How do I use the limit definition to find the derivative on $a^x$?

Question

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?

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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.

Answer 1

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

Answer 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.

Answer 3

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