Derivative of ex: Step-by-Step Proof and Examples

This tutorial presents the derivative of the natural exponential function \( e^x \) using the limit definition. We also cover the derivative of composite exponential functions of the form \( e^{u(x)} \), including detailed examples and solutions.

1. Proof of the Derivative of ex Using the Limit Definition

The definition of the derivative of a function \( f(x) \) is:

\[ f'(x) = \lim_{h \to 0} \frac{f(x+h) - f(x)}{h} \]

Let \( f(x) = e^x \). Then:

\[ \begin{aligned} f'(x) &= \lim_{h \to 0} \frac{e^{x+h} - e^x}{h} \\ &= \lim_{h \to 0} \frac{e^x e^h - e^x}{h} \quad \text{(using } e^{x+h} = e^x e^h \text{)} \\ &= \lim_{h \to 0} \frac{e^x (e^h - 1)}{h} \\ &= e^x \lim_{h \to 0} \frac{e^h - 1}{h} \end{aligned} \]

We now calculate the limit \( \lim_{h \to 0} \frac{e^h - 1}{h} \). Let \( y = e^h - 1 \), so that \( \lim_{h \to 0} y = 0 \). Then:

\[ e^h = y + 1 \implies h = \ln(y + 1) \] \[ \lim_{h \to 0} \frac{e^h - 1}{h} = \lim_{y \to 0} \frac{y}{\ln(y+1)} \] \[ = \lim_{y \to 0} \frac{1}{\frac{1}{y} \ln(y+1)} = \lim_{y \to 0} \frac{1}{\ln((y+1)^{1/y})} \]

Using the definition of the Euler constant \( e \):

\[ e = \lim_{m \to 0} (1 + m)^{1/m} \implies \lim_{y \to 0} (y+1)^{1/y} = e \] \[ \lim_{h \to 0} \frac{e^h - 1}{h} = \frac{1}{\ln e} = 1 \] \[ \boxed{f'(x) = e^x} \]

Conclusion: Any function of the form \( f(x) = k e^x \), where \( k \) is a constant, has the same derivative multiplied by \( k \).

2. Derivative of the Composite Function y = eu(x)

For a function \( u(x) \), we use the chain rule:

\[ \frac{d}{dx} e^{u(x)} = \frac{d}{du} e^{u} \cdot \frac{du}{dx} = e^{u(x)} \frac{du}{dx} \] \[ \boxed{\frac{d}{dx} e^{u(x)} = e^{u(x)} \frac{du}{dx}} \] ---

Example 1

Find the derivatives of:

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

Solution

  1. \[ u(x) = x^3 - 2x + 3 \implies \frac{du}{dx} = 3x^2 - 2 \] \[ \frac{d}{dx} f(x) = e^{u(x)} \frac{du}{dx} = (3x^2 - 2) e^{x^3 - 2x + 3} \]
  2. \[ u(x) = \sqrt{x^2 + 1} \implies \frac{du}{dx} = \frac{x}{\sqrt{x^2 + 1}} \] \[ \frac{d}{dx} g(x) = e^{u(x)} \frac{du}{dx} = \frac{x}{\sqrt{x^2 + 1}} \, e^{\sqrt{x^2 + 1}} \]
  3. \[ u(x) = \frac{x}{x-2} \implies \frac{du}{dx} = -\frac{2}{(x-2)^2} \] \[ \frac{d}{dx} h(x) = e^{u(x)} \frac{du}{dx} = -\frac{2}{(x-2)^2} \, e^{\frac{x}{x-2}} \]

3. Additional References