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.

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    $\begingroup$ I mean, $\ln$ appears in the derivative, so how do you expect to be able to evaluate this without $\ln$? $\endgroup$ – Don Thousand Oct 1 at 4:06
  • $\begingroup$ Isn't the next step the exact reason why we have L'Hopital's rule? $\endgroup$ – Axion004 Oct 1 at 4:08
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    $\begingroup$ See this answer. @Axion004 you can't use L'Hospital's here, that's literally the point. $\endgroup$ – Don Thousand Oct 1 at 4:08
  • $\begingroup$ What do you mean by without using $e$ and $\ln$? Do you accept the definition of $e$ as $\lim_{n \to \infty} (1+1/n)^n$? $\endgroup$ – Toby Mak Oct 1 at 4:09
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    $\begingroup$ @Axion004 How are you gonna use L'hopital without knowing the derivative of $a^x$ already :) $\endgroup$ – 79037662 Oct 1 at 4:11

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

  • $\begingroup$ Depends on what definition of $e$ one is using, but I agree. $\endgroup$ – Don Thousand Oct 1 at 4:10
  • $\begingroup$ You assume $f'(0) = \lim_{h \to 0} \frac{f(0+h) - f(0)}{h}$ by comparing to the resulting limit definition, and substituting $x=0$ $\endgroup$ – gatosec Oct 1 at 4:22
  • $\begingroup$ Is that a problem then? $\endgroup$ – Toby Mak Oct 1 at 4:23
  • $\begingroup$ No, it's a very clever move. $\endgroup$ – gatosec Oct 1 at 4:24
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    $\begingroup$ One either writes $a\ln x$ or $(\ln x)a$. It is not true that $a^x=e^{\ln (ax)}$. $\endgroup$ – Jack Oct 1 at 13:40

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.


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 $f(x \ln(a)) = e^{x \ln(a)}$.

  • $\begingroup$ I think your proof is still fine, because at least you haven't used implicit differentiation which is typically taught after the derivative of $a^x$. $\endgroup$ – Toby Mak Oct 1 at 23:41
  • $\begingroup$ @TobyMak, I can't seen any other way to do it without over-complicating it honestly. $\endgroup$ – gatosec Oct 1 at 23:45
  • $\begingroup$ 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 $x$. $\endgroup$ – Toby Mak Oct 1 at 23:48

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