Derivatives of Exponential Functions

∫ Derivatives of Exponential & Logarithmic Functions

Part 1 of 7 — The Natural Exponential Function

🔑 Key Concept: The function $e^x$ is its own derivative: $\frac{d}{dx}[e^x] = e^x$. This remarkable property makes $e$ the most natural base for calculus.

📖 The Derivative of $e^x$

$\boxed{\frac{d}{dx}[e^x] = e^x}$

This is the ONLY function equal to its own derivative (up to constant multiples).

With Chain Rule

$\frac{d}{dx}[e^{g(x)}] = e^{g(x)} \cdot g'(x)$

Why $e$?

$e \approx 2.71828$ is the base for which $\lim_{h \to 0}\frac{e^h - 1}{h} = 1$.

The limit definition of derivative gives: $\frac{d}{dx}[e^x] = \lim_{h \to 0}\frac{e^{x+h} - e^x}{h} = e^x \lim_{h \to 0}\frac{e^h - 1}{h} = e^x \cdot 1 = e^x$

General Exponential Functions: $a^x$

$\boxed{\frac{d}{dx}[a^x] = a^x \ln a}$

Derivation

Write $a^x = e^{x \ln a}$, then: $\frac{d}{dx}[a^x] = \frac{d}{dx}[e^{x\ln a}] = e^{x\ln a} \cdot \ln a = a^x \ln a$

Note: When $a = e$: $\frac{d}{dx}[e^x] = e^x \ln e = e^x \cdot 1 = e^x$ ✓

AP Tip

There is no formula sheet on the AP Calculus exam, so memorize $\frac{d}{dx}[a^x] = a^x \ln a$!

Exponential Derivatives 🎯

📌 Product & Quotient Rules with $e^x$

Products

$\frac{d}{dx}[xe^x] = e^x + xe^x = e^x(1 + x)$

$\frac{d}{dx}[x^2 e^{-x}] = 2xe^{-x} + x^2(-e^{-x}) = xe^{-x}(2 - x)$

$\frac{d}{dx}[e^x \sin x] = e^x \sin x + e^x \cos x = e^x(\sin x + \cos x)$

Quotients

$\frac{d}{dx}\left[\frac{e^x}{x}\right] = \frac{xe^x - e^x}{x^2} = \frac{e^x(x-1)}{x^2}$

Factor $e^{g(x)}$!

Since $e^{g(x)} > 0$ always, factoring it out simplifies analysis:

$f(x) = x^2e^{-x}$ has $f'(x) = xe^{-x}(2-x)$

Critical points: $x = 0$ and $x = 2$ (not from $e^{-x}$ since it's never zero!)

Exponential Derivative Practice 🧮

1) $\frac{d}{dx}[e^{3x}]$ at $x = 0$: (integer)

2) $\frac{d}{dx}[2^x]$ at $x = 0$: (enter as ln2)

3) $\frac{d}{dx}[xe^x]$ at $x = 1$: (enter as 2e)

Identify Derivatives 🔽

Exit Quiz ✅

∫ Derivatives of Exp & Log — Part 2

Part 2 of 7 — Natural Logarithm Derivatives

📖 The Derivative of $\ln x$

$\boxed{\frac{d}{dx}[\ln x] = \frac{1}{x}, \quad x > 0}$

With the chain rule:

$\frac{d}{dx}[\ln(g(x))] = \frac{g'(x)}{g(x)}$

Why $\ln|x|$?

$\frac{d}{dx}[\ln|x|] = \frac{1}{x}$ for all $x \neq 0$.

This extends the domain to negative $x$, which is important for integration.

General Logarithmic Derivatives

$\boxed{\frac{d}{dx}[\log_a x] = \frac{1}{x \ln a}}$

Since $\log_a x = \frac{\ln x}{\ln a}$:

$\frac{d}{dx}[\log_a x] = \frac{1}{\ln a} \cdot \frac{1}{x} = \frac{1}{x \ln a}$

AP Note: $\log_{10}$ and $\log_2$ rarely appear on the AP exam. Focus on $\ln x$.

Logarithmic Derivatives 🎯

📌 Simplify with Log Properties First

When the inside of a logarithm is a product, quotient, or power, use log properties to split it before differentiating. Each piece then needs only the basic $\frac{g'}{g}$ rule.

Example: $f(x) = \ln\left(\frac{x^2\sqrt{x+1}}{(2x-1)^3}\right)$

Step 1. Split: $f(x) = 2\ln x + \frac{1}{2}\ln(x+1) - 3\ln(2x-1)$

Step 2. Differentiate each term:

$f'(x) = \frac{2}{x} + \frac{1}{2(x+1)} - \frac{6}{2x-1}$

The last term needs the chain rule: $\frac{d}{dx}\ln(2x-1) = \frac{2}{2x-1}$, and $3 \cdot 2 = 6$.

Example: $f(x) = \ln(x\,e^{3x})$

$f(x) = \ln x + 3x \quad \Longrightarrow \quad f'(x) = \frac{1}{x} + 3$

AP Tip: Without the split, the first example needs the quotient rule inside the chain rule. Splitting first is faster and avoids algebra mistakes.

Logarithmic Derivative Practice 🧮

1) $\frac{d}{dx}[\ln(5x)]$ at $x = 1$: (integer)

2) $\frac{d}{dx}[\ln(x^2+1)]$ at $x = 0$: (integer)

3) $\frac{d}{dx}[x\ln x]$ at $x = 1$: (integer)

Log Rules in Derivatives 🔽

Exit Quiz ✅

∫ Derivatives of Exp & Log — Part 3

Part 3 of 7 — Combining with Product, Quotient, Chain Rules

📖 Product Rule Combinations

Pattern: $f(x) = p(x) \cdot e^{q(x)}$

$f'(x) = p'(x)e^{q(x)} + p(x) \cdot q'(x)e^{q(x)} = e^{q(x)}[p'(x) + p(x)q'(x)]$

Examples

Pattern: Factor out $e^{q(x)}$ since it's never zero. This simplifies finding critical points.

Quotient Rule with Exp/Log

Example 1: $f(x) = \frac{e^x}{x^2}$

$f'(x) = \frac{x^2 e^x - e^x \cdot 2x}{x^4} = \frac{e^x(x^2 - 2x)}{x^4} = \frac{e^x(x-2)}{x^3}$

Critical point at $x = 2$ (not $x = 0$ — that's not in the domain).

Example 2: $f(x) = \frac{\ln x}{x}$

$f'(x) = \frac{(1/x) \cdot x - \ln x \cdot 1}{x^2} = \frac{1 - \ln x}{x^2}$

$f'(x) = 0$ when $\ln x = 1$, i.e., $x = e$.

Maximum of $\frac{\ln x}{x}$ at $x = e$: value $= \frac{1}{e}$.

Example 3: $f(x) = \frac{e^x - e^{-x}}{2}$

$f'(x) = \frac{e^x - (-e^{-x})}{2} = \frac{e^x + e^{-x}}{2}$

Combined Rules 🎯

📌 Nested Chain Rule

Double Chain Rule

$\frac{d}{dx}[e^{\sin(2x)}] = e^{\sin(2x)} \cdot \cos(2x) \cdot 2 = 2\cos(2x)e^{\sin(2x)}$

$\frac{d}{dx}[\ln(\ln x)] = \frac{1}{\ln x} \cdot \frac{1}{x} = \frac{1}{x \ln x}$

$\frac{d}{dx}[e^{e^x}] = e^{e^x} \cdot e^x = e^{x + e^x}$

Implicit with Exp/Log

$e^{xy} = x + y$:

$e^{xy}(y + x\frac{dy}{dx}) = 1 + \frac{dy}{dx}$

$(xe^{xy} - 1)\frac{dy}{dx} = 1 - ye^{xy}$

$\frac{dy}{dx} = \frac{1 - ye^{xy}}{xe^{xy} - 1}$

Combined Practice 🧮

1) $\frac{d}{dx}[xe^x]$ at $x = 0$: (integer)

2) $\frac{d}{dx}[\frac{\ln x}{x}]$ at $x = e$: (integer)

Rule Selection 🔽

Exit Quiz ✅

∫ Derivatives of Exp & Log — Part 4

Part 4 of 7 — Growth, Decay & Applications

📖 Exponential Growth & Decay

The model $y = Ce^{kt}$ is the solution to:

$\frac{dy}{dt} = ky$

• $k > 0$: Growth (populations, investments)
• $k < 0$: Decay (radioactive decay, drug elimination)
• $C = y(0)$: initial amount

Verification

If $y = Ce^{kt}$: $\frac{dy}{dt} = Cke^{kt} = k \cdot Ce^{kt} = ky$ ✓

Half-Life

For decay: $y = Ce^{kt}$ with $k < 0$.

Half-life $t_{1/2}$: $\frac{C}{2} = Ce^{kt_{1/2}} \implies e^{kt_{1/2}} = \frac{1}{2}$

$t_{1/2} = \frac{\ln(1/2)}{k} = \frac{-\ln 2}{k}$

Doubling Time

For growth: $2C = Ce^{kt_d} \implies t_d = \frac{\ln 2}{k}$

Rate Proportional to Amount

Continuous Growth of a Balance

$A = A_0e^{rt}$

where $A_0$ = starting amount, $r$ = continuous growth rate, $t$ = time.

$\frac{dA}{dt} = rA_0 e^{rt} = rA$

The balance grows at a rate proportional to its current size, the same $\frac{dy}{dt} = ky$ model as population growth.

Example

A balance starts at $1000 and satisfies $\frac{dA}{dt} = 0.05A$:

$A = 1000e^{0.05t}$

$\frac{dA}{dt} = 50e^{0.05t}$

At $t = 10$: $\frac{dA}{dt} = 50e^{0.5} \approx 82.44$, i.e. $82.44/year

Newton's Law of Cooling

$T(t) = T_s + (T_0 - T_s)e^{-kt}$

$\frac{dT}{dt} = -k(T_0 - T_s)e^{-kt} = -k(T - T_s)$

Rate of cooling is proportional to temperature difference.

Growth & Decay 🎯

📌 AP Exam Applications

Rate of Change Analysis

Given $P(t) = 1000e^{0.03t}$:

(a) Find the rate of change at $t = 10$: $P'(t) = 30e^{0.03t}$ $P'(10) = 30e^{0.3} \approx 40.5$ per year

(b) When does the population reach 2000? $2000 = 1000e^{0.03t}$ $\ln 2 = 0.03t$ $t = \ln 2/0.03 \approx 23.1$ years

(c) At that moment, what is the rate of growth? $P'(23.1) = 30e^{0.03(23.1)} = 30e^{\ln 2} = 30(2) = 60$ per year

AP Pattern: When population doubles, its rate of growth also doubles (because rate is proportional to amount).

Growth/Decay Practice 🧮

1) $y = 500e^{-0.1t}$. $dy/dt$ at $t = 0$: (integer)

2) Doubling time when $k = \ln 2$: (integer)

Model Identification 🔽

Exit Quiz ✅

∫ Derivatives of Exp & Log — Part 5

Part 5 of 7 — Inverse Function Theorem

📖 Inverse Function Derivative Theorem

If $f$ and $g$ are inverse functions ($g = f^{-1}$), then:

$\boxed{g'(x) = \frac{1}{f'(g(x))}}$

Intuition

If $f$ stretches by factor $m$ at a point, its inverse compresses by factor $1/m$.

Verification with $e^x$ and $\ln x$

Let $f(x) = e^x$, $g(x) = \ln x$.

$f'(x) = e^x$

$g'(x) = \frac{1}{f'(g(x))} = \frac{1}{e^{\ln x}} = \frac{1}{x}$ ✓

Example: $f(x) = x^3 + x$

$f(1) = 2$, so $f^{-1}(2) = 1$.

$(f^{-1})'(2) = \frac{1}{f'(f^{-1}(2))} = \frac{1}{f'(1)} = \frac{1}{3(1)^2 + 1} = \frac{1}{4}$

Table-Based Inverse Problems (AP Style)

Given a table for $f$:

Find $(f^{-1})'(7)$:

$f(2) = 7$, so $f^{-1}(7) = 2$.

$(f^{-1})'(7) = \frac{1}{f'(f^{-1}(7))} = \frac{1}{f'(2)} = \frac{1}{5}$

Find $(f^{-1})'(4)$:

$f(1) = 4$, so $f^{-1}(4) = 1$.

$(f^{-1})'(4) = \frac{1}{f'(1)} = \frac{1}{3}$

AP Tip: These table problems appear almost every year. The recipe: find where $f =$ the given value, then take $1/f'$ at that input.

Inverse Function Derivatives 🎯

📌 Practice Problems

Problem 1

$f(x) = x^5 + 3x^3 + 2x$. Find $(f^{-1})'(6)$.

$f(1) = 1 + 3 + 2 = 6$ ✓

$f'(x) = 5x^4 + 9x^2 + 2$

$f'(1) = 5 + 9 + 2 = 16$

$(f^{-1})'(6) = 1/16$

Problem 2

$f(x) = e^{2x}$. Find $(f^{-1})'(e^4)$.

$f(2) = e^4$, so $f^{-1}(e^4) = 2$.

$f'(x) = 2e^{2x}$

$(f^{-1})'(e^4) = \frac{1}{f'(2)} = \frac{1}{2e^4}$

Inverse Derivative Practice 🧮

1) $f(x) = x^3$, $(f^{-1})'(8) =$ (enter as fraction)

2) $f(x) = e^x$, $(f^{-1})'(1) =$ (integer)

Inverse Function Concepts 🔽

Exit Quiz ✅

∫ Derivatives of Exp & Log — Part 6

Part 6 of 7 — Graphing & Analysis

📖 Graphing Exponential Functions

$f(x) = e^{-x^2/2}$ (Normal Distribution Shape)

First derivative: $f'(x) = -xe^{-x^2/2}$

• $f'(x) = 0$ at $x = 0$
• $f'(x) > 0$ for $x < 0$ (increasing)
• $f'(x) < 0$ for $x > 0$ (decreasing)

Second derivative: $f''(x) = (x^2 - 1)e^{-x^2/2}$

• $f''(x) = 0$ at $x = \pm 1$ (inflection points)
• Concave up for $|x| > 1$, concave down for $|x| < 1$

Summary: Bell-shaped curve, max at $x = 0$, inflection at $\pm 1$.

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

$f'(x) = e^{-x}(1 - x)$: max at $x = 1$, $f(1) = 1/e$

$f''(x) = e^{-x}(x - 2)$: inflection at $x = 2$

As $x \to \infty$: $f(x) \to 0$ (exponential dominates polynomial)

Graphing Logarithmic Functions

$f(x) = x - \ln x$ (for $x > 0$)

$f'(x) = 1 - 1/x = (x-1)/x$

$f'(x) = 0$ at $x = 1$: minimum $f(1) = 1$

$f''(x) = 1/x^2 > 0$: always concave up

$f(x) = \frac{\ln x}{x}$

$f'(x) = \frac{1 - \ln x}{x^2}$

Max at $x = e$: $f(e) = 1/e \approx 0.368$

As $x \to \infty$: $f(x) \to 0$ (log grows slower than any power)

Growth Rate Comparison

$\lim_{x \to \infty} \frac{\ln x}{x^p} = 0 \quad \text{for any } p > 0$

$\lim_{x \to \infty} \frac{x^p}{e^x} = 0 \quad \text{for any } p > 0$

Hierarchy: $\ln x \ll x^p \ll e^x$ as $x \to \infty$

Graphing Analysis 🎯

📌 Complete Analysis Example

$f(x) = x^2 \ln x$ (for $x > 0$)

Domain: $x > 0$

First derivative: $f'(x) = 2x\ln x + x = x(2\ln x + 1)$

$f'(x) = 0$: $x = 0$ (not in domain) or $\ln x = -1/2$ → $x = e^{-1/2} = 1/\sqrt{e}$

Second derivative: $f''(x) = 2\ln x + 3$

$f''(x) = 0$: $\ln x = -3/2$ → $x = e^{-3/2}$

At $x = 1/\sqrt{e}$: $f = (1/e)(-1/2) = -1/(2e)$ (minimum)

Behavior: $f(x) \to 0$ as $x \to 0^+$ (since $x^2$ dominates $|\ln x|$)

Graphing Practice 🧮

1) $f(x) = xe^{-x}$ max value: (enter as 1/e)

2) $f(x) = \ln x / x$ max value: (enter as 1/e)

Function Behavior 🔽

Exit Quiz ✅

∫ Derivatives of Exp & Log — Part 7

Part 7 of 7 — AP Exam Review & Comprehensive Practice

📖 Complete Formula Summary

Inverse Connections

$\frac{d}{dx}[\ln(e^x)] = 1 \quad \text{and} \quad \frac{d}{dx}[e^{\ln x}] = 1$

Integration Preview

$\int e^x \, dx = e^x + C$ $\int \frac{1}{x} \, dx = \ln|x| + C$ $\int a^x \, dx = \frac{a^x}{\ln a} + C$

AP-Style Free Response

Problem: $f(x) = x^2 e^{-x}$ for $x \geq 0$

(a) Find critical points: $f'(x) = xe^{-x}(2 - x) = 0$ $x = 0$ (min) and $x = 2$ (max, $f(2) = 4/e^2$)

(b) Inflection points: $f''(x) = e^{-x}(x^2 - 4x + 2) = 0$ $x = 2 \pm \sqrt{2}$

(c) End behavior: $\lim_{x \to \infty} x^2 e^{-x} = 0$ (exp dominates)

(d) Tangent line at $x = 1$: $f(1) = e^{-1}$ and $f'(1) = 1 \cdot e^{-1}(2 - 1) = e^{-1}$, so $y = \frac{1}{e} + \frac{1}{e}(x - 1) = \frac{x}{e}$

Comprehensive Review 🎯

📌 Common AP Mistakes

❌ Confusing $\frac{d}{dx}[a^x]$ and $\frac{d}{dx}[x^a]$

• $\frac{d}{dx}[2^x] = 2^x \ln 2$ (exponential rule)
• $\frac{d}{dx}[x^2] = 2x$ (power rule)

❌ Forgetting $\ln a$ in $\frac{d}{dx}[a^x]$

$\frac{d}{dx}[3^x] \neq 3^x$ — you need the $\ln 3$ factor!

❌ Domain errors with $\ln x$

$\ln x$ requires $x > 0$. Don't forget to check domain.

❌ Misapplying the chain rule

$\frac{d}{dx}[e^{x^2}] = 2xe^{x^2}$, NOT $e^{x^2}$!

Final Challenge 🧮

1) $\frac{d}{dx}[e^x \cdot \ln x]$ at $x = 1$: (enter as e)

2) $\frac{d}{dx}[3^x]$ at $x = 0$: (enter as ln3)

Final Review 🔽

Final Exit Quiz ✅