Differentiating Exponential Functions with Base e

Key Questions

• $y ' = - {e}^{\frac{1}{x}} / \left({x}^{2}\right)$

Explanation :

Using Chain Rule,

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

then, $y ' = {e}^{f} \left(x\right) \cdot f ' \left(x\right)$

Similarly following for the $y = {e}^{\frac{1}{x}}$

$y ' = {e}^{\frac{1}{x}} \cdot \left(\frac{1}{x}\right) '$

$y ' = {e}^{\frac{1}{x}} \cdot \left(- \frac{1}{x} ^ 2\right)$

$y ' = - {e}^{\frac{1}{x}} / \left({x}^{2}\right)$

• This is one of the favorite function to take the derivatives of.
$y ' = {e}^{x}$

If you wish to find this derivative by the limit definition, then here is how we find it. First, we have to know the following property of $e$:
${\lim}_{h \to 0} \frac{{e}^{h} - 1}{h} = 1$.
(Note: This means that the slope of $y = {e}^{x}$ at $x = 0$ is $1$.)

By the limit definition of the derivative, we have
y'=lim_{h to 0}{e^{x+h}-e^x}/h =lim_{h to 0}{e^x cdot e^h-e^x}/h
by factoring out ${e}^{x}$,
$= {\lim}_{h \to 0} \frac{{e}^{x} \left({e}^{h} - 1\right)}{h} = {e}^{x} {\lim}_{h \to 0} \frac{{e}^{h} - 1}{h}$
by the property of $e$ mentioned above,
$= {e}^{x} \cdot 1 = {e}^{x}$

Hence, the derivative of ${e}^{x}$ is itself.