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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 $0$:

$$\lim_{h\to0}\frac{a^h-1}{h}\tag{1}$$

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:

$$\lim_{n\to\infty}n(\sqrt[n]a-1)\tag{2}$$

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.

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  • $\begingroup$ 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
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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.

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  • $\begingroup$ 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
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    $\begingroup$ @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
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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, $\;h\to 0\iff x=h\log a\to 0\;$

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  • $\begingroup$ 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
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Well, historically that is just the definition of the natural logarithm, so define

$$l(a)=\lim_{n\to\infty}n(\sqrt[n]{a}-1)$$

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

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  • $\begingroup$ 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
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    $\begingroup$ 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
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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: $$ \frac{d}{dx}a^x=a^x\ln(a) $$

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