Chapter 4 — Front Matter
Purpose and Overview
Chapter 3 established what a derivative is: the instantaneous rate of change of a function at a point, defined as a limit of difference quotients. That definition is the foundation of everything in differential calculus. It is rigorous, conceptually clear, and exact. But applying the definition from scratch every time a derivative is needed is slow, increasingly laborious as functions grow complex, and ultimately impractical in applied work.
Mathematicians recognized this limitation early and derived a collection of differentiation rules — systematic shortcuts that are themselves proven from the limit definition — that allow you to differentiate most functions quickly and reliably. These rules are not arbitrary tricks. Each one encodes a precise limit calculation that has already been carried out in full generality.
This chapter teaches those rules. You will learn how to differentiate constants, power functions, polynomials, products, quotients, and compositions of functions. By the end, you should be able to differentiate a wide class of algebraic functions, identify which rule applies to a given function before computing, and interpret every derivative you find as a meaningful rate of change.
The goal is fluency without mindlessness. A student who can execute differentiation rules but has no idea what the result means has not finished the problem. Every derivative this chapter asks you to find is a rate of change. Its sign tells you whether something is increasing or decreasing. Its magnitude tells you how sensitive the output is to changes in the input. Its units connect the abstract symbol to a real measurement. These rules are tools; context gives them meaning.
Opening Question
A company's daily revenue from selling \(x\) units is modeled by \(R(x) = 120x - 0.5x^2\), where revenue is measured in dollars. At what rate is revenue changing when 80 units are sold? Is revenue increasing or decreasing at that production level? And at what production level does revenue stop growing? To answer these questions precisely, you need \(R'(x)\). The rules in this chapter let you find it in a single step — and interpret the result as the marginal revenue that drives economic decisions.
14Sections
11Objectives
12Key Terms
15+Worked Examples
Why This Chapter Matters
Differentiation rules are the working vocabulary of applied calculus. Physics uses derivatives of position to find velocity and acceleration. Chemistry uses them to study reaction rates. Economics uses them for marginal cost, marginal revenue, and marginal profit. Biology uses them for growth and population dynamics. Engineering uses them to analyze stress, heat flow, and signal behavior. Machine learning uses them in gradient descent — the algorithm that trains neural networks by minimizing a loss function.
None of these applications can pause for a from-scratch limit calculation every time a derivative is needed. Differentiation rules make calculus operational. This chapter gives you those rules and, more importantly, develops the judgment to use them correctly.
Learning Objectives
- 1Differentiate constant functions using the Constant Rule.
- 2Apply the Power Rule to positive integers, negative integers, and rational exponents.
- 3Apply the Constant Multiple Rule to scale derivatives.
- 4Apply the Sum and Difference Rules to differentiate multi-term functions.
- 5Apply the Product Rule to differentiate products of two functions.
- 6Apply the Quotient Rule to differentiate quotients of two functions.
- 7Apply the Chain Rule to differentiate composite functions.
- 8Combine multiple differentiation rules in a single problem.
- 9Compute second and higher-order derivatives.
- 10Interpret derivative notation in context with correct units.
- 11Identify and avoid the most common differentiation mistakes.
Sections 4.1 – 4.6
From Definition to Rules: Constant, Power, and Polynomial Differentiation
Why Rules Exist · Constant Rule · Power Rule · Constant Multiple · Sum/Difference · Polynomials
Section 4.1
Why Differentiation Rules Are Needed
In Chapter 3, you learned that the derivative of \(f\) at \(x\) is defined as
\[f'(x) = \lim_{h \to 0} \frac{f(x+h) - f(x)}{h}\]
This definition is the rigorous foundation of differential calculus. It is exact, and it is the reason every rule in this chapter is trustworthy. But applying the limit definition from scratch every time is slow and error-prone. To differentiate \(f(x) = x^5\) from the definition alone, you would need to expand \((x+h)^5\) using the binomial theorem, subtract \(x^5\), divide by \(h\), simplify, and then evaluate the limit. That calculation is manageable for \(x^5\) but becomes overwhelming for \(x^{20}\), or for a product like \((3x^2 + 1)(x^4 - 5x)\), or for a composition like \((x^3 + 2)^{10}\).
Differentiation rules solve this problem. Each rule was derived from the limit definition — but once proven to hold in full generality, it can be applied directly to any function matching its form. Using the Power Rule, the derivative of \(x^{20}\) is \(20x^{19}\) — one step, no expansion needed.
This efficiency is not a shortcut that sacrifices understanding. The rules encode the underlying limit behavior; they simply compress the work of calculations that are guaranteed to give the same result every time. The real skill this chapter develops is rule identification: looking at a function before computing and asking, "What is the structure here?" Is it a constant? A power? A product? A quotient? A composition? The structure determines the rule. Identifying structure before computing saves errors and builds mathematical judgment.
Strategy — Read Before You Differentiate
Before writing a single derivative symbol, examine the function. Ask: Is this a sum? A product? A quotient? A composition? The outermost operation tells you which rule to start with. Rushing to compute without identifying structure is the root of most differentiation errors.
Section 4.2
Derivative of a Constant
The simplest differentiation rule concerns constant functions — functions of the form \(f(x) = c\), where \(c\) is any fixed number. A constant function is a horizontal line. Its output does not change as \(x\) changes. Therefore, its rate of change is zero everywhere. This geometric fact is what the Constant Rule formalizes.
The Constant Rule
\[\frac{d}{dx}[c] = 0\]
For any constant \(c\), the derivative is zero at every point.
Derivation: If \(f(x) = c\), then \(f(x+h) = c\) as well, since a constant does not depend on \(x\). Therefore:
\[f'(x) = \lim_{h \to 0} \frac{c - c}{h} = \lim_{h \to 0} \frac{0}{h} = \lim_{h \to 0} 0 = 0\]
Example 4.2.1
Derivatives of constants
- ProblemFind the derivative of each: (a) \(f(x) = 7\) (b) \(g(x) = -\pi\) (c) \(h(x) = \sqrt{5}\)
- ThinkEach is a constant function. The Constant Rule applies to all of them, regardless of whether the constant is an integer, a negative number, or an irrational number.
- Calculate(a) \(f'(x) = 0\) (b) \(g'(x) = 0\) (c) \(h'(x) = 0\)
- InterpretNone of these functions change as \(x\) changes. Their graphs are horizontal lines with slope zero. There is nothing to accumulate or vary — the rate of change is zero everywhere.
Answer: All three derivatives equal 0.
Critical Distinction
The Constant Rule applies when the entire function value is a constant. It does not mean that a constant coefficient disappears. In \(f(x) = 5x^2\), the number 5 is a coefficient multiplying a variable term — the Constant Multiple Rule (Section 4.4) handles that. Only when the whole function equals a fixed number, like \(f(x) = 5^3 = 125\), does the Constant Rule apply directly.
Section 4.3
Power Rule
The Power Rule is the most-used differentiation rule in first-semester calculus. It applies to any function of the form \(f(x) = x^n\), where \(n\) is a real number.
The Power Rule
\[\frac{d}{dx}[x^n] = nx^{n-1}\]
The exponent becomes the coefficient, and the exponent is reduced by one. This holds for integer, negative integer, and rational exponents.
Why it works (integer case): For a positive integer \(n\), expanding \((x+h)^n\) with the binomial theorem yields \(x^n + nx^{n-1}h + \text{(terms with } h^2 \text{ and higher)}\). Subtracting \(x^n\), dividing by \(h\), and taking the limit as \(h \to 0\) leaves only \(nx^{n-1}\). The same result extends to all real-number exponents through related arguments.
The Power Rule applies to positive integer powers (\(x^2, x^3, x^{10}\)), negative integer powers (\(x^{-1}, x^{-3}\)), rational exponents (\(x^{1/2}, x^{2/3}\)), and zero (\(\frac{d}{dx}[x^0] = \frac{d}{dx}[1] = 0\), consistent with the Constant Rule). It does not apply to exponential functions like \(2^x\) or \(e^x\), where the variable is in the exponent rather than the base.
| Expression | Rewrite as a Power | Derivative |
| \(\sqrt{x}\) | \(x^{1/2}\) | \(\tfrac{1}{2}x^{-1/2}\) |
| \(\sqrt[3]{x^2}\) | \(x^{2/3}\) | \(\tfrac{2}{3}x^{-1/3}\) |
| \(\dfrac{1}{x^4}\) | \(x^{-4}\) | \(-4x^{-5}\) |
| \(\dfrac{1}{\sqrt{x}}\) | \(x^{-1/2}\) | \(-\tfrac{1}{2}x^{-3/2}\) |
Example 4.3.1
Power Rule with integer exponents
- ProblemDifferentiate: (a) \(f(x) = x^4\) (b) \(g(x) = x^7\) (c) \(h(x) = x\)
- Calculate
(a) \(f'(x) = 4x^3\)
(b) \(g'(x) = 7x^6\)
(c) \(\dfrac{d}{dx}[x^1] = 1 \cdot x^0 = 1\). The derivative of \(x\) is 1, which matches the constant slope of the line \(y = x\).
- InterpretEach result is a polynomial of one lower degree. The derivative of \(x^4\) tells you the rate at which \(x^4\) is growing for any value of \(x\).
Answers: (a) 4x³ (b) 7x⁶ (c) 1
Example 4.3.2
Power Rule with negative and rational exponents
- ProblemDifferentiate: (a) \(f(x) = x^{-3}\) (b) \(g(x) = \sqrt{x}\)
- RepresentWrite (b) as a fractional exponent: \(g(x) = x^{1/2}\).
- Calculate
(a) \(f'(x) = -3x^{-4} = -\dfrac{3}{x^4}\)
(b) \(g'(x) = \dfrac{1}{2}x^{-1/2} = \dfrac{1}{2\sqrt{x}}\)
- CheckFor (a): \(f(x) = 1/x^3\) decreases for \(x > 0\), so a negative derivative is expected. For (b): \(g'(x) > 0\) for all \(x > 0\), consistent with \(\sqrt{x}\) being an increasing function. Notice \(g'(x) \to \infty\) as \(x \to 0^+\), reflecting the steep slope of \(y = \sqrt{x}\) near the origin.
Answers: (a) −3x⁻⁴ = −3/x⁴ (b) 1/(2√x)
Algebra Note — Rewrite Before Differentiating
Before applying the Power Rule, always rewrite radical and fraction expressions as powers with rational exponents. \(\sqrt[3]{x^2} = x^{2/3}\), \(\;\dfrac{1}{x^5} = x^{-5}\). This keeps the Power Rule mechanics clean and avoids sign errors.
Section 4.4
Constant Multiple Rule
When a function is multiplied by a constant, the derivative is also multiplied by that constant. The constant factor simply passes through differentiation unchanged.
The Constant Multiple Rule
\[\frac{d}{dx}[c \cdot f(x)] = c \cdot f'(x)\]
A constant coefficient in a function remains a constant coefficient in its derivative.
Derivation: From the limit definition, \(\dfrac{d}{dx}[c \cdot f(x)] = \displaystyle\lim_{h \to 0} \dfrac{c \cdot f(x+h) - c \cdot f(x)}{h} = c \cdot \displaystyle\lim_{h \to 0} \dfrac{f(x+h) - f(x)}{h} = c \cdot f'(x)\). The constant factor passes through the limit operator because limits respect scalar multiplication.
Example 4.4.1
Constant Multiple Rule
- ProblemDifferentiate: (a) \(f(x) = 5x^3\) (b) \(g(x) = -2x^6\) (c) \(h(x) = \tfrac{3}{4}x^8\)
- Calculate
(a) \(f'(x) = 5 \cdot 3x^2 = 15x^2\)
(b) \(g'(x) = -2 \cdot 6x^5 = -12x^5\)
(c) \(h'(x) = \tfrac{3}{4} \cdot 8x^7 = 6x^7\)
Answers: (a) 15x² (b) −12x⁵ (c) 6x⁷
Example 4.4.2
Velocity from a position model
- ProblemA falling object's height is \(s(t) = 16t^2\) feet after \(t\) seconds. Find the velocity function.
- ThinkVelocity is the derivative of position: \(v(t) = s'(t)\). Apply the Constant Multiple Rule and Power Rule.
- Calculate\(v(t) = s'(t) = 16 \cdot 2t = 32t\) feet per second.
- InterpretVelocity increases linearly with time. Each additional second adds 32 ft/s to the speed — the object accelerates uniformly, consistent with free fall near Earth's surface.
Answer: v(t) = 32t ft/sec
Section 4.5
Sum and Difference Rules
Functions often consist of several terms added or subtracted. The Sum and Difference Rules allow you to differentiate each term independently and add or subtract the results.
Sum and Difference Rules
\[\frac{d}{dx}[f(x) + g(x)] = f'(x) + g'(x)\]
\[\frac{d}{dx}[f(x) - g(x)] = f'(x) - g'(x)\]
Differentiation is a linear operation: it distributes over addition and subtraction and scales with constant multiplication. The Sum, Difference, and Constant Multiple Rules all follow from this single property of limits.
Example 4.5.1
Sum and Difference Rules
- ProblemDifferentiate \(f(x) = x^4 + 3x^2 - 7x + 2\).
- ThinkDifferentiate each term independently, combining the Power Rule, Constant Multiple Rule, and Constant Rule.
- Calculate
\(\dfrac{d}{dx}[x^4] = 4x^3\)
\(\dfrac{d}{dx}[3x^2] = 6x\)
\(\dfrac{d}{dx}[-7x] = -7\)
\(\dfrac{d}{dx}[2] = 0\)
\(\therefore f'(x) = 4x^3 + 6x - 7\)
- InterpretAt \(x = 1\): \(f'(1) = 4 + 6 - 7 = 3\). The function is increasing at a rate of 3 units of output per unit of input at that point.
Answer: f′(x) = 4x³ + 6x − 7
Section 4.6
Derivatives of Polynomial Functions
A polynomial function is a sum of terms of the form \(a_n x^n + a_{n-1}x^{n-1} + \cdots + a_1 x + a_0\), where all exponents are non-negative integers. Combining the Constant Rule, Power Rule, Constant Multiple Rule, and Sum/Difference Rules, you can differentiate any polynomial term by term. There is no special rule for polynomials — you already have all the tools.
One important pattern: the derivative of a polynomial of degree \(n\) is a polynomial of degree \(n - 1\). Each differentiation reduces the degree by one. Constant terms vanish; linear terms become constants; quadratics become linear; and so on.
Example 4.6.1
Differentiating a full polynomial
- ProblemDifferentiate \(p(x) = 2x^5 - 3x^4 + x^3 - 9x + 4\).
- Calculate\(p'(x) = 10x^4 - 12x^3 + 3x^2 - 9\). The constant term 4 contributes zero.
Answer: p′(x) = 10x⁴ − 12x³ + 3x² − 9
Example 4.6.2
Marginal revenue
- ProblemRevenue is \(R(x) = 120x - 0.5x^2\) dollars, where \(x\) is units sold. (a) Find \(R'(x)\). (b) Evaluate \(R'(80)\) and interpret it. (c) Find where \(R'(x) = 0\).
- Think\(R'(x)\) is the marginal revenue — the approximate additional revenue from selling one more unit.
- Calculate
(a) \(R'(x) = 120 - x\)
(b) \(R'(80) = 120 - 80 = 40\)
(c) \(120 - x = 0 \Rightarrow x = 120\)
- InterpretWhen 80 units are sold, revenue is increasing at approximately $40 per additional unit. Revenue is maximized near \(x = 120\) units: for \(x < 120\), selling more increases revenue (\(R'(x) > 0\)); for \(x > 120\), selling more decreases revenue (\(R'(x) < 0\)). This diminishing-returns pattern is built into the quadratic model.
Answers: (a) R′(x) = 120 − x (b) $40/unit (c) x = 120 units
- Differentiate \(f(x) = 9\)
- Differentiate \(g(x) = x^{10}\)
- Differentiate \(h(x) = x^{-4}\)
- Differentiate \(f(x) = \sqrt[5]{x^3}\)
- Differentiate \(g(x) = \dfrac{1}{x^6}\)
- Differentiate \(f(x) = 6x^5\)
- Differentiate \(g(x) = -\tfrac{2}{3}x^9\)
- Differentiate \(f(x) = 4x^3 - 7x^2 + 5x - 1\)
- Differentiate \(p(x) = 3x^4 - x^3 + 2x - 8\)
- Differentiate \(h(x) = 5\sqrt{x} - \dfrac{3}{\sqrt{x}}\)
Sections 4.7 – 4.8
Products and Quotients: When Functions Are Multiplied or Divided
Product Rule · Quotient Rule · When to Simplify First
Section 4.7
Product Rule
Suppose \(h(x) = f(x) \cdot g(x)\) — the product of two functions. The natural question is: can you differentiate by simply multiplying the individual derivatives? The answer is no. This is one of the most important facts in differential calculus.
Consider \(f(x) = x^2\) and \(g(x) = x^3\). Their product is \(h(x) = x^5\), so \(h'(x) = 5x^4\). But the product of the derivatives is \(f'(x) \cdot g'(x) = 2x \cdot 3x^2 = 6x^3 \neq 5x^4\). Multiplying derivatives gives the wrong answer. The correct rule is more nuanced.
The Product Rule
\[\frac{d}{dx}[f(x) \cdot g(x)] = f'(x) \cdot g(x) + f(x) \cdot g'(x)\]
The derivative of a product is the derivative of the first times the second, plus the first times the derivative of the second. Each function takes a turn being differentiated while the other is held fixed, and the results are added.
Example 4.7.1
Applying the Product Rule
- ProblemDifferentiate \(h(x) = (3x^2 + 1)(x^4 - 5x)\).
- IdentifyLet \(f(x) = 3x^2 + 1\), so \(f'(x) = 6x\). Let \(g(x) = x^4 - 5x\), so \(g'(x) = 4x^3 - 5\).
- Calculate
\(h'(x) = (6x)(x^4 - 5x) + (3x^2 + 1)(4x^3 - 5)\)
\(= 6x^5 - 30x^2 + 12x^5 - 15x^2 + 4x^3 - 5\)
\(= 18x^5 + 4x^3 - 45x^2 - 5\)
- CheckYou could expand \(h(x) = 3x^6 - 15x^3 + x^4 - 5x\) and differentiate term by term: \(h'(x) = 18x^5 - 45x^2 + 4x^3 - 5\). Same result. For polynomials, expanding is sometimes easier. For more complex functions, the Product Rule is essential.
Answer: h′(x) = 18x⁵ + 4x³ − 45x² − 5
Example 4.7.2
Product Rule in a force model
- ProblemForce on an object: \(F(t) = (2t + 1)(t^2 - 3)\) newtons, \(t\) in seconds. Find \(F'(t)\) and evaluate at \(t = 2\).
- Calculate
\(f(t) = 2t+1,\; f'(t) = 2\)
\(g(t) = t^2-3,\; g'(t) = 2t\)
\(F'(t) = 2(t^2-3) + (2t+1)(2t) = 2t^2 - 6 + 4t^2 + 2t = 6t^2 + 2t - 6\)
\(F'(2) = 6(4) + 4 - 6 = 22\) newtons per second
- InterpretAt \(t = 2\) seconds, the force is increasing at a rate of 22 newtons per second.
Answer: F′(t) = 6t² + 2t − 6; F′(2) = 22 N/sec
Section 4.8
Quotient Rule
Suppose \(h(x) = \dfrac{f(x)}{g(x)}\). As with products, you cannot simply divide the derivatives. The Quotient Rule gives the correct formula.
The Quotient Rule
\[\frac{d}{dx}\left[\frac{f(x)}{g(x)}\right] = \frac{f'(x) \cdot g(x) - f(x) \cdot g'(x)}{[g(x)]^2}\]
Valid wherever \(g(x) \neq 0\). A common memory aid: "low d-high minus high d-low, over the square of what's below." That is, denominator times derivative of numerator, minus numerator times derivative of denominator, all divided by the denominator squared.
Order Matters — Do Not Reverse
The numerator is always \(f'g - fg'\), not \(fg' - f'g\). Reversing the order flips the sign of the entire result. The mnemonic "low d-high minus high d-low" enforces the correct order.
Example 4.8.1
Applying the Quotient Rule
- ProblemDifferentiate \(h(x) = \dfrac{x^2 + 1}{2x - 3}\).
- Identify\(f(x) = x^2+1,\; f'(x) = 2x\). \(\quad g(x) = 2x-3,\; g'(x) = 2\).
- Calculate
\[h'(x) = \frac{(2x)(2x-3) - (x^2+1)(2)}{(2x-3)^2} = \frac{4x^2 - 6x - 2x^2 - 2}{(2x-3)^2} = \frac{2x^2 - 6x - 2}{(2x-3)^2}\]
- CheckThe domain excludes \(x = 3/2\), where the denominator is zero. The derivative is also undefined there, which is consistent.
Answer: (2x² − 6x − 2) / (2x − 3)²
Example 4.8.2
Drug concentration model
- ProblemConcentration of a drug \(t\) hours after administration: \(C(t) = \dfrac{100t}{t^2 + 4}\) mg/L. Find \(C'(t)\) and evaluate at \(t = 2\).
- Identify\(f(t) = 100t,\; f'(t) = 100\). \(\quad g(t) = t^2+4,\; g'(t) = 2t\).
- Calculate
\[C'(t) = \frac{100(t^2+4) - 100t(2t)}{(t^2+4)^2} = \frac{100t^2 + 400 - 200t^2}{(t^2+4)^2} = \frac{400 - 100t^2}{(t^2+4)^2}\]
\[C'(2) = \frac{400 - 400}{64} = 0\]
- InterpretAt \(t = 2\) hours, the rate of change of concentration is zero — this is a local maximum. Before \(t = 2\), \(C'(t) > 0\): concentration is rising. After \(t = 2\), \(C'(t) < 0\): the body is metabolizing the drug and concentration falls. The derivative tells the complete clinical story of drug absorption and clearance.
Answer: C′(t) = (400 − 100t²) / (t² + 4)²; C′(2) = 0 (local maximum)
Simplify Before You Apply
Always check whether a quotient can be simplified before differentiating. For example, \(\dfrac{x^3 + 2x}{x} = x^2 + 2\) needs only the Power Rule. Save the Quotient Rule for cases where the numerator and denominator are both non-trivial and cannot be simplified first.
- Differentiate \((x^2 + 3)(x - 1)\)
- Differentiate \((2x^3 - x)(x^2 + 4)\)
- Differentiate \(x^4(x - 2)\)
- Differentiate \(\dfrac{x^2 + 1}{x - 2}\)
- Differentiate \(\dfrac{3x - 1}{x^2 + 4}\)
- Differentiate \(\dfrac{x^3}{2x + 1}\)
Sections 4.9 – 4.10
The Chain Rule and Combining Rules
Composite Functions · Outer and Inner · Leibniz Form · Multi-Rule Problems
Section 4.9
Chain Rule
The Chain Rule is arguably the most important differentiation rule you will learn. It handles composite functions — functions of the form \(h(x) = f(g(x))\), where one function is applied to the output of another. Without the Chain Rule, differentiation is restricted to functions you can write as explicit sums, products, or quotients. With it, you can differentiate functions of enormous complexity built by nesting one function inside another.
Consider \(h(x) = (x^2 + 3)^4\). This is a power function applied to an inner expression \(x^2 + 3\). The Power Rule alone gives \(4(x^2+3)^3\) — but that is incomplete. Differentiating a composite function means accounting for how fast both the outer and the inner function are changing. The Chain Rule captures both.
The Chain Rule
\[\frac{d}{dx}[f(g(x))] = f'(g(x)) \cdot g'(x)\]
In Leibniz notation: if \(y = f(u)\) and \(u = g(x)\), then \[\frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx}\]
Differentiate the outer function (leaving the inner function unchanged), then multiply by the derivative of the inner function.
| Composite Expression | Outer Function \(f\) | Inner Function \(g\) |
| \((x^2 + 1)^5\) | \(u^5\) | \(x^2 + 1\) |
| \(\sqrt{3x - 2}\) | \(\sqrt{u}\) | \(3x - 2\) |
| \((x^3 - 4)^{-2}\) | \(u^{-2}\) | \(x^3 - 4\) |
How to Apply the Chain Rule
Step 1: Identify the outer and inner functions.
Step 2: Differentiate the outer function with respect to the inner, leaving the inner function unchanged.
Step 3: Multiply the result by the derivative of the inner function.
Example 4.9.1
Chain Rule: power of an expression
- ProblemDifferentiate \(h(x) = (x^2 + 3)^4\).
- IdentifyOuter: \(f(u) = u^4\), so \(f'(u) = 4u^3\). Inner: \(g(x) = x^2 + 3\), so \(g'(x) = 2x\).
- Calculate
\[h'(x) = 4(x^2+3)^3 \cdot 2x = 8x(x^2+3)^3\]
- CheckThe factor \(2x\) is the derivative of the inner function. Without it, the answer would be incomplete.
Answer: 8x(x² + 3)³
Example 4.9.2
Chain Rule: square root of an expression
- ProblemDifferentiate \(f(x) = \sqrt{5x^3 - 2}\).
- Rewrite\(f(x) = (5x^3 - 2)^{1/2}\)
- Calculate
Outer: \(\tfrac{1}{2}u^{-1/2}\) Inner: \(5x^3 - 2\), derivative \(15x^2\)
\[f'(x) = \frac{1}{2}(5x^3-2)^{-1/2} \cdot 15x^2 = \frac{15x^2}{2\sqrt{5x^3-2}}\]
Answer: 15x² / (2√(5x³ − 2))
Example 4.9.3
Chain Rule: negative power
- ProblemDifferentiate \(g(x) = \dfrac{1}{(3x+1)^3} = (3x+1)^{-3}\).
- Calculate
Outer: \(-3u^{-4}\) Inner: \(3x+1\), derivative \(3\)
\[g'(x) = -3(3x+1)^{-4} \cdot 3 = \frac{-9}{(3x+1)^4}\]
- Interpret\(g'(x) < 0\) everywhere the function is defined, so this function is always decreasing.
Answer: −9 / (3x + 1)⁴
The Most Common Chain Rule Error
Forgetting the inner derivative. The derivative of \((x^2+1)^5\) is NOT \(5(x^2+1)^4\). It is \(5(x^2+1)^4 \cdot 2x\). Whenever you see a function raised to a power, nested inside a root, or otherwise composed, you must multiply by the derivative of the inside. If the inner function were just \(x\), the inner derivative would be 1 and nothing would change — but any other inner function produces a non-trivial factor that must be included.
Leibniz Notation and the Chain Rule
The form \(\dfrac{dy}{dx} = \dfrac{dy}{du} \cdot \dfrac{du}{dx}\) looks like fraction cancellation. This is not a rigorous proof, but it is a useful memory device: the "\(du\)" terms appear to cancel, leaving \(\dfrac{dy}{dx}\). The key insight is that the inner rate of change \(\dfrac{du}{dx}\) multiplies the outer rate \(\dfrac{dy}{du}\).
Section 4.10
Combining Differentiation Rules
Real functions often require more than one rule. A product of two composite functions requires both the Product Rule and the Chain Rule. A quotient whose numerator is a composition requires both the Quotient Rule and the Chain Rule. The key skill is reading the outermost structure of a function first, then working inward.
Strategy: Identify the outermost operation — sum, product, quotient, or composition. Apply the rule for that structure first. As you differentiate each piece, apply further rules as needed inside. When finished, simplify and factor the result, especially if you will later need to solve \(h'(x) = 0\).
Example 4.10.1
Product Rule with Chain Rule
- ProblemDifferentiate \(h(x) = x^3(2x-1)^5\).
- IdentifyOutermost structure: product. Apply the Product Rule. The second factor \((2x-1)^5\) is a composition requiring the Chain Rule.
- Calculate
\(f(x) = x^3,\; f'(x) = 3x^2\)
\(g(x) = (2x-1)^5,\; g'(x) = 5(2x-1)^4 \cdot 2 = 10(2x-1)^4\)
\(h'(x) = 3x^2(2x-1)^5 + x^3 \cdot 10(2x-1)^4\)
- FactorBoth terms share \(x^2(2x-1)^4\):
\[h'(x) = x^2(2x-1)^4[3(2x-1) + 10x] = x^2(2x-1)^4(16x - 3)\]
- CheckFactoring is good mathematical practice here — especially important if we later need to find where \(h'(x) = 0\).
Answer: x²(2x − 1)⁴(16x − 3)
Example 4.10.2
Quotient Rule with Chain Rule
- ProblemDifferentiate \(f(x) = \dfrac{(x+1)^3}{x^2+1}\).
- IdentifyOutermost structure: quotient. The numerator \((x+1)^3\) is a composition.
- Calculate
Numerator derivative (Chain Rule): \(3(x+1)^2\)
Denominator derivative: \(2x\)
\[f'(x) = \frac{3(x+1)^2(x^2+1) - (x+1)^3(2x)}{(x^2+1)^2}\]
- FactorBoth numerator terms share \((x+1)^2\):
\[f'(x) = \frac{(x+1)^2[3(x^2+1) - 2x(x+1)]}{(x^2+1)^2} = \frac{(x+1)^2(x^2 - 2x + 3)}{(x^2+1)^2}\]
Answer: (x + 1)²(x² − 2x + 3) / (x² + 1)²
- Differentiate \((x^2 - 5)^4\)
- Differentiate \((3x+2)^{10}\)
- Differentiate \(\sqrt{x^3+1}\)
- Differentiate \((x^2+x+1)^5\)
- Differentiate \(\dfrac{1}{(x^2-4)^3}\)
- Differentiate \(x^3(x+1)^4\)
- Differentiate \(\dfrac{(2x-1)^3}{x^2+1}\)
- Differentiate \(x^2\sqrt{3x-1}\)
Sections 4.11 – 4.12
Higher-Order Derivatives and Notation
Second Derivatives · Acceleration · Notation Systems · Applied Context
Section 4.11
Higher-Order Derivatives
You can differentiate a function more than once. The derivative of the derivative is the second derivative. Differentiating again yields the third derivative, and so on. These higher-order derivatives carry meaningful information about how a function changes.
The first derivative measures the rate of change of \(f\). The second derivative measures the rate of change of the first derivative — how quickly the rate of change itself is changing. In motion: if \(s(t)\) is position, then \(s'(t) = v(t)\) is velocity, and \(s''(t) = a(t)\) is acceleration. The second derivative also governs the concavity of a graph, a topic developed fully in Chapter 7.
| Order | Function Notation | Leibniz Notation | Meaning |
| 1st | \(f'(x)\) | \(\dfrac{dy}{dx}\) | Rate of change of \(f\) |
| 2nd | \(f''(x)\) | \(\dfrac{d^2y}{dx^2}\) | Rate of change of \(f'\); concavity; acceleration |
| 3rd | \(f'''(x)\) | \(\dfrac{d^3y}{dx^3}\) | Rate of change of \(f''\) |
| \(n\)-th | \(f^{(n)}(x)\) | \(\dfrac{d^ny}{dx^n}\) | Repeated differentiation |
Example 4.11.1
Finding second and third derivatives
- ProblemFind \(f'(x)\), \(f''(x)\), and \(f'''(x)\) for \(f(x) = 4x^4 - 3x^3 + x^2 - 5\).
- Calculate
\(f'(x) = 16x^3 - 9x^2 + 2x\)
\(f''(x) = 48x^2 - 18x + 2\)
\(f'''(x) = 96x - 18\)
- NoteEach differentiation reduces the degree by 1. The fourth derivative of this degree-4 polynomial would be a constant (96); the fifth derivative would be zero.
f′ = 16x³ − 9x² + 2x f″ = 48x² − 18x + 2 f‴ = 96x − 18
Example 4.11.2
Position, velocity, and acceleration
- ProblemA particle's position is \(s(t) = t^3 - 6t^2 + 9t\) meters, \(t\) in seconds. Find velocity and acceleration. At \(t = 3\) s, is the particle speeding up or slowing down?
- Calculate
\(v(t) = s'(t) = 3t^2 - 12t + 9\)
\(a(t) = v'(t) = s''(t) = 6t - 12\)
At \(t = 3\):
\(v(3) = 27 - 36 + 9 = 0\) m/s
\(a(3) = 18 - 12 = 6\) m/s²
- InterpretThe particle momentarily stops at \(t = 3\) s (velocity = 0). The positive acceleration means it immediately begins moving again in the positive direction. This is a turning point: the particle reverses direction at \(t = 3\).
v(t) = 3t² − 12t + 9; a(t) = 6t − 12; turning point at t = 3
Section 4.12
Derivative Notation in Context
Calculus uses several notation systems for derivatives. You will encounter all of them in this book, in other courses, and in applications. Understanding what each notation means — and when each is preferred — is part of mathematical fluency.
Function notation \(f'(x)\) emphasizes that the derivative is itself a function of \(x\). It is compact and common in pure mathematics.
Leibniz notation \(\dfrac{dy}{dx}\) makes the variables explicit. It is widely used in physics, engineering, and applied mathematics because it shows both what is being differentiated and with respect to which variable. The operator \(\dfrac{d}{dx}\) means "differentiate with respect to \(x\)."
Evaluating at a point: \(f'(a)\) or \(\left.\dfrac{dy}{dx}\right|_{x=a}\) both mean the value of the derivative when \(x = a\).
Non-standard variables: If time is the independent variable and position is \(s\), write \(\dfrac{ds}{dt}\) for velocity. If cost is \(C\) and quantity is \(q\), write \(\dfrac{dC}{dq}\) for marginal cost. Leibniz notation makes the variable relationship transparent in applied contexts.
Example 4.12.1
Translating notation and interpreting units
- ProblemCost function \(C(q) = 200 + 3q + 0.01q^2\) dollars, \(q\) units. Find marginal cost at \(q = 100\) and interpret.
- Calculate
\(C'(q) = \dfrac{dC}{dq} = 3 + 0.02q\)
\(C'(100) = 3 + 0.02(100) = 5\) dollars per unit
- InterpretWhen 100 units are being produced, one additional unit costs approximately $5 more to manufacture. The marginal cost is increasing — each additional unit beyond 100 will cost slightly more than the last, due to the \(0.01q^2\) term.
C′(100) = $5 per unit (marginal cost)
- Find \(f'(x)\) and \(f''(x)\) for \(f(x) = x^5 - 4x^3 + 2x\).
- Find the third derivative of \(g(x) = 2x^4 - x^3 + 6x - 1\).
- A particle's position is \(s(t) = t^4 - 4t^3 + 6t^2 - 4t\). Find velocity \(v(t)\) and acceleration \(a(t)\).
- For what values of \(x\) is \(f''(x) = 0\) if \(f(x) = x^4 - 6x^2 + 5\)?
- A company's revenue is \(R(x) = 80x - 0.2x^2\). Find \(R'(x)\), evaluate \(R'(100)\), and interpret in dollars per unit.
Sections 4.13 – 4.14
Common Mistakes and the Road Ahead
Ten Recurring Errors · Prevention · Preparing for Transcendental Derivatives
Section 4.13
Common Differentiation Rule Mistakes
The rules in this chapter are systematic, but students make a predictable set of errors when applying them. The mistakes below are not random — each reflects a specific misunderstanding. Studying them carefully is itself part of learning calculus.
Mistake 1 — Multiplying derivatives instead of using the Product Rule
\(\dfrac{d}{dx}[(x^2)(x^3)] = (2x)(3x^2) = 6x^3\)
Correct: \(x^2 \cdot x^3 = x^5\), so \(\dfrac{d}{dx}[x^5] = 5x^4\). Or apply the Product Rule: \(2x \cdot x^3 + x^2 \cdot 3x^2 = 2x^4 + 3x^4 = 5x^4\). The product of derivatives never equals the derivative of the product in general.
Mistake 2 — Forgetting the inner derivative in the Chain Rule
\(\dfrac{d}{dx}[(3x^2-1)^6] = 6(3x^2-1)^5\)
Correct: multiply by the derivative of the inner function \(3x^2-1\), which is \(6x\). Answer: \(6(3x^2-1)^5 \cdot 6x = 36x(3x^2-1)^5\). The inner derivative is the most frequently forgotten factor in calculus.
Mistake 3 — Reversing the Quotient Rule subtraction
\(\dfrac{d}{dx}\!\left[\dfrac{f}{g}\right] = \dfrac{fg' - gf'}{g^2}\)
Correct: the numerator is always \(f'g - fg'\). Reversing flips the sign of the entire derivative. Remember "low d-high minus high d-low" — denominator first, then numerator.
Mistake 4 — Applying the Power Rule to exponential functions
\(\dfrac{d}{dx}[2^x] = x \cdot 2^{x-1}\)
Correct: the Power Rule applies when the variable is the base and the exponent is a constant (like \(x^n\)). In \(2^x\), the variable is the exponent. This is an exponential function; its derivative is \(2^x \ln 2\), derived in Chapter 5.
Mistake 5 — Differentiating a numerical constant with the Power Rule
\(\dfrac{d}{dx}[5^3] = 3 \cdot 5^2 = 75\)
Correct: \(5^3 = 125\) is a constant — a fixed number with no variable. The Constant Rule applies: its derivative is 0. The Power Rule requires a variable base like \(x^n\), not a number raised to a power.
Mistake 6 — Squaring only part of the Quotient Rule denominator
Denominator of Quotient Rule written as \(g(x)^2\) applied to only one term
Correct: the denominator of the Quotient Rule is \([g(x)]^2\) — the entire denominator function, squared. Do not square only a term or a coefficient inside \(g(x)\).
Mistake 7 — Confusing composite and product structure
Applying the Product Rule to \((x^2+1)^5\) by treating it as \((x^2+1) \cdot (x^2+1)^4\)
Correct: \((x^2+1)^5\) is a composition — a power function applied to the inner expression \(x^2+1\). The Chain Rule applies, not the Product Rule (unless you deliberately split it that way as a check). Recognize composition by asking: "Is one function's output the input of another?"
Mistake 8 — Failing to simplify before differentiating
Applying the Quotient Rule to \(\dfrac{x^3+2x}{x}\) without simplifying
Correct: simplify first. \(\dfrac{x^3+2x}{x} = x^2 + 2\), differentiable in one step as \(2x\). Always check whether a function simplifies before choosing a method. Unnecessary complexity invites errors.
Mistake 9 — Dropping units or mixing up variable names
Writing \(C'(x)\) when the variable is \(q\) or \(t\)
Correct: keep variable names explicit and consistent. If the problem uses \(\dfrac{dC}{dq}\), maintain that notation throughout. In applied problems, confusing variable names causes interpretive errors and incorrect unit analysis.
Mistake 10 — Assuming a product of two positive functions has a positive derivative
Reasoning: "both factors are positive, so the derivative must be positive."
Correct: the Product Rule computes \(f'g + fg'\). The sign of each term depends on the derivatives of \(f\) and \(g\), not just their values. Even if both \(f > 0\) and \(g > 0\), the derivative can be negative if one function is decreasing fast enough. Always evaluate \(f'g + fg'\) rather than reasoning about signs informally.
Section 4.14
Preparing for Transcendental Derivatives
This chapter covered differentiation rules for algebraic functions: functions built from powers, products, quotients, and compositions of polynomial and rational expressions. But the most important functions in science, engineering, finance, and data science extend far beyond algebra.
Exponential functions model growth and decay: compound interest, radioactive decay, population, bacteria. Logarithmic functions model scale, information theory, and entropy. Trigonometric functions model oscillation, waves, rotation, and alternating current. These are the transcendental functions you studied in Trigonometry and Precalculus, and calculus applies fully to all of them.
Chapter 5 develops the derivatives of these function families. The Chain Rule you learned in Section 4.9 will be essential throughout. Differentiating \(e^{3x^2}\) requires recognizing a composition and multiplying by the inner derivative. Differentiating \(\sin(x^2)\) requires the same pattern. Every Chain Rule problem you have practiced in this chapter is preparation for Chapter 5.
Before moving forward, make sure you are comfortable with:
Readiness Checklist for Chapter 5
□ Identifying whether a function is a constant, power, product, quotient, or composition
□ Applying the Power, Product, Quotient, and Chain Rules independently
□ Combining rules for multi-structure functions
□ Computing second derivatives
□ Reading and writing derivative notation correctly, including with non-standard variables
□ Interpreting every derivative result as a rate of change with appropriate units
- Identify the structure of each function (sum, product, quotient, or composition) and state which rule(s) are needed. Do not differentiate.
(a) \(f(x) = (2x^3 + 1)^8\) (b) \(g(x) = \dfrac{x^2+5}{x-1}\) (c) \(h(x) = x^4(x+2)^3\) (d) \(k(x) = \sqrt{x^2-9}\)
- A function \(y = (u)^{1/2}\) where \(u = x^3 + 4\). Write \(\dfrac{dy}{dx}\) using the Leibniz form of the Chain Rule: \(\dfrac{dy}{dx} = \dfrac{dy}{du} \cdot \dfrac{du}{dx}\).
- Explain why \(\dfrac{d}{dx}[3^x]\) cannot be computed using the Power Rule. What feature of the function prevents it?
Chapter 4 — Back Matter
Chapter Summary
- Sec 4.1Differentiation rules are derived from the limit definition and applied to classes of functions. The key skill is identifying function structure — sum, product, quotient, or composition — before differentiating.
- Sec 4.2Constant Rule: \(\dfrac{d}{dx}[c] = 0\). A constant function is a horizontal line; its rate of change is zero everywhere.
- Sec 4.3Power Rule: \(\dfrac{d}{dx}[x^n] = nx^{n-1}\). Applies to integer, negative, and rational exponents. Rewrite radicals and fractions as powers before differentiating.
- Sec 4.4Constant Multiple Rule: \(\dfrac{d}{dx}[cf(x)] = c \cdot f'(x)\). A constant coefficient passes through differentiation unchanged.
- Sec 4.5Sum and Difference Rules: differentiation distributes over addition and subtraction. Differentiate each term independently.
- Sec 4.6Any polynomial can be differentiated term by term. Differentiation reduces degree by one. The derivative of a degree-\(n\) polynomial is degree \(n-1\).
- Sec 4.7Product Rule: \(\dfrac{d}{dx}[fg] = f'g + fg'\). The derivative of a product is NOT the product of the derivatives.
- Sec 4.8Quotient Rule: \(\dfrac{d}{dx}[f/g] = \dfrac{f'g - fg'}{g^2}\). Order in the numerator matters. Simplify the quotient first when possible.
- Sec 4.9Chain Rule: \(\dfrac{d}{dx}[f(g(x))] = f'(g(x)) \cdot g'(x)\). Differentiate the outer function (leaving the inner unchanged), then multiply by the inner derivative. The most common error is forgetting the inner derivative.
- Sec 4.10Complex functions require combining rules. Identify the outermost structure first. Apply the appropriate rule, then work inward. Factor the derivative when possible.
- Sec 4.11Higher-order derivatives are found by repeated differentiation. \(f''(x)\) is the second derivative; \(s''(t) = a(t)\) is acceleration. Second derivatives relate to concavity.
- Sec 4.12Function notation (\(f'(x)\)) and Leibniz notation (\(\dfrac{dy}{dx}\)) express the same derivative. Leibniz notation is preferred in applied problems because it makes variable relationships explicit.
- Sec 4.13Ten common errors include: multiplying derivatives instead of using the Product Rule; forgetting the Chain Rule inner factor; reversing the Quotient Rule; applying the Power Rule to exponential functions; and failing to simplify before differentiating.
- Sec 4.14The Chain Rule is the essential tool for Chapter 5. Students should be comfortable identifying composite functions and differentiating them before moving to transcendental derivatives.
Key Terms Review
Constant Rule
\(\dfrac{d}{dx}[c] = 0\). The derivative of any constant function is zero; a constant has no rate of change.
Power Rule
\(\dfrac{d}{dx}[x^n] = nx^{n-1}\). The exponent becomes the coefficient; the exponent is reduced by one.
Constant Multiple Rule
\(\dfrac{d}{dx}[cf] = cf'\). A constant coefficient passes through differentiation.
Sum Rule
\(\dfrac{d}{dx}[f+g] = f'+g'\). Differentiation distributes over addition.
Product Rule
\(\dfrac{d}{dx}[fg] = f'g + fg'\). Derivative of the first times the second, plus the first times the derivative of the second.
Quotient Rule
\(\dfrac{d}{dx}[f/g] = \dfrac{f'g - fg'}{g^2}\). Low d-high minus high d-low, over the square of what's below.
Chain Rule
\(\dfrac{d}{dx}[f(g(x))] = f'(g(x)) \cdot g'(x)\). Derivative of the outer times derivative of the inner.
Composite function
A function formed by applying one function to the output of another: \(f(g(x))\). The Chain Rule differentiates composite functions.
Outer/Inner function
In \(f(g(x))\), \(f\) is the outer function and \(g\) is the inner function. Identifying them is the first step in applying the Chain Rule.
Second derivative
The derivative of the first derivative, \(f''(x)\). Measures the rate of change of the rate of change; related to acceleration and concavity.
Higher-order derivative
Any derivative beyond the first. Denoted \(f^{(n)}(x)\) or \(\dfrac{d^ny}{dx^n}\).
Marginal analysis
The use of derivatives to study how small unit changes in input affect output. Central in economics: marginal cost, revenue, and profit.
Rule Identification Practice
For each function, identify which rule or combination of rules is needed to differentiate it. Do not compute the derivative — only identify the structure.
- \(f(x) = 7x^4\)
- \(g(x) = 3x^2 + 5x - 1\)
- \(h(x) = (x^2+1)(x-3)\)
- \(k(x) = \dfrac{2x+1}{x^2-4}\)
- \(m(x) = (4x^3-1)^7\)
- \(p(x) = x^2(3x-1)^5\)
- \(q(x) = \dfrac{(x+2)^4}{x^3+1}\)
- \(r(x) = \sqrt{x^2-9}\)
- \(s(x) = \left(\dfrac{x+1}{x-1}\right)^3\)
- \(t(x) = \dfrac{1}{(2x^2+3)^4}\)
Differentiation Skill Practice
- \(f(x) = x^9\)
- \(g(x) = x^{-4}\)
- \(h(x) = x^{3/4}\)
- \(k(x) = \dfrac{1}{x^5}\)
- \(f(x) = \sqrt[4]{x^3}\)
- \(g(x) = \dfrac{1}{\sqrt[3]{x^2}}\)
- \(f(x) = 4x^5 - 3x^3 + 2x - 7\)
- \(g(x) = -6x^4 + x^2 - 9\)
- \(h(x) = \tfrac{5}{2}x^6 - \tfrac{1}{3}x^3 + 4\)
- \(f(x) = 3x^{-2} + 4x^{-1} - 5\)
- \(g(x) = 2\sqrt{x} - \dfrac{3}{\sqrt{x}}\)
- \(p(x) = 6x^{5/2} - 4x^{3/2} + x^{1/2}\)
- \((2x+1)(x^3-4)\)
- \((x^2-3)(x^2+3)\)
- \((3x^3-x)(2x^2+5)\)
- \(x^4(x-2)\)
- \((x^2+1)(x^2-1)\) (simplify first)
- \((4x^2-1)(3x^3+2x)\)
- \(\dfrac{x^2+1}{x-2}\)
- \(\dfrac{3x-1}{x^2+4}\)
- \(\dfrac{x^3}{2x+1}\)
- \(\dfrac{x^2-4}{x+2}\) (simplify first)
- \(\dfrac{x^4-1}{x^2+1}\)
- \(\dfrac{5x^2-3}{x^3-8}\)
- \((x^2-5)^4\)
- \((3x+2)^{10}\)
- \(\sqrt{x^3+1}\)
- \((x^2+x+1)^5\)
- \(\dfrac{1}{(x^2-4)^3}\)
- \((4x^3-2x+1)^6\)
- \(x^3(x+1)^4\)
- \(\dfrac{(2x-1)^3}{x^2+1}\)
- \((x^2+1)^3(x-2)^2\)
- \(\dfrac{x}{(x^2+1)^2}\)
- \(\left(\dfrac{x+1}{x-1}\right)^2\)
- \(x^2\sqrt{3x-1}\)
- \(\dfrac{\sqrt{x+1}}{2x-3}\)
- \((x^3-2x)^4(x+1)^2\)
Applications
- Marginal Cost. A company's cost function is \(C(x) = 400 + 2x + 0.003x^2\), where \(C\) is in dollars and \(x\) is units produced.
(a) Find the marginal cost function \(C'(x)\).
(b) Find and interpret \(C'(50)\).
(c) Find and interpret \(C'(200)\).
(d) Is marginal cost increasing or decreasing as production increases?
- Falling Object. Height of a dropped object: \(h(t) = 100 - 4.9t^2\) meters, \(t\) seconds.
(a) Find velocity \(v(t) = h'(t)\).
(b) Find velocity at \(t = 3\) seconds.
(c) Find acceleration \(a(t) = h''(t)\) and interpret its constant value.
(d) When does the object reach the ground?
- Drug Concentration. Using \(C(t) = \dfrac{100t}{t^2+4}\) from Example 4.8.2:
(a) For which values \(t > 0\) is \(C'(t) > 0\)? What does this mean clinically?
(b) For which values \(t > 0\) is \(C'(t) < 0\)?
- Expanding Oil Spill. Radius at time \(t\) minutes: \(r(t) = 3t^{1/2}\) meters. Area of the spill: \(A = \pi r^2 = \pi(3t^{1/2})^2 = 9\pi t\).
(a) Find \(\dfrac{dA}{dt}\) directly from the simplified form.
(b) Now write \(A = \pi(3\sqrt{t})^2\) and use the Chain Rule on the form \(A = \pi r^2\) with \(r = 3t^{1/2}\), using \(\dfrac{dA}{dt} = \dfrac{dA}{dr} \cdot \dfrac{dr}{dt}\). Do both methods agree?
(c) Interpret \(\dfrac{dA}{dt}\) at \(t = 4\) minutes.
- Profit Function. A firm's profit is \(P(x) = 50x - 0.4x^2 - 200\) dollars, \(x\) units.
(a) Find the marginal profit \(P'(x)\).
(b) At what quantity is \(P'(x) = 0\)? What does this production level represent?
(c) Is profit increasing or decreasing at \(x = 80\)?
Error Analysis
Each problem shows incorrect student work. Identify the mistake and show the correct derivative.
Problem A
Differentiate \((x^3+2)^4\): Student writes \(f'(x) = 4(x^3+2)^3\)
What rule was misapplied? What factor is missing? Show the correct derivative.
Problem B
Differentiate \(\dfrac{x^2+1}{x-1}\): Student writes numerator as \((x-1) - (x^2+1)\)
Identify the specific error in the Quotient Rule setup. Show the correct numerator and final answer.
Problem C
Differentiate \((2x+1)(x^2-3)\): Student writes \(f'(x) = (2)(2x) = 4x\)
What rule did the student ignore? Apply the correct rule and show all steps.
Problem D
Differentiate \(f(x) = 5^3\): Student writes \(f'(x) = 3 \cdot 5^2 = 75\)
Why does the Power Rule not apply here? What is the correct derivative?
Problem E
Differentiate \(x^2 \cdot x^4\) using the Product Rule: Student writes \(f'(x) = (2x)(4x^3) = 8x^4\)
What is the correct answer? Also differentiate by simplifying first. What do you notice?
Chapter 4 Checkpoint
Chapter 4 Checkpoint
Complete without assistance — assesses all key skills and concepts from Chapter 4
Part A — Rule Identification
For each function, state which rule(s) are needed. Do not compute.
- \(f(x) = 8x^5 - 3x + 1\)
- \(g(x) = (x+2)(x^2-1)\)
- \(h(x) = \dfrac{x^3-1}{x+1}\)
- \(k(x) = (5x^2-3)^9\)
- \(m(x) = x^4(x-1)^3\)
Part B — Differentiation
- Differentiate \(f(x) = 3x^4 - 5x^3 + 2x - 9\).
- Differentiate \(g(x) = \dfrac{x^2+3}{x-1}\).
- Differentiate \(h(x) = (2x^3+x)^5\).
- Differentiate \(k(x) = x^2(3x-1)^4\). Factor your answer.
- Differentiate \(m(x) = \dfrac{(x+1)^3}{x^2+4}\).
Part C — Higher Derivatives
- Find \(f''(x)\) for \(f(x) = x^5 - 4x^3 + x\).
- A particle's position is \(s(t) = 2t^3 - 9t^2 + 12t\). Find velocity and acceleration, then find when the particle is momentarily at rest.
Part D — Applications
- A company's revenue function is \(R(x) = 80x - 0.2x^2\). Find the marginal revenue \(R'(x)\), evaluate at \(x = 100\), and interpret the result in dollars per unit. At what production level is revenue maximized?
- An oil spill radius: \(r(t) = 3t^{1/2}\) meters. Find \(\dfrac{dr}{dt}\) and interpret its value at \(t = 9\) minutes with correct units.
Part E — Conceptual Reflection
- Explain in your own words why the derivative of a product is not the product of the derivatives. Construct a specific numerical example to support your explanation.
- Explain when the Chain Rule is needed and how you identify a composite function. Describe the two steps of applying it.
- A student differentiates \(f(x) = 2^x\) and writes \(f'(x) = x \cdot 2^{x-1}\). What mistake was made? What additional knowledge is needed to differentiate this function correctly?
Looking Ahead
Chapter 5: Derivatives of Trigonometric, Exponential, and Logarithmic Functions
Chapter 4 gave you the complete toolkit for differentiating algebraic functions — functions built from powers, products, quotients, and compositions. Chapter 5 extends this toolkit to the major transcendental function families: trigonometric, exponential, and logarithmic functions. These are the functions you know from precalculus, and calculus applies fully to every one of them.
You will learn that \(\dfrac{d}{dx}[\sin x] = \cos x\), that \(\dfrac{d}{dx}[e^x] = e^x\) (the remarkable self-reproducing property of the natural exponential), and that \(\dfrac{d}{dx}[\ln x] = \dfrac{1}{x}\). The Chain Rule you practiced in Section 4.9 will be essential throughout Chapter 5 — every composed transcendental derivative requires it.
MGU Library Connection
- Derivative Rules Formula Sheet — Appendix F of this textbook
- Common Derivatives Reference — Appendix G of this textbook
- Precalculus Readiness Review — Appendix A (function families and composition)
- MGU Physics: Mechanics — acceleration as the second derivative of position; Newton's Second Law as a differential equation
- MGU Economics: Microeconomic Reasoning — marginal cost, marginal revenue, and profit optimization using derivatives
- MGU Data Science Foundations — gradient descent in machine learning requires differentiating compositions of many functions using the Chain Rule at every layer
- MGU Calculus II — implicit differentiation, integration by substitution (the integral analog of the Chain Rule), and Taylor series all depend on Chain Rule fluency