1

I was wondering on how to use the limit definition to find the derivative of the function $f(x) = a^x$ without using the constant $e$ and the logarithm $\ln(x)$ but only using the the definition:

$$\lim_{h \to 0}\frac{a^{x+h} - a^x}{h}= \lim_{h \to 0}\frac{a^{x} \cdot a^{h} - a^x}{h}= \lim_{h \to 0}\frac{a^{x}(a^{h} - 1)}{h}= a^x \cdot \lim_{h \to 0}\frac{a^{h} - 1}{h}$$

And here we have an indeterminate form $\frac{0}{0}$ when $h \to 0$.

How can I get past this loop hole?

Trying to substitute $f(x) = a^x$ by $f(x \ln(a)) = e^{x \ln(a)}$ is NOT an acceptable demonstration as we are invoking results we are deliberately trying to prove.

emandret
  • 956
  • 1
    I mean, $\ln$ appears in the derivative, so how do you expect to be able to evaluate this without $\ln$? – Rushabh Mehta Oct 01 '19 at 04:06
  • Isn't the next step the exact reason why we have L'Hopital's rule? – Axion004 Oct 01 '19 at 04:08
  • 1
    See this answer. @Axion004 you can't use L'Hospital's here, that's literally the point. – Rushabh Mehta Oct 01 '19 at 04:08
  • What do you mean by without using $e$ and $\ln$? Do you accept the definition of $e$ as $\lim_{n \to \infty} (1+1/n)^n$? – Toby Mak Oct 01 '19 at 04:09
  • 1
    @Axion004 How are you gonna use L'hopital without knowing the derivative of $a^x$ already :) – 79037662 Oct 01 '19 at 04:11
  • "Trying to substitute $a^x$ by $e^{x\ln a}$ is NOT an acceptable demonstration" Then you would need to write down your definition of function $a^x$ because writing $a^x=e^{x\ln a}$ is one way to define the exponential function with base $a$. –  Oct 01 '19 at 13:36
  • @Jack the problem is that you introduce $e$, as you point it out, I could define $a^x = 10^{\log ax}$ – emandret Oct 01 '19 at 13:40
  • No a problem. One path to define $a^x$ for general $a>0$ is: (1) define the constant $e$ (2) define the exponential function $e^x$ (3) define $a^x:=e^{a\ln(x)}$. There are of course other ways to define the function $x\mapsto a^x$. In order to avoid any circularity of finding $a^x$, one needs to give the definition of this expression first. What is yours? –  Oct 01 '19 at 13:43
  • @Jack you could to the same with decimal logarithm, not necessarily by using the natural logarithm – emandret Oct 01 '19 at 13:45
  • Moreover, it is plainly wrong to write $a^x=e^{\ln(ax)}$. The correct one is $a^x=x\ln(a)$. –  Oct 01 '19 at 13:46
  • Again, what is your definition of $a^x$? –  Oct 01 '19 at 13:48
  • Not the same. I am not talking about identities of expressions but definition of the function $a^x$. One cannot define $a^x=10^{x\log_{10}a}$ if the expression $10^x$ has not been defined yet. –  Oct 01 '19 at 13:51
  • The definition of $a^x$ is $a,x \in \mathbb{R} > 0$ – emandret Oct 01 '19 at 13:58
  • When I wrote $a^x = e^{ \ln ax}$, I was meaning $e^{\ln(a)x}$ and not $e^{\ln(ax)}$ – emandret Oct 01 '19 at 14:01
  • That is not a definition: you simply say that $a$ and $x$ are two positive real numbers but what is $a^x$? –  Oct 01 '19 at 14:18
  • An exponential function. – emandret Oct 01 '19 at 14:19
  • @TobyMak please continue your demonstration! It is not complete. I will delete this question and ask it again if people post incomplete answers. – emandret Oct 01 '19 at 19:20
  • My proof is complete: I was just finding $f'(0)$ which can be derived from the derivative of $e^x$. – Toby Mak Oct 01 '19 at 23:44
  • @TobyMak You mentioned: "Therefore we have that $f'(x) = \frac{d}{dx} e^x \cdot \ln a$, and since $ \frac{d}{dx} e^x = e^x$, $f'(0) = e^0 \cdot \ln a = \ln a$." but $f'(x)$ is not $\frac{d}{dx} e^x \cdot \ln a$ but $\frac{d}{dx} e^{x \ln(a)}$ – emandret Oct 01 '19 at 23:49
  • No, the derivative changes by a factor of $\ln a$. This comes from the horizontal transformation. – Toby Mak Oct 01 '19 at 23:53
  • You did not use the chain rule to find the derivative then. – emandret Oct 01 '19 at 23:54
  • No, just transformations. – Toby Mak Oct 01 '19 at 23:54
  • How do you find derivative using transformations ? I can't find any defined method. – emandret Oct 01 '19 at 23:56
  • Let us continue this discussion in chat. – Toby Mak Oct 01 '19 at 23:56

3 Answers3

3

Using only limits you have:

$$f'(0) = \lim_{h \to 0} \frac{f(0+h) - f(0)}{h}$$ $$= \lim_{h \to 0} \frac{a^h-1}{h}$$ $$\therefore f'(x) = a^x \times f'(0)$$

However, you cannot prove that $f'(0) = \ln a$ without using the property that $e^x$ is its own derivative.

If you accept the fact as described in this answer, use the fact that $a^x = e^{x \ln a} = f(x \ln a)$. This means that $a^x$ is a horizontal transformation of $e^x$, compressed by a factor of $\ln a$ (and stretched when $\ln a < 1, a < e$). Since the vertical dimension is not transformed, using $\text{slope} = \frac{\text{rise}}{\text{run}}$ gives:

$$f'(x) = \frac{\Delta y}{\frac{1}{\ln a} \cdot \Delta x} \left(e^x \right) = \ln a \times\frac{\mathrm{d}}{\mathrm{d}x} \left(e^x \right)$$

when $\Delta y$ and $\Delta x$ are small.

Since $ \frac{d}{dx} e^x = e^x$, therefore we have that $f'(0) = e^0 \cdot \ln a = \ln a$.

Toby Mak
  • 17,073
  • Depends on what definition of $e$ one is using, but I agree. – Rushabh Mehta Oct 01 '19 at 04:10
  • You assume $f'(0) = \lim_{h \to 0} \frac{f(0+h) - f(0)}{h}$ by comparing to the resulting limit definition, and substituting $x=0$ – emandret Oct 01 '19 at 04:22
  • Is that a problem then? – Toby Mak Oct 01 '19 at 04:23
  • Okay, I see what you mean. The first part of this proof was borrowed from a textbook called Haese's Additional Maths. – Toby Mak Oct 01 '19 at 04:25
  • Thank you for the reference. Is this limit substitution proof method named? – emandret Oct 01 '19 at 04:28
  • Proving that there exists $e$ such that $\lim \frac{e^h-1}h=1$ seems to me like precisely the hardest part of the original question. – Jack M Oct 01 '19 at 04:28
  • There is no specific name in the textbook. – Toby Mak Oct 01 '19 at 04:49
  • @TobyMak AFAIK there is no definitive answer to this question without assuming that $a^x = e^{\ln ax}$, we are using hints we are trying to prove. – emandret Oct 01 '19 at 13:18
  • 1
    One either writes $a\ln x$ or $(\ln x)a$. It is not true that $a^x=e^{\ln (ax)}$. –  Oct 01 '19 at 13:40
  • Sorry, there was a slight typo there. If another part of the proof is not fine, please let me know. – Toby Mak Oct 01 '19 at 23:39
  • I do not understand where the $\frac{1}{\ln(a)}$ comes from. Can you explain me @TobyMak – emandret Oct 02 '19 at 21:46
  • Since $f(x)$ is just $e^x$ compressed by a factor of $\ln a$, then the 'run' of 'rise over run' changes by $\frac{1}{\ln a}$. In other words, for two points close together, the horizontal distance between them is only $\frac{1}{\ln a}$ as much as $e^x$. – Toby Mak Oct 03 '19 at 08:58
2

It’s not directly using $\ln$ in the limit itself

Consider $$f’(x) = a^x(f’(0))$$ $$\frac{f’(x)}{f(x)} = f’(0)$$ Taking definite integral $$\displaystyle\int_0^1 \frac{df(x)}{f(x)dx} dx = f’(0)$$ $$f’(0) = \ln a$$ so we have the desired result.

Certainly not a dog
  • 1,462
1

We are going down the rabbit hole and simply not exiting it:

$$f'(x) = a^x \cdot \lim_{h \to 0}\frac{a^{h} - 1}{h}$$

Notice that:

$$f'(0) = \lim_{h \to 0}\frac{a^{h} - 1}{h}$$

Substitute by $f'(0)$:

$$f'(x) = a^x \cdot f'(0)$$

Let $f(x \ln(a)) = e^{x \ln(a)}$ and differentiate using the chain rule for $f'(x \ln(a))$:

$$\frac{d}{dx} e^{x \ln(a)} = \ln(a) \cdot e^{x \ln(a)}$$ $$\frac{d}{dx} e^{x \ln(a)} = \ln(a) \cdot a^x$$

Since $f'(0) = f'(0 \ln(a))$, substitute with the derivative:

$$f'(x) = a^x \cdot (\ln(a) \cdot a^0)$$ $$f'(x) = a^x \cdot \ln(a)$$

The result is here, but I am not happy with it because we used the chain rule to differentiate $f(x \ln(a)) = e^{x \ln(a)}$.

emandret
  • 956
  • I think your proof is still fine, because at least you haven't used implicit differentiation which is typically taught after the derivative of $a^x$. – Toby Mak Oct 01 '19 at 23:41
  • @TobyMak, I can't seen any other way to do it without over-complicating it honestly. – emandret Oct 01 '19 at 23:45
  • All of these are standard techniques, so it is hard to know which techniques you want and which techniques you don't want. I don't think you will get a satisfactory answer unless you provide a clear definition of $a^x$, $e^x$ and $\ln x$. With your definition, $a^x$ would not be defined for something like $x = \pi$ since $\pi$ is irrational, but the differentiation must hold for all $x$. – Toby Mak Oct 01 '19 at 23:48