I've been trying to think for the past few days how one could differentiate $a^x$ based on the definition that $a^n$ is repeated multiplication, $a^{n/m}=(\sqrt[m]a)^n$, and $a^x$ is the completion of the above function by continuity.

With a bit of algebra, the problem quickly reduces to finding the derivative at $0$:


And that limit's really got me stumped. Since you'll obviously have to use the definition in some way, I thought I'd replace $h$ with $\frac{1}{n}$ and use the $n$-th root definition:


Of course, just because $(2)$ exists doesn't automatically imply $(1)$ exists, but it might be a first step. Even $(2)$ has me stumped, though.

  • $\begingroup$ Interesting. But why do you say (2) doesn't imply (1) (in term of existence), please? $\endgroup$ – Abhimanyu Arora Feb 16 '14 at 11:07
  • $\begingroup$ @AbhimanyuArora Just because there exists a sequence $x_n\to a$ with $f(x_n)\to b$ doesn't imply that the limit of $f(x)$ as $x\to a$ is $b$. It does if $f$ is continuous, but I don't think we can assume that this function is continuous here. $\endgroup$ – Jack M Feb 16 '14 at 11:11
  • $\begingroup$ There is some confusion in what you regard as "first principles", see comments for DonAntonio's answer. Is the limit of $\frac{e^x-1}{x}$ a first principle you'd accept? $\endgroup$ – JiK Feb 16 '14 at 11:14
  • $\begingroup$ @JackM:Thanks for clarifying. You got me thinking here. But $f(x_n)\rightarrow b$ means that b is the limit (by definition), or am I mistaking something with my understanding? $\endgroup$ – Abhimanyu Arora Feb 16 '14 at 11:16
  • $\begingroup$ To avoid confusion, I'd suggest explicitely stating that you "don't know" what $e^x$ or $\log$ are, if don't want to use them (without proofs for the properties used). $\endgroup$ – JiK Feb 16 '14 at 11:16

I must appreciate effort put by OP to define the exponential function $a^{x}$ for $a > 0$ by extending the algebraical definition when $x$ is rational to the case where $x$ is irrational by using continuity argument. While there is nothing wrong with this approach it turns out to be one of the difficult routes to a theory of logarithmic and exponential function.

Now back to the question at hand. Differentiation by first principle of $f(x) = a^{x}$ involves the evaluation of limit $$L(a) = \lim_{h \to 0}\frac{a^{h} - 1}{h}$$ The challenge here is not to find $L(a)$ but to prove that this limit exists. Clearly the limit wont exist unless we have $\lim_{h \to 0}a^{h} = 1$. So as a part of definition of $a^{x}$ we must ensure that we have established $\lim_{h \to 0}a^{h} = 1$.

Note that if $a = 1$ then the limit is $0$ trivially. So let $a \neq 1$ and then there are two cases $a > 1$ and $0 < a < 1$. Clearly by putting $a = 1/b$ we can see that $L(1/a) = L(b) = -L(a)$ (note while proving this we will need $\lim_{h \to 0}a^{h} = 1$) and hence it is sufficient to consider the case $a > 1$.

Now inequalities come to the rescue. From this answer we have $$\frac{a^{r} - 1}{r} > \frac{a^{s} -1 }{s}$$ where $r, s$ are positive rationals and $r > s$. Note that by continuity arguments the inequality can be extended to positive irrational values of $r, s$ with $r > s$ but then the inequality weakens to $\geq$. There are ways to make this inequality strict for irrationals $r, s$ but we won't need the strict version here. Clearly from the above we can see that the function $g(h) = (a^{h} - 1)/h$ is an increasing function of $h$ for $h > 0$. Clearly since $a > 1$ it follows that $g(h) > 0$ for all $h > 0$. Now as $h \to 0^{+}$ the function $g(h)$ decreases but is bounded below by $0$ hence tends to a limit $L(a)$.

If $h \to 0^{-}$ then we can put $h = -k$ and see that $$\lim_{h \to 0^{-}}\frac{a^{h} - 1}{h} = \lim_{k \to 0^{+}}\frac{1 - a^{k}}{-ka^{k}} = \lim_{k \to 0^{+}}\frac{a^{k} - 1}{k} = L(a)$$ It now follows that $g(h)$ tends to a limit as $h \to 0$ which we have denoted by $L(a)$.

By further careful considerations it can be shown that $a > 1$ implies that $L(a) > 0$ and since $L(1/a) = -L(a)$ we have $L(a) < 0$ if $0 < a < 1$. It can be further established using inequalities that $L(a) $ is a strictly increasing function of $a$ for $a > 0$. This function $L(a)$ is traditionally written as $\log a$. Simple properties like $\log(ab) = \log a + \log b$ are provable very easily using this definition. Using this we also get $\log(a^{n}) = n\log a$ for any integer $n$ which shows that range of this $\log $ function is $(-\infty, \infty)$.

It is now a simple matter to show that $(a^{x})' = a^{x}\log a$. Next we can define $e$ by $\log e = 1$ and then $(e^{x})' = e^{x}$ and we can prove that $e^{\log a} = a$ for $a > 0$ and $\log (e^{a}) = a$ for all $a$. Thus $\log x$ and $e^{x}$ are inverses and $(\log x)' = 1/x$ by rule for differentiation of inverse functions. I hope you can proceed along these lines to develop full theory of exponential and logarithmic functions.

  • $\begingroup$ Thanks for your detailed answer. When you say 'Clearly the limit wont exist unless we have $\lim_{h \to 0}a^{h} = 1$.', why is it the case...is it related to L'hospital's rule? $\endgroup$ – Abhimanyu Arora Feb 16 '14 at 13:22
  • 1
    $\begingroup$ @AbhimanyuArora: if $\lim_{h \to 0}a^{h} \neq 1$ then the numerator is non-zero and denominator is $0$ so that $\lim_{h \to 0}(a^{h} - 1)/h$ does not exist. $\endgroup$ – Paramanand Singh Feb 16 '14 at 13:26
  • $\begingroup$ Doesn't your other answer only show that $g$ is increasing as an integer function, not a real number function? $\endgroup$ – Jack M Feb 16 '14 at 13:42
  • $\begingroup$ @JackM: My linked answer proves the result for rational $r , s$ with $r > s$. Please see the linked answer carefully (especially last few paragraphs there). The extension to irrational $r, s$ is done by continuity argument. Basically we take sequences $r_{n}, s_{n}$ of rationals tending to $r, s$ and then $g(r_{n}) > g(s_{n})$. Then take limits as $n \to \infty$. This gives $g(r) \geq g(s)$ and not the strict inequality. But this is sufficient here. $\endgroup$ – Paramanand Singh Feb 16 '14 at 13:46

What about using $\;a^x=e^{x\log a}\;?$ Then

$$\frac{a^h-1}h=\frac{e^{h\log a}-1}h\;\;\stackrel{\text{subst.}\;h\log a\to x}=\;\;\frac{e^x-1}{\frac x{\log a}}\xrightarrow[x\to 0]{}\log a$$

since, in the above substitution, $\;h\to 0\iff x=h\log a\to 0\;$

  • $\begingroup$ I'd say this goes pretty wildly against my wanting to do it "from first principles". For instance, one reason I wanted to do this was to justify the definition of $e$ as the unique base that has $e^x=e^x$. $\endgroup$ – Jack M Feb 16 '14 at 11:05
  • $\begingroup$ Doesn't the OP want to find the derivative starting from $a^{n/m}$, not in some other way? $\endgroup$ – JiK Feb 16 '14 at 11:06
  • $\begingroup$ @JackM Many of the students that ask question like this one usually consider the limit $\;\frac{e^h-1}h\xrightarrow[h\to 0]{}1\;$ to be a "basic, elementary one" (you can check other questions close to this one). Anyway, any "first principles" proof of this basic limit will work for your case, and downvoting the answer seems a little rude at this stage. $\endgroup$ – DonAntonio Feb 16 '14 at 11:08
  • $\begingroup$ No @JiK. The OP did write explicitly that his problem reduces to finsd the above left hand limit. $\endgroup$ – DonAntonio Feb 16 '14 at 11:08
  • $\begingroup$ @DonAntonio I wasn't the one who downvoted it, don't worry. In any case, do you have a proof for that limit? I'd Google, but, well... googling math symbols... yeah. $\endgroup$ – Jack M Feb 16 '14 at 11:09

Well, historically that is just the definition of the natural logarithm, so define


and conclude that this function has the usual properties of a logarithm, like $l(ab)=l(a)+l(b)$.

  • $\begingroup$ Would one need to use Taylor series expansion to prove the properties of logarithms you mention? $\endgroup$ – Abhimanyu Arora Feb 16 '14 at 11:12
  • 1
    $\begingroup$ No, just the limit rules, esp. $\lim_{n\to\infty}\sqrt[n]a=1$. $l(ab)=\lim_{n\to\infty}\sqrt[n]a\cdot n(\sqrt[n]b-1)+\lim_{n\to\infty} n(\sqrt[n]a-1)$. $\endgroup$ – Dr. Lutz Lehmann Feb 16 '14 at 11:14
  • $\begingroup$ The proof that this limit exists as $n \to \infty$ requires considerations of inequalities like the one I have used in my answer and is not a trivial exercise. But once the existence is done the result $l(ab) =l(a) + l(b)$ is almost trivial as you have shown. $\endgroup$ – Paramanand Singh Feb 16 '14 at 12:14
  • $\begingroup$ @ParamanandSingh Yes, you are right, existence of $l(a)$ has to be secured first. My point was that there is no further "underlying" truth behind this limit being the logarithm, it already is the most elementary characterization. $\endgroup$ – Dr. Lutz Lehmann Feb 16 '14 at 13:04
  • $\begingroup$ @LutzL: fully agree with you. $\endgroup$ – Paramanand Singh Feb 16 '14 at 13:05

An interesting way to prove it is the following:

Assuming you can prove $\frac{d}{dx}\ln(x)=\frac{1}{x}$ you can write: $$ x=\ln(x)\Rightarrow\ln(a^x)=x\ln(a). $$ Thus: $$ \frac{d}{dx}\ln(a^x)=\frac{d}{dx}x\ln(a)=\ln(a), $$ But for the chain rule, $\frac{d}{dy}\ln(y)=\frac{1}{y}\frac{d}{dy}y$, hence: $$ \frac{d}{dx}\ln(a^x)=\frac{1}{a^x}\frac{d}{dx}a^x. $$ Therefore: $$ \frac{1}{a^x}\frac{d}{dx}a^x=\ln(a). $$ Finally: $$ \frac{d}{dx}a^x=a^x\ln(a) $$


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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