# Differentiate e^(ax) using first principles?

Dec 12, 2016

$f ' \left(x\right) = a {e}^{a x}$

#### Explanation:

The definition of the derivative of $y = f \left(x\right)$ is

$f ' \left(x\right) = {\lim}_{h \rightarrow 0} \frac{f \left(x + h\right) - f \left(x\right)}{h}$

So Let $f \left(x\right) = {e}^{a x}$ then;

$\setminus \setminus \setminus \setminus \setminus f \left(x + h\right) = {e}^{a \left(x + h\right)}$
$\therefore f \left(x + h\right) = {e}^{a x + a h}$
$\therefore f \left(x + h\right) = {e}^{a x} {e}^{a h}$

And so the derivative of $y = f \left(x\right)$ is given by:

$\setminus \setminus \setminus \setminus \setminus f ' \left(x\right) = {\lim}_{h \rightarrow 0} \frac{\left({e}^{a x} {e}^{a h}\right) - \left({e}^{a x}\right)}{h}$
$\therefore f ' \left(x\right) = {\lim}_{h \rightarrow 0} \frac{{e}^{a x} \left({e}^{a h} - 1\right)}{h}$
$\therefore f ' \left(x\right) = {e}^{a x} {\lim}_{h \rightarrow 0} \frac{\left({e}^{a h} - 1\right)}{h}$
$\therefore f ' \left(x\right) = {e}^{a x} {\lim}_{h \rightarrow 0} \frac{a \left({e}^{a h} - 1\right)}{a h}$
$\therefore f ' \left(x\right) = a {e}^{a x} {\lim}_{h \rightarrow 0} \frac{\left({e}^{a h} - 1\right)}{a h}$

And the clever readers who already know the answer can hopefully spot that we are almost there if we can show that

${\lim}_{h \rightarrow 0} \frac{\left({e}^{a h} - 1\right)}{a h} = 1$

Now depending upon how you have defined $e$ and the level of calculus that you are at, then this limit can be show to be true. So I wont prove the limit, but just accept it. Wikipedia explains some of the definitions for e using the limit definition.

Once the limit has been established, then the result is evident giving:

$f ' \left(x\right) = a {e}^{a x}$