$ \frac{\text{d}}{\text{d}x}f(x) = \lim_{h \rightarrow 0}\frac{f(x+ h) - f(x) }{h}$
$\frac{\text{d}}{\text{d}x}f(x) = \lim_{h \rightarrow 0}\frac{e^{(x + h)} - e^{x} } {h}$
= $\lim_{h \rightarrow 0}\frac{e^{x} .e^{h} -e^{x} } {h}$
= $\lim_{h \rightarrow 0}\frac{e^{x}(e^{h} - 1) } {h}$
= $\lim_{h \rightarrow 0}e^{x} .\lim_{h \rightarrow 0}\frac{e^{h} - 1}{h}$
$ \frac{\text{d}}{\text{d}x}(e^{x}) = e^{x}$. 1 ( Since $\lim_{h \rightarrow 0}\frac{e^{h} - 1}{h}$ = 1 )
$ \frac{\text{d}}{\text{d}x}(e^{x}) = e^{x}$
Find the derivative of the following.
Example 1: f(x) = $e^{-x}$
Solution : f(x) = $e^{-x}$
Here we will use the derivative of exponential function and chain rule,
Let u = - x
$\frac{\text{d}u}{\text{d}x} = -1 $
f(x) = $e^{u}$
$\frac{\text{d}}{\text{d}x}(e^{u}) = e^{u}. \frac{\text{d}u}{\text{d}x}$
$\frac{\text{d}}{\text{d}x}(e^{-x}) = e^{-x}.(-1) $
$\frac{\text{d}}{\text{d}x}(e^{-x}) = - e^{-x} $
Example 2: f(x) = $e^{3x}$
Solution : f(x) = $e^{3x}$
Here we will use the derivative of exponential function and chain rule,
Let u = 3x
$\frac{\text{d}u}{\text{d}x} = 3 $
f(x) = $e^{u}$
$\frac{\text{d}}{\text{d}x}(e^{u}) = e^{u}. \frac{\text{d}u}{\text{d}x}$
$\frac{\text{d}}{\text{d}x}(e^{3x}) = e^{3x}.(3) $
$\frac{\text{d}}{\text{d}x}(e^{3x}) = 3 e^{3x} $
Example 3: f(x) = $e^{ax + b}$
Solution : f(x) = $e^{ax + b}$
Here we will use the derivative of exponential function and chain rule,
Let u = ax + b
$\frac{\text{d}u}{\text{d}x} = a $
f(x) = $e^{u}$
$\frac{\text{d}}{\text{d}x}(e^{u}) = e^{u}. \frac{\text{d}u}{\text{d}x}$
$\frac{\text{d}}{\text{d}x}(e^{ax + b}) = e^{ax + b}.(a) $
$\frac{\text{d}}{\text{d}x}(e^{ax + b}) = a e^{ax + b} $
Example 4: f(x) = $x^{2}.e^{x}$
Solution : f(x) = $x^{2}.e^{x}$
Here we will use the derivative of exponential function and product rule
Let u = $x^{2}$
$\frac{\text{d}u}{\text{d}x} = 2x $
v = $e^{x}$
$\frac{\text{d}v}{\text{d}x} = e^{x} $
Product Rule :$ \frac{\text{d}}{\text{d}x}[u.v] = v.\frac{\text{d}u}{\text{d}x} + u.\frac{\text{d}v}{\text{d}x}$
$ \frac{\text{d}}{\text{d}x}[x^{2}.e^{x}] = 2x.e^{x} + x^{2}.e^{x} $