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
\( f(x) = 2^{-x^4+5x-4} \)
\( g(x) = 3^{\sqrt{x^4+2x}} \)
\( h(x) = 5^{ \left(\dfrac{2x}{3x+2}\right)} \)
Solution to Example 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
\( \displaystyle \dfrac{d}{dx} f(x) = ( \ln 2) 2^u \dfrac{du}{dx} = ( \ln 2) \; 2^{-x^4+5x-4} \times (-4 x^3 + 5 ) \)
\( = ( \ln 2) \; (-4 x^3 + 5 ) \; 2^{-x^4+5x-4} \)
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
\( \displaystyle \dfrac{d}{dx} g(x) = (\ln 3) 3^u \dfrac{du}{dx} = (\ln 3) 3^{\sqrt{x^2+1}} \times \dfrac{2 x^3 + 1}{\sqrt{x^4+2x}} \)
\( = ( \ln 3) \dfrac{2 x^3 + 1}{\sqrt{x^4+2x}} \; 3^{\sqrt{x^2+1}} \)
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
\( \displaystyle \dfrac{d}{dx} h(x) = ( \ln 5 ) \; 5^u \; \dfrac{du}{dx} \)
\( = ( \ln 5) \; \dfrac{4}{\left(3x+2\right)^2} \; 5^{\left(\frac{2x}{3x+2}\right)} \)