Derivative of Exponential Function

Theorem
Let $$\exp x$$ be the exponential function.

Then $$D_x \left({\exp x}\right) = \exp x$$.

Corollary 1
Let $$c \in \reals$$.

Then $$D_x \left({\exp \left({c x}\right)}\right) = c \exp \left({c x}\right)$$.

Corollary 2
Let $$a \in \reals; a > 0$$.

Let $$a^x$$ be $$a$$ to the power of $$x$$.

Then $$D_x \left({a^x}\right) = a^x \ln a$$.

Proof
We have:

$$ $$

From one of the definitions of the exponential function, $$\frac{\exp h - 1}{h} = \frac{\lim_{n \to \infty} (1 + \frac{h}{n})^{n} - 1}{h} = \frac{\lim_{n \to \infty} \sum_{k=0}^{n} {{n}\choose{k}}1^{n-k}(\frac{h}{n})^{k} - 1}{h}.$$

We can now carry the limit outside of the ratio to yield $$\lim_{n \to \infty} \frac{\sum_{k=0}^{n} {{n}\choose{k}}1^{n-k}(\frac{h}{n})^{k} - 1}{h} = \lim_{n \to \infty} {{n}\choose{1}}\frac{1}{n} + \sum_{k=2}^{n}{{n}\choose{k}}\frac{h^{k-1}}{n^k} = 1 + \lim_{n \to \infty}\sum_{k=2}^{n}{{n}\choose{k}}\frac{h^{k-1}}{n^k} = 1 + h\lim_{n \to \infty}\sum_{k=2}^{n}{{n}\choose{k}}\frac{h^{k-2}}{n^k}.$$

The right summand clearly converges to zero as h gets arbitrarily small, and so $$\lim_{h \to \infty}\frac{\exp h - 1}{h} = 1.$$

From Combination Theorem for Functions, $$\lim_{h \to 0} \frac {\exp x \left({\exp h - 1}\right)} {h} = \exp x \left({\lim_{x \to 0} \frac {\exp h - 1} {h}}\right).$$.

The result follows.

Alternative proof
We use the definition of the exponential function as the inverse of the natural logarithm function.

Let $$y = \exp x$$.

Then from Derivative of an Inverse Function, we have:

$$D_x \exp x = \frac 1 {D_x \ln x} = \frac 1 {1/y} = y = \exp x$$.

Proof of Corollary 1
Follows directly from the Derivative of Function of Constant Multiple:

$$D_x \left({\exp \left({c x}\right)}\right) = c D_{c x} \left({\exp \left({c x}\right)}\right) = c \exp \left({c x}\right)$$.

Proof of Corollary 2
From the definition, $$a^x = e^{x \ln a}$$.

Thus from Corollary 1, $$D_x \left({a^x}\right) = D_x \left({e^{x \ln a}}\right) = \ln a e^{x \ln a} = a^x \ln a$$.