Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

When applying the definition of a derivative to $\frac{d}{dx}b^x$ and a little algebra one arrives to

$$b^x\times\lim\limits_{h \to 0}\frac{b^h - 1}{h}$$

where of course that limit equals $\ln(b)$. I know this limit can be evaluated with L'Hospitals rule, but that involves using the derivative what I am just about to proove, so that would be a circular proof. I suspect one has to reduce this limit to something that relates to the definition of e as $$\lim\limits_{n \to \infty}(1+1/n)^n$$ but I can not do it.

How to show that that limit equals $\ln(b)$, so the proof of the derivative of $b^x$ is complete ?

Note: I found that by substituting for $s = b^h-1$, etc, one can separate $\ln(b)$, but than the another indeterminate limit remains: $\lim\limits_{s \to 0}(s/\ln(1+s))$, which must be $1$, and again I do not want to solve it using L'Hospitals rule, but I cannot otherwise.

share|cite|improve this question
Definitely a duplicate question. – Graphth Oct 21 '12 at 15:44
Both the limit with $h \to 0$ and $s \to 0$ are "elementary" limits which you can derive, for example, with a Taylor expansion of the function. For the limit with $s$ try to bound it from above with something that approaches 1: what do you know about $\log(1 + s)$ in $(0,1)$? – Andy Oct 21 '12 at 15:45

As I said in the comment, your substitution is good. Now if we take $\log(s +1)$ can we bound it by something resembling $s$? Recall that $s+1 < e^s$ for all $s > 0$ (try to prove it). Then you can use the fact that if $\alpha \leq \lim_{x \to c} f(x) \leq \lim_{x \to c} g(x) = \alpha$ and the limit of $f$ exists, then $\lim_{x \to c} f(x) = \alpha$ (which is a special case of the comparison theorem).

share|cite|improve this answer

As I recall we wrote $b^x=e^{x \ln b}$ and used the chain rule: $\frac {d\ b^x}{dx}=\frac {d\ e^{x \ln b}}{dx}=e^{x \ln b}\ln b=\ln b \ b^x$. Have you proved the chain rule yet?

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.