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I know that the derivative of $b^x$ is just $b^x \log{(b)}$, and I've seen it being derived using chain rule and such (not that I understand how it's done, I just learned about $e$ today so using the chain rule to derive $b^x$ is outside my scope as of now). In any case I was trying to find out how you could derive $b^x$ using the definition of the derivative, $[f(x+h) - f(x)]/h$. I couldn't find it on the internet so I was wondering if someone here could show me. Thanks in advance!

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  • $\begingroup$ Actually I think it's $b^x\log(b)$ not $b^x\log(x)$. Does that help? $\endgroup$ – Gregory Grant Apr 21 '15 at 20:33
  • $\begingroup$ Aha, sorry, brain fart lol. And not really I'm afraid. $\endgroup$ – inglavatin Apr 21 '15 at 20:34
  • $\begingroup$ I guess you need to show $\lim_{h\rightarrow 0}\frac{b^h-1}{h}=\log(b)$. I'm not sure how difficult that is, but I don't think it's trivial. $\endgroup$ – Gregory Grant Apr 21 '15 at 20:37
  • $\begingroup$ How do you define $\log b$ without knowing $e$? $\endgroup$ – celtschk Apr 21 '15 at 20:40
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    $\begingroup$ @celtschk Area below the hyperbola, as usual. $\endgroup$ – Alamos Apr 21 '15 at 20:43
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$$ \frac{d}{dx} b^x = \lim_{h \to 0} \frac{b^{x+h}-b^x}{h} = b^x \lim_{h \to 0} \frac{b^h-1}{h}. $$ At this point, you have to evaluate this limit. You have a couple of options here. Obviously L'Hôpital's not going to do. One way is, if your definition of $e^{y}$ is $$ \exp{y} = \sum_{k=0}^{\infty} \frac{y^k}{k!}, $$ then write $b^h = \exp{(h\log{b})} $, and you can evaluate the limit using the first few terms of the series.

On the other hand, if your definition is $$ \exp{y} = \lim_{n \to \infty} \left( 1 + \frac{y}{n} \right)^n, $$ you have a bit more of a problem. Essentially what this comes down to is the rearrangement $$ y = \left( 1+\frac{x}{n} \right)^n \to x = n(y^{1/n}-1) $$ Now, set $h=1/n$ and this looks like your limit, so you conclude that the limit is $\log{b}$. (I would like to see a cleaner version of this, should anyone have one to offer: the rearrangement to get the inverse doesn't look completely justified the way I normally do it.)

And if your definition of $\log{x}$ is an integral, and $\exp{x}$ as its inverse, then you're probably best proving it equivalent to one of the above first...

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  • $\begingroup$ Nevermind ... I was just nitpicking. Or even pettifogging ;) $\endgroup$ – String Apr 21 '15 at 20:58
  • $\begingroup$ @String I for one don't mind those being discreetly edited away by others... still, a good knock to one's ego :-) $\endgroup$ – Chappers Apr 21 '15 at 21:00
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Here's how one might do it with the definition $\log b = \int_1^b \frac{\mathrm dx}{x}$. However I'm not sure if the exchange of derivative and limit in the first step is allowed.

On one hand we have: $$\frac{\mathrm d}{\mathrm db}\lim_{h\to 0}\frac{b^h-1}{h} = \lim_{h\to 0}\frac{\mathrm d}{\mathrm db}\frac{b^h-1}{h} = \lim_{h\to 0}\frac{hb^{h-1}}{h} = \frac{1}{b}$$ On the other hand we have: $$\frac{\mathrm d}{\mathrm db}\log b = \frac{\mathrm d}{\mathrm db}\int_1^b\frac{\mathrm dx}{x} = \frac1b$$ Therefore we have $$\lim_{h\to 0}\frac{b^h-1}{h} = \log b+C$$ To determine $C$ we note that with $b=1$, we have on one hand $$\lim_{h\to 0}\frac{1^h-1}{h} = 0$$ and on the other hand $$\int_1^1\frac{\mathrm dx}{x} = 0$$ and therefore $C=0$.

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  • $\begingroup$ Hang on, how are you interchanging the limits in $d/db$ and $\lim_{h \to 0}$? $\endgroup$ – Chappers Apr 21 '15 at 21:35

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