 # Proof of Derivative of $e^x$

The proof of the derivative of the natural exponential $e^x$ is presented using the limit definition of the derivative. The derivative of a composite function of the form $e^{u(x)}$ is also presented including examples with their solutions.

## Proof of the Derivative of $e^x$ Using the Definition of the Derivative

The definition of the derivative $f'$ of a function $f$ is given by the limit $f'(x) = \lim_{h \to 0} \dfrac{f(x+h)-f(x)}{h}$ Let $f(x) = e^x$ and write the derivative of $e^x$ as follows
$f'(x) = \lim_{h \to 0} \dfrac{e^{x+h}-e^x}{h}$
Use the formula $e^{x+h} = e^x e^h$ to rewrite the derivative of $e^x$ as
$f'(x) = \lim_{h \to 0} \dfrac{e^x e^h - e^x}{h}$
Factor $e^x$ out in the numerator
$f'(x) = \lim_{h \to 0} \dfrac{e^x (e^h - 1)}{h}$
Since $e^x$ does not depend on $h$, the above may be rewriten as
$f'(x) = e^x \lim_{h \to 0} \dfrac{ e^h - 1}{h}$
We now need to find the limit $\lim_{h \to 0} \dfrac{ e^h - 1}{h}$.
Let $y = e^h - 1$
and note that
$\lim_{h \to 0} y = 0$
We now express h in terms of y
$e^h = y + 1$
take the ln of both sides
$ln(e^h) = ln(y + 1)$
simplify the left side using the rule: $ln(e^x) = x$
$h = \ln(y + 1)$
With the above substitution, we can write
$\lim_{h \to 0} \dfrac{ e^h - 1}{h} = \lim_{y \to 0} \dfrac{ y}{\ln(y+1)}$
Rewrite the term on the right as follows
$= \lim_{y \to 0} \dfrac{ 1}{\dfrac{1}{y}\ln(y+1)}$
Use power rule of logarithms ( $a \ln y = \ln y^a$ ) to rewrite the above limit as
$= \lim_{y \to 0} \dfrac{ 1}{\ln(y+1)^{\dfrac{1}{y}}}$
Use the limit rule of a quotient and limit of a composite function to rewrite the above as
$= \dfrac{ \lim_{y \to 0} 1}{\lim_{y \to 0} \ln(y+1)^{\dfrac{1}{y}}} = \dfrac{1}{\ln \left(\lim_{y \to 0} (y+1)^{\dfrac{1}{y}} \right)}$
One of the definitions of the
Euler Constant e is
$e = \lim_{m \to 0} ( 1 + m) ^{\dfrac{1}{m}}$
Hence the limit we are looking for is given by
$\lim_{h \to 0} \dfrac{ e^h - 1}{h} = \dfrac{1}{\ln \left(\lim_{y \to 0} (y+1)^{\dfrac{1}{y}} \right)} = \dfrac{1}{\ln e} = 1$
which finally gives
$f'(x) = e^x \lim_{h \to 0} \dfrac{ e^h - 1}{h} = e^x \times 1 = e^x$
Conclusion:
$\dfrac{d}{dx} e^x = e^x$
Note that any function of the form $f(x) = k e^x$, where k is a constant, is equal to its derivative.

## Derivative of the Composite Function $y = e^{u(x)}$

We now consider the composite exponential of another function u(x). Use the chain rule of differentiation to write
$\displaystyle \dfrac{d}{dx} e^{u(x)} = \dfrac{d}{du} e^{u(x)} \dfrac{d}{dx} u$
Simplify
$= e^u \dfrac{d}{dx} u$
Conclusion
$\displaystyle \dfrac{d}{dx} e^{u(x)} = e^u \dfrac{d}{dx} u$

Example 1
Find the derivative of the composite exponential functions

1. $f(x) = e^{x^3-2x+3}$
2. $g(x) = e^{\sqrt{x^2+1}}$
3. $h(x) = e^{ \left(\dfrac{x}{x-2}\right)}$

Solution to Example 1

1. Let $u(x) = x^3-2x+3$ and therefore $\dfrac{d}{dx} u = \dfrac{d}{dx} (x^3-2x+3) = 3x^2-2$
Apply the rule for the composite exponential function found above
$\displaystyle \dfrac{d}{dx} f(x) = e^u \dfrac{d}{dx} u = e^{x^3-2x+3} \times (3x^2-2)$
$= (3x^2-2) e^{x^3-2x+3}$

2. Let $u(x) = \sqrt{x^2+1}$ and therefore $\dfrac{d}{dx} u = \dfrac{d}{dx} \sqrt{x^2+1} = \dfrac{x}{\sqrt{x^2+1}}$.
Apply the above rule of differentiation for the composite exponential function
$\displaystyle \dfrac{d}{dx} g(x) = e^u \dfrac{d}{dx} u = e^{\sqrt{x^2+1}} \times \dfrac{x}{\sqrt{x^2+1}}$
$= \dfrac{x}{\sqrt{x^2+1}} \; e^{\sqrt{x^2+1}}$

3. Let $u(x) = \dfrac{x}{x-2}$ and therefore $\dfrac{d}{dx} u = -\dfrac{2}{\left(x-2\right)^2}$
Apply the rule of differentiation for the composite exponential function obtained above
$\displaystyle \dfrac{d}{dx} h(x) = e^u \dfrac{d}{dx} u = e^{ \left(\dfrac{x}{x-2}\right)} \times ( -\dfrac{2}{\left(x-2\right)^2} )$
$= -\dfrac{2}{\left(x-2\right)^2} \; e^{\left(\dfrac{x}{x-2}\right)}$