Calculus of the exponential function

Differentiation of exponential functions

The derivatives of $f(x) = e^x$ and $g(x) = a^x$ have a minor difference. Let's differentiate them using the first principle.

Derivative of $f(x) = e^x$

From the first principle:

$f'(x) = \lim_{h \to 0} \frac{f(x + h) - f(x)}{h}$

$f'(x) = \lim_{h \to 0} \frac{e^{x+h} - e^x}{h}$

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

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

As $h \to 0$, $\frac{e^h - 1}{h} \approx 1$ (see Table).

Therefore, $f'(x) = e^x$.

Derivative of $g(x) = a^x$

From the first principle:

$g'(x) = \lim_{h \to 0} \frac{g(x + h) - g(x)}{h}$

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

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

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

As $h \to 0$, $\frac{a^h - 1}{h} \approx k$, where $k$ is a constant depending on $a$.

Table: values of $\frac{a^h - 1}{h}$ as $h \to 0$

$h \to 0$	0.1	0.01	0.001	0.0001	0.00001
$\frac{e^h - 1}{h}$	1.05171	1.00501	1.00050	1.00005	1.000005
$\frac{2^h - 1}{h}$	0.7177	0.6956	0.6934	0.69317	0.69315
$\frac{4^h - 1}{h}$	1.48698	1.39595	1.3873	1.3864	1.3863
$\frac{10^h - 1}{h}$	2.5893	2.3293	2.3052	2.3029	2.3026

The derivative of $f(x) = e^x$ is $\frac{dy}{dx} = e^x$, and for $g(x) = a^x$ is $g'(x) = ka^x$.

Derivative of $f(x) = e^{g(x)}$

To differentiate $y = e^{g(x)}$ (where $g(x)$ is a function of $x$), use the chain rule:

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

Example: derivatives of exponential functions

Find the derivative of:

1. $f(x) = e^{2x+3}$
2. $g(x) = e^{1-x^2}$
3. $h(x) = e^{3x}$
4. $g(x) = e^{3-2x}$

Solution:

1. $f'(x) = \frac{d}{dx}(e^{2x+3}) = e^{2x+3} \cdot \frac{d}{dx}(2x + 3) = 2e^{2x+3}$
2. $g'(x) = \frac{d}{dx}(e^{1-x^2}) = e^{1-x^2} \cdot \frac{d}{dx}(1 - x^2) = -2xe^{1-x^2}$
3. $h'(x) = \frac{d}{dx}(e^{3x}) = e^{3x} \cdot \frac{d}{dx}(3x) = 3e^{3x}$
4. $g'(x) = \frac{d}{dx}(e^{3-2x}) = e^{3-2x} \cdot \frac{d}{dx}(3 - 2x) = -2e^{3-2x}$