Proof of the Derivative of the Exponential Function \( a^x \)

This lesson presents a complete proof of the derivative of the exponential function \( a^x \), where \( a>0 \) and \( a\neq1 \). 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

\[ y=a^x , \qquad a>0,\; a\neq1 \]

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 \( y=a^x \):

\[ \boxed{\frac{d}{dx}\left(a^x\right)=(\ln a)\,a^x} \]

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,

\[ \boxed{\frac{d}{dx}\left(a^{u(x)}\right)=(\ln a)\,a^{u(x)}\,\frac{du}{dx}} \]

Example: Differentiating Composite Exponential Functions

Find the derivatives:

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

Solutions

  1. Let \[ u(x)=-x^4+5x-4 \] \[ \frac{du}{dx}=-4x^3+5 \] \[ f'(x)=(\ln2)\,2^{-x^4+5x-4}(-4x^3+5) \]
  2. Let \[ u(x)=\sqrt{x^4+2x} \] \[ \frac{du}{dx}=\frac{2x^3+1}{\sqrt{x^4+2x}} \] \[ g'(x)=(\ln3)\,3^{\sqrt{x^4+2x}}\frac{2x^3+1}{\sqrt{x^4+2x}} \]
  3. Let \[ u(x)=\frac{2x}{3x+2} \] \[ \frac{du}{dx}=\frac{4}{(3x+2)^2} \] \[ h'(x)=(\ln5)\,5^{\frac{2x}{3x+2}}\frac{4}{(3x+2)^2} \]

Further Reading