\left\{\log(x + h) + \dfrac{\log\left(1 + \dfrac{h}{x}\right)}{\dfrac{h}{x}}\right\}\notag\\ &= x^{x}\left\{\log x + \lim_{v \to 0}\frac{\log(1 + v)}{v}\right\}\notag\\ &= x^{x}(\log x + 1)\notag \end{align} In the above we have used the substitutions $$t = (x + h)\log(x + h) - x\log x, v = h/x$$ so that both $t, v$ tend to $0$ as $h \to 0$. Also the following standard limits are used $$\lim_{x \to 0}\frac{e^{x} - 1}{x} = 1,\,\lim_{x \to 0}\frac{\log(1 + x)}{x} = 1$$ Note that without the use of these standard limits it is not possible get derivatives from first principles for any function of type $a(x)^{b(x)}$.

Paramanand Singh
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  • Awesome... that's quite a bit of algebra, but the steps are much clearer to me now. Thank you very much! :) – Cristopher Jun 18 '15 at 23:50
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$$\displaystyle\lim_{h\to0}\frac{e^{(x+h)\ln(x+h)}-e^{x\ln(x)}}{h} = \displaystyle\lim_{h\to0}e^{x \ln(x)}\frac{e^{(x+h)\ln(x+h)- x \ln(x)}-1}{h}$$

$$ = \displaystyle\lim_{h\to0}e^{x \ln(x)}\frac{e^{x\ln(\frac{x+h}{x})+h \ln(x+h)}-1}{h} \quad = \quad \displaystyle\lim_{h\to0}e^{x \ln(x)}\frac{e^{x\ln(\frac{x+h}{x})} e^{h \ln(x+h)}-1}{h} $$

$$\displaystyle\lim_{h\to0}e^{x \ln(x)}\frac{e^{x\ln(1+\frac{h}{x})}e^{h \ln(x+h)}-1}{h} =^{taylor} \displaystyle\lim_{h\to0}e^{x \ln(x)}\frac{e^{x \frac{h}{x}}(x+h)^h-1}{h} $$

$$x^x \displaystyle\lim_{h\to0}\frac{e^h (x+h)^h-1}{h} $$

That's how far I'm able to get without using chain rules or similar. Can anyone help me conclude?

AnalysisStudent0414
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  • I suggest the following definition for the natural logarithm: $\log x = \lim_{n \to \infty} n (\sqrt[n]{x} - 1)$. – John Colanduoni Jun 14 '15 at 23:43
  • @AnalysisStudent0414, nice! Thanks for trying. Now, let's try to finish the proof... :) – Cristopher Jun 15 '15 at 04:03
  • @JohnColanduoni Yet another limit? haha. This is more complicated than I thought... but your suggestion is interesting. I'm not sure how to use that definition though... – Cristopher Jun 15 '15 at 15:19
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    @Christopher The final limit in the above answer is tantalizingly close; if you switch variables from $h$ to $n = 1/h$, the limit becomes $\lim_{n \to \infty} n (e^{1/n} \sqrt[n]{x + 1/n} - 1)$. If not for the $+ 1/n$ under the square root, we would already be done; as it would equal $\log(e x) = \log x + 1$. – John Colanduoni Jun 15 '15 at 18:22