# How do you use the definition of a derivative to find the derivative of f(x) =e^x?

May 7, 2016

Using $\frac{\mathrm{df}}{\mathrm{dx}} = {\lim}_{\Delta x \to 0} \frac{f \left(x + \Delta x\right) - f \left(x\right)}{\Delta x}$

#### Explanation:

$\frac{d}{\mathrm{dx}} {e}^{x} = {\lim}_{\Delta x \to 0} \frac{{e}^{x + \Delta x} - {e}^{x}}{\Delta x} = {\lim}_{\Delta x \to 0} {e}^{x} \left(\frac{{e}^{\Delta x} - 1}{\Delta x}\right)$

As $\Delta x \to 0$, ${e}^{\Delta x} \to {1}^{+}$

The ratio approximates to $1$ hence

The derivative of ${e}^{x}$ is ${e}^{x}$.

May 7, 2016

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

${\lim}_{h \to 0} \frac{f \left(x + h\right) - f \left(x\right)}{h} = \frac{{e}^{x + h} - {e}^{x}}{h}$

from power rules

$= {\lim}_{h \to 0} \frac{{e}^{x} {e}^{h} - {e}^{x}}{h}$

you can use taylor series to have an excellent approximation of ${e}^{h}$ when $h \to 0$ which is

${e}^{h} = h + 1$ when $h \to 0$

$= {\lim}_{h \to 0} \frac{{e}^{x} \left(h + 1\right) - {e}^{x}}{h} = {\lim}_{h \to 0} \frac{h {e}^{x} + {e}^{x} - {e}^{x}}{h}$

which is ${\lim}_{h \to 0} {e}^{x} = {e}^{x}$

at your level when you're blocked, think to use taylor series, it can be very useful.

Maybe you will tell me "Yeah you're right but you used taylor series, which are created by using formula with derivate of ${e}^{x}$ so you use what you need to proove in the proof"

${e}^{x}$ is created by his taylor series not the opposite. You can have the series without the taylor formula, so the proof is not perfect but good.