This lesson presents a complete proof of the derivative of the exponential function \( a^x \), where \( a > 0 \) and \( a \neq 1 \). We also derive the formula for the composite exponential function \( a^{u(x)} \) using the chain rule and provide worked examples.
Proof of the Derivative of \( a^x \)
Let:
\[ y = a^x, \qquad a > 0, \; a \neq 1 \]Take the natural logarithm of both sides:
\[ \ln y = \ln(a^x) \]Using the logarithmic identity \( \ln(a^x) = x \ln a \), we obtain:
\[ \ln y = x \ln a \]Differentiate both sides with respect to \( x \):
\[ \frac{d}{dx}(\ln y) = \frac{d}{dx}(x \ln a) \]Apply the chain rule to the left side:
\[ \frac{1}{y} \frac{dy}{dx} = \ln a \]Multiply both sides by \( y \):
\[ \frac{dy}{dx} = y \ln a \]Substitute \( y = a^x \):
Formula:
\[ \frac{d}{dx}\left(a^x\right) = (\ln a)\,a^x \]
Derivative of the Composite Function \( y = a^{u(x)} \)
Let \( u = u(x) \). By the chain rule:
\[ \frac{d}{dx}\left(a^{u(x)}\right) = \frac{d(a^u)}{du} \frac{du}{dx} \]Using the previous result:
\[ \frac{d(a^u)}{du} = (\ln a)a^u \]Therefore:
Chain Rule Formula:
\[ \frac{d}{dx}\left(a^{u(x)}\right) = (\ln a)\,a^{u(x)}\,\frac{du}{dx} \]
Examples with Solutions
Worked Examples: Differentiating Composite Exponential Functions
Find the derivatives of the following functions:
- \( f(x) = 2^{-x^4+5x-4} \)
- \( g(x) = 3^{\sqrt{x^4+2x}} \)
- \( h(x) = 5^{\frac{2x}{3x+2}} \)
Solutions:
-
Let \( u(x) = -x^4 + 5x - 4 \). Then \( \frac{du}{dx} = -4x^3 + 5 \).
Applying the formula: \[ f'(x) = (\ln 2)\,2^{-x^4+5x-4}(-4x^3+5) \] -
Let \( u(x) = \sqrt{x^4+2x} \). Then \( \frac{du}{dx} = \frac{4x^3+2}{2\sqrt{x^4+2x}} = \frac{2x^3+1}{\sqrt{x^4+2x}} \).
Applying the formula: \[ g'(x) = (\ln 3)\,3^{\sqrt{x^4+2x}}\frac{2x^3+1}{\sqrt{x^4+2x}} \] - Let \( u(x) = \frac{2x}{3x+2} \). Using the quotient rule: \[ \frac{du}{dx} = \frac{2(3x+2) - 2x(3)}{(3x+2)^2} = \frac{6x+4-6x}{(3x+2)^2} = \frac{4}{(3x+2)^2} \] Applying the formula: \[ h'(x) = (\ln 5)\,5^{\frac{2x}{3x+2}}\frac{4}{(3x+2)^2} \]