First principles derivation

The derivative of the exponential function given by f\left(x\right)=a^x can be developed by first principles:

f'\left(x\right)	=	\lim_{h\rightarrow 0}\frac{f(x+h)-f(x)}{h}
	=	\lim_{h\rightarrow 0}\frac{a^{x+h}-a^{x}}{h}
	=	\lim_{h\rightarrow 0}\frac{a^{x}{\left (a^{h}-1 \right )}}{h}
	=	a^{x}\times \lim_{h\rightarrow 0}\frac{a^h-1}{h}

Thus, provided the quantity \frac{a^h-1}{h} converges as h\rightarrow 0, the derivative exists as the product of the original function itself and some constant whose value depends on the base a.

In fact, for the function f\left(x\right)=2^x, as h\rightarrow 0, the quantity \frac{2^h-1}{h} does converge to an irrational number that is approximately 0.693147. This means that f\left(x\right)=2^x, then f'\left(x\right)\approx 0.693147\times 2^x.

The table shows derivative results for four values of a:

a	\lim_{h\rightarrow 0}\frac{a^h-1}{h}	f'(x)
2	0.693147...	\approx 0.693147\times 2^x
3	1.098612...	\approx 1.098612\times 3^x
4	1.386294...	\approx 1.386294\times 4^x
5	1.609438	\approx 1.609438\times 5^x

The results suggest that there is a base between a=2 and a=3 where the quantity in the middle column - namely \lim_{h\rightarrow 0}\frac{a^h-1}{h}, is exactly 1.

This would reveal the existence of a function that is its own derivative!

Your turn now. We can see that this function must exist somewhere between 2 and 3. Pick three values between 2 and 3 and see if you can get closer to this function.

In fact...

This remarkable and irrational base has the approximate value of 2.718281828459045 and has been denoted by the letter e (standing for the word exponential, and first discovered by the mathematician Leonard Euler)

Thus with the base e\approx 2.718, we have the important result:

If f\left(x\right)=e^x, then f'\left(x\right)=e^x.

Applet

The following applet shows the function f\left(x\right)=a^x and its derivative for a range of values between a=2 and a=3 . Satisfy yourself that the curves of the function and the differential function coincide when a\approx 2.72.

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Limit of (a^h-1)/h as h→0 (investigation)