Chapter 10: Derivatives of Exponential Functions

Derivatives of Exponential Functions

Introduction

10.1 The Natural Logarithm Base $e$

Before studying the derivatives of exponential functions, we introduce $e$, one of the most important constants in mathematics.

Definition of the Natural Logarithm Base $e$
$$e = \lim_{n \to \infty} \left(1 + \dfrac{1}{n}\right)^n$$

An equivalent definition is:

$$e = \lim_{h \to 0} (1 + h)^{1/h} \qquad (h > -1)$$

$e \approx 2.71828\ldots$ (an irrational number)

Convergence of the sequence (1 + 1/n)^n to e As n increases, the values of the sequence approach e ≈ 2.718 from below. (1 + 1/n)ⁿ Convergence n y e ≈ 2.718 2 3 1 2 3 5 10 50 100 As n → ∞, (1 + 1/n)ⁿ → e
Figure 1: Values of $\left(1 + \dfrac{1}{n}\right)^n$. As $n$ increases, the values approach $e \approx 2.718$ from below. (The horizontal axis uses a logarithmic scale, and the vertical axis is truncated.)
An Important Limit Involving $e$
$$\lim_{h \to 0} \dfrac{e^h - 1}{h} = 1$$
Proof

From $e = \lim_{h \to 0}(1+h)^{1/h}$ and the continuity of the logarithm function, we obtain:

$$\ln e = \lim_{h \to 0} \dfrac{\ln(1+h)}{h}$$

Since $\ln e = 1$, it follows that:

$$\lim_{h \to 0} \dfrac{\ln(1+h)}{h} = 1$$

Let $t = e^h - 1$. Then $h = \ln(1+t)$, and as $h \to 0$, we have $t \to 0$.

$$\lim_{h \to 0} \dfrac{e^h - 1}{h} = \lim_{t \to 0} \dfrac{t}{\ln(1+t)} = \dfrac{1}{\lim_{t \to 0}\dfrac{\ln(1+t)}{t}} = \dfrac{1}{1} = 1$$

10.2 The Derivative of $e^x$

Theorem
$$\dfrac{d}{dx}e^x = e^x$$

The function $e^x$ is special because its derivative is equal to the function itself.

Proof

By the definition of the derivative:

$$\dfrac{d}{dx}e^x = \lim_{h \to 0} \dfrac{e^{x+h} - e^x}{h}$$

Step 1: Use the exponent law $e^{x+h} = e^x \cdot e^h$.

$$= \lim_{h \to 0} \dfrac{e^x \cdot e^h - e^x}{h}$$

Step 2: Factor out $e^x$.

$$= \lim_{h \to 0} e^x \cdot \dfrac{e^h - 1}{h}$$

Step 3: Since $e^x$ does not depend on $h$, it can be taken outside the limit.

$$= e^x \cdot \lim_{h \to 0} \dfrac{e^h - 1}{h}$$

Step 4: Apply the fundamental limit formula $\displaystyle \lim_{h \to 0} \dfrac{e^h - 1}{h} = 1$.

$$= e^x \cdot 1 = e^x$$
The exponential function y = e^x and its tangent lines The graph of y = e^x. The slopes of the tangent lines at x = 0 and x = 1 are e^0 = 1 and e^1 = e, respectively. y = eˣ (the function and its derivative are identical) x y -2 -1 1 2 1 2 3 Special Property Slope of the tangent = function value x=0: slope=1=e⁰ x=1: slope=e≈2.72
Figure 2: The graph of $y = e^x$ and its tangent lines at $x = 0$ and $x = 1$. The slope of each tangent equals the function value at that point: $e^0 = 1$ and $e^1 = e$.
Why Is $e^x$ Special?

The function $e^x$ is special because differentiating it gives the same function again.

Every function satisfying $f'(x) = f(x)$ is of the form $f(x) = Ce^x$, where $C$ is a constant.

10.3 The Derivative of $a^x$ ($a > 0$)

Theorem

For $a > 0$:

$$\dfrac{d}{dx}a^x = a^x \ln a$$
Proof

Rewrite $a^x$ using $e$.

Step 1: Since $a = e^{\ln a}$, we have:

$$a^x = (e^{\ln a})^x = e^{x \ln a}$$

Step 2: Apply the chain rule. Let $u = x \ln a$.

$$\dfrac{d}{dx}a^x = \dfrac{d}{dx}e^{x \ln a} = e^{x \ln a} \cdot \dfrac{d}{dx}(x \ln a)$$

Step 3: Since $\ln a$ is a constant:

$$= e^{x \ln a} \cdot \ln a$$

Step 4: Substitute $a^x$ for $e^{x \ln a}$.

$$= a^x \ln a$$
Example 1: Differentiating $2^x$
$$\dfrac{d}{dx}2^x = 2^x \ln 2 \approx 2^x \cdot 0.693$$
Example 2: Differentiating $10^x$
$$\dfrac{d}{dx}10^x = 10^x \ln 10 \approx 10^x \cdot 2.303$$
Example 3: Differentiating $3^{2x}$

Apply the chain rule. Let $u = 2x$.

$$\dfrac{d}{dx}3^{2x} = 3^{2x} \ln 3 \cdot 2 = 2 \cdot 3^{2x} \ln 3$$

10.4 Combining Exponential Functions with the Chain Rule

Differentiating Composite Exponential Functions

If $g(x)$ is differentiable and $a > 0$, then:

$$\dfrac{d}{dx}e^{g(x)} = e^{g(x)} \cdot g'(x)$$ $$\dfrac{d}{dx}a^{g(x)} = a^{g(x)} \ln a \cdot g'(x)$$
Example 1: Differentiating $e^{3x}$
$$\dfrac{d}{dx}e^{3x} = e^{3x} \cdot 3 = 3e^{3x}$$
Example 2: Differentiating $e^{x^2}$
$$\dfrac{d}{dx}e^{x^2} = e^{x^2} \cdot 2x = 2xe^{x^2}$$
Example 3: Differentiating $e^{\sin x}$
$$\dfrac{d}{dx}e^{\sin x} = e^{\sin x} \cdot \cos x$$
Example 4: Differentiating $e^{-x^2}$

Since $g(x) = -x^2$ and $g'(x) = -2x$:

$$\dfrac{d}{dx}e^{-x^2} = e^{-x^2} \cdot (-2x) = -2xe^{-x^2}$$

This derivative is important in the study of the Gaussian function.

The Gaussian Function and Its Derivative The Gaussian function e^(-x^2) reaches its maximum at x = 0. Its derivative, -2x e^(-x^2), is an odd function that passes through the origin. Gaussian Function y = e⁻ˣ² and Its Derivative x 1 y = e⁻ˣ² y' = -2xe⁻ˣ² Maximum at x=0; derivative passes through 0
Figure 3: The Gaussian function $y = e^{-x^2}$ (blue) and its derivative $y = -2xe^{-x^2}$ (red dashed line). The derivative is zero at $x = 0$ and has extrema at $x = \pm\dfrac{1}{\sqrt{2}}$.

10.5 Combining Exponential Functions with Products and Quotients

Example 1: Differentiating $xe^x$

Apply the product rule:

$$\dfrac{d}{dx}(xe^x) = 1 \cdot e^x + x \cdot e^x = e^x(1 + x) = (1+x)e^x$$
Example 2: Differentiating $x^2 e^{-x}$

Apply the product rule:

$$\dfrac{d}{dx}(x^2 e^{-x}) = 2x \cdot e^{-x} + x^2 \cdot (-e^{-x})$$ $$= e^{-x}(2x - x^2) = x(2-x)e^{-x}$$
Example 3: Differentiating $\dfrac{e^x}{x}$

For $x \neq 0$, apply the quotient rule:

$$\dfrac{d}{dx}\dfrac{e^x}{x} = \dfrac{e^x \cdot x - e^x \cdot 1}{x^2} = \dfrac{e^x(x-1)}{x^2}$$
Example 4: Differentiating $\dfrac{e^x - e^{-x}}{2}$ (the hyperbolic sine function $\sinh x$)
$$\dfrac{d}{dx}\dfrac{e^x - e^{-x}}{2} = \dfrac{e^x - (-1)e^{-x}}{2} = \dfrac{e^x + e^{-x}}{2} = \cosh x$$

10.6 Summary of Derivative Formulas for Exponential Functions

Derivative Formulas for Exponential Functions
Function Derivative
$e^x$ $e^x$
$e^{g(x)}$ $e^{g(x)} \cdot g'(x)$
$a^x$ ($a > 0$) $a^x \ln a$
$a^{g(x)}$ ($a > 0$) $a^{g(x)} \ln a \cdot g'(x)$

Frequently Asked Questions

Why is $(e^x)' = e^x$?

By the definition of the derivative, $(e^x)' = e^x \cdot \lim_{h \to 0}\dfrac{e^h - 1}{h}$. From the definition of $e$, we know that $\lim_{h \to 0}\dfrac{e^h - 1}{h} = 1$. Therefore, $(e^x)' = e^x$. This is the distinctive property of $e^x$: its derivative is equal to the function itself.

What happens when the base is not $e$? What is $(a^x)'$?

Assume $a > 0$. Rewriting $a^x$ as $e^{x \ln a}$ and applying the chain rule gives $(a^x)' = a^x \ln a$. When $a = e$, we have $\ln e = 1$, so $(e^x)' = e^x$. If $a > 1$, then $\ln a > 0$ and the function is increasing; if $0 < a < 1$, then $\ln a < 0$ and the function is decreasing.

How is the fact that the derivative of $e^x$ equals itself applied?

The functions satisfying $f'(x) = kf(x)$, meaning that the rate of change is proportional to the function itself, are given by $f(x) = Ce^{kx}$, where $C$ is a constant. This relationship appears in population growth, radioactive decay, charging and discharging in RC circuits, and compound interest. It is an important model of natural growth and decay.

Practice Problems

Exercise 1: Basic

Differentiate the following functions.

  1. $e^{5x}$
  2. $e^{-2x}$
  3. $e^{x^3}$
  4. $3^x$
Exercise 2: Combined Rules

Differentiate the following functions.

  1. $x^3 e^x$
  2. $e^x \sin x$
  3. $\dfrac{e^x}{x^2}$
  4. $e^{x^2 + 2x}$
Exercise 3

Consider $f(x) = xe^{-x}$.

  1. Find $f'(x)$.
  2. Find the value of $x$ for which $f'(x) = 0$.
  3. Find the extremum of $f(x)$.
Show Answers

Answers to Exercise 1

  1. $5e^{5x}$
  2. $-2e^{-2x}$
  3. $e^{x^3} \cdot 3x^2 = 3x^2 e^{x^3}$
  4. $3^x \ln 3$

Answers to Exercise 2

  1. $3x^2 e^x + x^3 e^x = x^2(3+x)e^x$
  2. $e^x \sin x + e^x \cos x = e^x(\sin x + \cos x)$
  3. $\dfrac{e^x \cdot x^2 - e^x \cdot 2x}{x^4} = \dfrac{e^x(x-2)}{x^3}$
  4. $e^{x^2+2x} \cdot (2x+2) = 2(x+1)e^{x^2+2x}$

Answers to Exercise 3

  1. $f'(x) = 1 \cdot e^{-x} + x \cdot (-e^{-x}) = e^{-x}(1-x) = (1-x)e^{-x}$
  2. $f'(x) = 0 \Leftrightarrow 1-x = 0 \Leftrightarrow x = 1$
  3. For $x < 1$, $f'(x) > 0$ (increasing); for $x > 1$, $f'(x) < 0$ (decreasing).
    Therefore, $f(x)$ has a local maximum at $x = 1$.
    Maximum value: $f(1) = 1 \cdot e^{-1} = \dfrac{1}{e}$