How to solve the derivative of $b^x$ using the definition

Question

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!

Answer 1

$$ \frac{d}{dx} b^x = \lim_{h \to 0} \frac{b^{x+h}-b^x}{h} = b^x \lim_{h \to 0} \frac{b^h-1}{h}. $$ 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...

Answer 2

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 $C=0$.