# Proof of the Derivative of $a^x$

The proof of the derivative of the exponential function $a^x$, such that $a \gt 0$ and $a \ne 1$, is presented. The derivative of a composite function of the form $a^{u(x)}$ is also presented. Examples with their solutions are included.

## Proof of the Derivative of $a^x$

Let $y = a^x$ , such that $a \gt 0$ and $a \ne 1$
Take the $\ln$ of both sides of the equation $y = a^x$ to obtain
$\ln y = \ln a^x \qquad (1)$
Use the property of $\; \ln \;$ given by $\; \ln a^x = x \ln a \;$ in $(1)$ above and write $\ln y = \ln a^x$ as
$\ln y = x \ln a$
Take the derivative, with respect to $x$, of both sides of the above equation.
$\dfrac{d}{dx} (\ln y) = \dfrac{d}{dx} ( x \ln a) \qquad (2)$
Use the chain rule of differentiation to write the left side of $(2)$ as $\dfrac{d}{dx} (\ln y) = \dfrac{1}{y} \dfrac{dy}{dx}$ and the derivative of the right side of $(2)$ is given by $\dfrac{d}{dx} ( x \ln a) = \ln a$ Substitute the two results above in $(2)$ to obtain
$\dfrac{1}{y} \dfrac{dy}{dx} = \ln a$
Multiply both sides of the above by $y$
$y \dfrac{1}{y} \dfrac{dy}{dx} = y \ln a$
Simplify to obtain
$\dfrac{dy}{dx} = y \; \ln a$
Substitute $y$ by $a^x$ to write the derivative of $a^x$ as
$\dfrac{d \left(a^x \right)}{dx} = (\ln a) \; a^x \qquad (I)$

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

We now consider the composite exponential of another function u(x). Use the chain rule of differentiation to write
$\displaystyle \dfrac{d a^{u(x)}}{dx} = \dfrac{d( a^{u})}{du} \dfrac{du}{dx} \qquad (3)$
Use the above result in $(I)$ to write $\dfrac{d( a^{u})}{du} = (\ln a) \; a^u$ and substitute in $(3)$ above to obtain the derivative of the composite function $y = a^{u(x)}$ as
$\displaystyle \dfrac{d \left(a^{u(x)} \right)}{dx} = (\ln a) \; a^u \; \dfrac{du}{dx} \qquad (II)$

Example 1
Find the derivative of the composite exponential functions

1. $f(x) = 2^{-x^4+5x-4}$
2. $g(x) = 3^{\sqrt{x^4+2x}}$
3. $h(x) = 5^{ \left(\dfrac{2x}{3x+2}\right)}$

Solution to Example 1

1. Let $u(x) = -x^4+5x-4$ and therefore $\dfrac{du}{dx} = -4 x^3 + 5$
Apply the rule for the composite exponential function in $(II)$ above
$\displaystyle \dfrac{d}{dx} f(x) = ( \ln 2) 2^u \dfrac{du}{dx} = ( \ln 2) \; 2^{-x^4+5x-4} \times (-4 x^3 + 5 )$
$= ( \ln 2) \; (-4 x^3 + 5 ) \; 2^{-x^4+5x-4}$

2. Let $u(x) = \sqrt{x^4+2x}$ and therefore $\dfrac{du}{dx} = \dfrac{1}{2} \dfrac{4 x^3 + 2}{\sqrt{x^4+2x}} = \dfrac{2 x^3 + 1}{\sqrt{x^4+2x}}$.
Apply the rule of differentiation for the composite exponential function in $(II)$ above
$\displaystyle \dfrac{d}{dx} g(x) = (\ln 3) 3^u \dfrac{du}{dx} = (\ln 3) 3^{\sqrt{x^2+1}} \times \dfrac{2 x^3 + 1}{\sqrt{x^4+2x}}$
$= ( \ln 3) \dfrac{2 x^3 + 1}{\sqrt{x^4+2x}} \; 3^{\sqrt{x^2+1}}$

3. Let $u(x) = \dfrac {2x}{3x+2}$ and therefore $\dfrac{du}{dx} = \dfrac{4}{\left(3x+2\right)^2}$
The use of the rule of differentiation for the composite exponential function obtained in $(II)$ above gives
$\displaystyle \dfrac{d}{dx} h(x) = ( \ln 5 ) \; 5^u \; \dfrac{du}{dx}$
$= ( \ln 5) \; \dfrac{4}{\left(3x+2\right)^2} \; 5^{\left(\frac{2x}{3x+2}\right)}$