v1 · padrão canônico

Lesson 52 — Differentiation Rules

The algebraic rules of differentiation — power, constant multiple, sum, product, quotient — and the derivatives of elementary functions. No more limits in practice.

Used in: 2.º ano do EM (16 anos) · Equiv. AP Calculus AB Unit 2 · Equiv. Calculus I §3.3–3.5 · Equiv. Math III japonês cap. 3

(fg)' = f'g + fg'

The **product rule**: the derivative of the product of two functions has *two* terms. The first differentiates $f$ and keeps $g$ ; the second keeps $f$ and differentiates $g$ . This pattern repeats in the quotient and chain rules — mastering it is key to differentiating any elementary function.

Choose your door

Rigorous notation, full derivation, hypotheses

Formal Definitions and Theorems

Table of Elementary Derivatives

Definition· Derivatives of Elementary Functions

Let $f$ be differentiable at $x$ . The derivatives below are direct consequences of the limit definition and are considered fundamental results:

Function $f(x)$	Derivative $f'(x)$	Condition
$c$ (constant)	$0$	$c \in \mathbb{R}$
$x^n$	$n x^{n-1}$	$n \in \mathbb{R}$
$\sin x$	$\cos x$	—
$\cos x$	$-\sin x$	—
$\tan x$	$\sec^2 x$	$x \neq \frac{\pi}{2} + k\pi$
$e^x$	$e^x$	—
$\ln x$	$\dfrac{1}{x}$	$x > 0$
$a^x$	$a^x \ln a$	$a > 0, a \neq 1$
$\log_a x$	$\dfrac{1}{x \ln a}$	$x > 0, a > 0, a \neq 1$

Operational Rules

Theorem· Differentiation Rules (Leibniz and Cauchy)

Let $f$ and $g$ be differentiable functions at $x$ , and $c \in \mathbb{R}$ be a constant. Then:

(c)' = 0

(R1)

what this means · Constant: the derivative of any constant number is zero, as there is no variation.

(x^n)' = n x^{n-1}

(R2)

what this means · Power Rule: the exponent comes down as a coefficient and decreases by 1. Valid for any real exponent.

(c f)' = c f'

(R3)

what this means · Constant Multiple: constants pass through the derivative unchanged.

(f \pm g)' = f' \pm g'

(R4)

what this means · Sum and Difference: the derivative distributes over sums and differences.

(fg)' = f'g + fg'

(R5)

what this means · Product Rule (Leibniz): the product generates two terms. Each differentiates one of the factors, keeping the other intact.

\left(\frac{f}{g}\right)' = \frac{f'g - fg'}{g^2}, \quad g \neq 0

(R6)

what this means · Quotient Rule: the numerator is f prime g minus f g prime; the denominator is g squared. The minus sign in the numerator is critical — do not reverse the order.

"If $f$ and $g$ are differentiable functions, then the derivative of the product $(fg)'$ exists and is given by $f'(x)g(x) + f(x)g'(x)$ ." — Active Calculus §2.3

Proof of the Product Rule

Theorem· Proof of the Product Rule from the Definition

Let $h(x) = f(x) g(x)$ . By definition:

$h'(x) = \lim_{\Delta x \to 0} \frac{f(x + \Delta x)\,g(x + \Delta x) - f(x)\,g(x)}{\Delta x}$

Add and subtract $f(x+\Delta x)\,g(x)$ in the numerator:

$= \lim_{\Delta x \to 0} \frac{[f(x+\Delta x) - f(x)]\,g(x+\Delta x) + f(x)\,[g(x+\Delta x) - g(x)]}{\Delta x}$

Separating the limits and using the continuity of $g$ (every differentiable function is continuous, so $g(x + \Delta x) \to g(x)$ ):

$= \lim_{\Delta x \to 0} \frac{f(x+\Delta x)-f(x)}{\Delta x} \cdot g(x) + f(x) \cdot \lim_{\Delta x \to 0} \frac{g(x+\Delta x)-g(x)}{\Delta x} = f'(x)g(x) + f(x)g'(x)$

$\blacksquare$

Tangent Line

Solved Examples

Example— 1· Derivative of a polynomial (direct application of rules R2–R4)

Problem: Calculate $f'(x)$ for $f(x) = 4x^5 - 3x^3 + 7x - 2$ .

Strategy: Polynomial = sum of monomials. Apply R4 (linearity), R2 (power), and R1 (constant) term by term.

Solution:

$f'(x) = (4x^5)' - (3x^3)' + (7x)' - (2)'$

Applying R3 (constant multiple) and R2 (power):

$= 4 \cdot 5x^4 - 3 \cdot 3x^2 + 7 \cdot 1 - 0$

$= 20x^4 - 9x^2 + 7$

Verification: Each exponent reduced by 1; each coefficient multiplied by the original exponent. Constant $-2$ disappeared (R1). Correct.

Source. Active Calculus §2.1 Exercise 2.1.3 — Boelkins · 2024 · CC-BY-NC-SA.

Example— 2· Product Rule — polynomial times trigonometric

Problem: Calculate $h'(x)$ for $h(x) = x^3 \sin x$ .

Strategy: $h = f \cdot g$ with $f(x) = x^3$ and $g(x) = \sin x$ . Apply R5: $(fg)' = f'g + fg'$ .

Solution:

$f(x) = x^3 \Rightarrow f'(x) = 3x^2$
$g(x) = \sin x \Rightarrow g'(x) = \cos x$

Applying R5:

$h'(x) = f'(x)g(x) + f(x)g'(x) = 3x^2 \sin x + x^3 \cos x$

Factored form (optional): $h'(x) = x^2(3 \sin x + x \cos x)$

Verification: Two terms (correct for product). First: differentiated $x^3$ , kept $\sin x$ . Second: kept $x^3$ , differentiated $\sin x$ . Both elementary rules applied correctly.

Source. Active Calculus §2.3 Exercise 2.3.2 — Boelkins · 2024 · CC-BY-NC-SA.

Example— 3· Quotient Rule — rational function

Problem: Calculate $k'(x)$ for $k(x) = \dfrac{x^2 + 1}{x - 3}$ , $x \neq 3$ .

Strategy: $k = f/g$ with $f(x) = x^2 + 1$ and $g(x) = x - 3$ . Apply R6.

Solution:

$f'(x) = 2x$
$g'(x) = 1$

Applying R6:

$k'(x) = \frac{f'(x)g(x) - f(x)g'(x)}{[g(x)]^2} = \frac{2x(x-3) - (x^2+1) \cdot 1}{(x-3)^2}$

$= \frac{2x^2 - 6x - x^2 - 1}{(x-3)^2} = \frac{x^2 - 6x - 1}{(x-3)^2}$

Attention: The numerator is $f'g - fg'$ , not $fg' - f'g$ . The minus sign is not commutative — reversing it creates the most frequent error in this rule.

Source. OpenStax Calculus Vol. 1 §3.3 Example 3.28 — OpenStax · 2016 · CC-BY-NC-SA.

Example— 4· Product of exponential with trigonometric

Problem: Calculate $p'(x)$ for $p(x) = e^x \cos x$ .

Strategy: Product of $f(x) = e^x$ and $g(x) = \cos x$ . Apply R5 with the derivatives from the table.

Solution:

$f'(x) = e^x$ (notable: $e^x$ differentiates to itself)
$g'(x) = -\sin x$

Applying R5:

$p'(x) = e^x \cos x + e^x \cdot (-\sin x) = e^x(\cos x - \sin x)$

Verification: At $x = 0$ : $p(0) = e^0 \cos 0 = 1$ and $p'(0) = e^0(\cos 0 - \sin 0) = 1 - 0 = 1$ . The tangent at $(0, 1)$ has a slope of 1, which makes geometric sense (the function grows with a slope of 1 at the origin).

Source. Active Calculus §2.2 Exercise 2.2.3 — Boelkins · 2024 · CC-BY-NC-SA.

Example— 5· Equation of the Tangent Line with Critical Point

Problem: Find the equation of the tangent line to the curve $y = \dfrac{x}{x^2 + 1}$ at the point where $x = 1$ .

Strategy: (a) Calculate $y(1)$ to get the point of tangency. (b) Calculate $y'(x)$ using the quotient rule. (c) Evaluate $y'(1)$ . (d) Write in point-slope form.

Solution:

(a) Point of tangency:

$y(1) = \frac{1}{1^2 + 1} = \frac{1}{2}$

Point: $\left(1, \dfrac{1}{2}\right)$ .

(b) Derivative by quotient rule with $f = x$ and $g = x^2 + 1$ :

$y'(x) = \frac{f'g - fg'}{g^2} = \frac{1 \cdot (x^2+1) - x \cdot 2x}{(x^2+1)^2} = \frac{x^2 + 1 - 2x^2}{(x^2+1)^2} = \frac{1 - x^2}{(x^2+1)^2}$

(c) Slope at $x = 1$ :

$y'(1) = \frac{1 - 1}{(1+1)^2} = \frac{0}{4} = 0$

(d) Equation of the tangent:

$y - \frac{1}{2} = 0 \cdot (x - 1) \implies y = \frac{1}{2}$

The tangent at $x = 1$ is horizontal — this makes sense, as $y = x/(x^2+1)$ has a local maximum at this point (the function increases from $x=0$ to $x=1$ and then decreases).

Source. OpenStax Calculus Vol. 1 §3.3 Exercise 176 — OpenStax · 2016 · CC-BY-NC-SA.

Exercise list

40 exercises · 10 with worked solution (25%)

Application 27Understanding 4Modeling 7Challenge 1Proof 1

Ex. 52.1Application
Calculate $(x^5)'$ .
Solve online Active Calculus · §2.1
Show solution
By the power rule R2: $(x^5)' = 5 x^{5-1} = 5x^4$ . The exponent 5 comes down as a coefficient and the new exponent is $5 - 1 = 4$ .
Ex. 52.2Application
Calculate the derivative of $f(x) = 3x^2 - 4x + 1$ .
Active Calculus · §2.1
Show solution
By linearity R4 and power rule R2: $(3x^2 - 4x + 1)' = 3 \cdot 2x - 4 \cdot 1 + 0 = 6x - 4$ . The constant term 1 has a derivative of zero.
Show step-by-step (with the why)
1. Recognize the structure. The function $f(x) = 3x^2 - 4x + 1$ is a polynomial of degree 2. By rule R4, the derivative distributes over sums and differences.
2. Differentiate each term separately.
  - Term $3x^2$ : by the constant multiple rule R3 and power rule R2, $(3x^2)' = 3 \cdot 2x^1 = 6x$ .
  - Term $-4x$ : $(-4x)' = -4 \cdot 1 = -4$ .
  - Constant term $1$ : $(1)' = 0$ by rule R1.
3. Combine the results. $f'(x) = 6x - 4 + 0 = 6x - 4$ .
Tip: for any polynomial, differentiate term by term. The constant always disappears, and the derivative lowers the degree by 1.
Ex. 52.3ApplicationAnswer key
Calculate $(\sqrt{x})'$ . Hint: write as $x^{1/2}$ and apply R2.
Solve online OpenStax Calculus Vol. 1 · §3.3
Show solution
Rewrite $\sqrt{x} = x^{1/2}$ . By rule R2: $(x^{1/2})' = \tfrac{1}{2} x^{1/2 - 1} = \tfrac{1}{2} x^{-1/2} = \dfrac{1}{2\sqrt{x}}$ .
Ex. 52.4Application
Calculate $\left(\dfrac{1}{x^2}\right)'$ .
Solve online OpenStax Calculus Vol. 1 · §3.3
Show solution
Rewrite $1/x^2 = x^{-2}$ . By rule R2: $(x^{-2})' = -2 x^{-3} = -2/x^3$ .
Show step-by-step (with the why)
1. Rewrite as a negative power. $1/x^2 = x^{-2}$ . The power rule R2 works for any real exponent, including negative ones.
2. Apply R2. The exponent $-2$ comes down as a coefficient: $(x^{-2})' = -2 \cdot x^{-2-1} = -2x^{-3}$ .
3. Rewrite in fractional form. $-2x^{-3} = -2/x^3$ .
Tip: functions of the form $1/x^n$ are negative powers. Always rewrite before differentiating.
Ex. 52.5ApplicationAnswer key
Calculate $f'(x)$ for $f(x) = 4x^5 - 3x^3 + 7x - 2$ .
Solve online Active Calculus · §2.1
Show solution
Apply R4, R3, and R2 term by term: $(4x^5)' = 20x^4$ , $(-3x^3)' = -9x^2$ , $(7x)' = 7$ , $(-2)' = 0$ . Sum: $f'(x) = 20x^4 - 9x^2 + 7$ .
Ex. 52.6Application
Calculate $\left(-\dfrac{1}{x^3}\right)'$ .
Solve online OpenStax Calculus Vol. 1 · §3.3
Show solution
Rewrite $-1/x^3 = -x^{-3}$ . Derivative: $(-x^{-3})' = -1 \cdot (-3x^{-4}) = 3x^{-4} = 3/x^4$ .
Ex. 52.7Application
Calculate $f'(x)$ for $f(x) = x^2\sqrt{x} - x\sqrt{x}$ .
Solve online Active Calculus · §2.1
Show solution
Rewrite: $x^2 \sqrt{x} = x^2 \cdot x^{1/2} = x^{5/2}$ and $x \sqrt{x} = x \cdot x^{1/2} = x^{3/2}$ . Then $f(x) = x^{5/2} - x^{3/2}$ and $f'(x) = \tfrac{5}{2}x^{3/2} - \tfrac{3}{2}x^{1/2}$ .
Ex. 52.8Application
Calculate $f'(x)$ for $f(x) = x^3 + 2x$ .
Solve online OpenStax Calculus Vol. 1 · §3.3
Show solution
$f(x) = x^3 + 2x$ . Differentiating term by term: $f'(x) = 3x^2 + 2$ .
Ex. 52.9Application
Calculate $\left(\dfrac{1}{\sqrt{x}}\right)'$ .
Solve online OpenStax Calculus Vol. 1 · §3.3
Show solution
Rewrite: $1/\sqrt{x} = x^{-1/2}$ . Derivative by the power rule: $(x^{-1/2})' = -\tfrac{1}{2} x^{-3/2}$ .
Ex. 52.10ApplicationAnswer key
Calculate $g'(x)$ for $g(x) = 3x^4 - 3x^2$ .
Solve online Active Calculus · §2.1
Show solution
$g(x) = 3x^4 - 3x^2$ . Differentiating: $g'(x) = 12x^3 - 6x$ .
Show step-by-step (with the why)
1. Identify the terms. $g(x) = 3x^4 - 3x^2$ . Two terms separated by rule R4.
2. Differentiate $3x^4$ . By the constant multiple rule R3 and power rule R2: $(3x^4)' = 3 \cdot 4x^3 = 12x^3$ .
3. Differentiate $-3x^2$ . $(-3x^2)' = -3 \cdot 2x = -6x$ .
4. Combine. $g'(x) = 12x^3 - 6x$ . You can factor: $6x(2x^2 - 1)$ .
Curiosity: the zeros of $g'(x) = 0$ are $x = 0$ and $x = \pm 1/\sqrt{2}$ . These are the points where the tangent to the graph of $g$ is horizontal — local extrema.
Ex. 52.11Application
Calculate $(\sin x \cdot \cos x)'$ .
Solve online Active Calculus · §2.2
Show solution
Product rule with $f = \sin x$ and $g = \cos x$ : $(\sin x \cos x)' = \cos x \cdot \cos x + \sin x \cdot (-\sin x) = \cos^2 x - \sin^2 x$ . By the double angle identity, this is equivalent to $\cos(2x)$ .
Show step-by-step (with the why)
1. Identify the factors. We have $h(x) = \sin x \cdot \cos x$ . Define $f(x) = \sin x$ and $g(x) = \cos x$ .
2. Calculate the individual derivatives. From the table: $f'(x) = \cos x$ and $g'(x) = -\sin x$ .
3. Apply the product rule R5. $h'(x) = f'g + fg' = \cos x \cdot \cos x + \sin x \cdot (-\sin x) = \cos^2 x - \sin^2 x$ .
4. Optional recognition. By the trigonometric identity, $\cos^2 x - \sin^2 x = \cos(2x)$ . This confirms that $(\sin x \cos x)' = (\tfrac{1}{2}\sin(2x))' = \cos(2x)$ . Consistent.
Tip: always check if the result can be simplified by trigonometric identities — often the result has a more compact form.
Ex. 52.12ApplicationAnswer key
Calculate $(x e^x)'$ .
Solve online Active Calculus · §2.1
Show solution
Product rule with $f = x$ and $g = e^x$ : $(xe^x)' = 1 \cdot e^x + x \cdot e^x = e^x + xe^x = e^x(1 + x)$ .
Ex. 52.13Application
Calculate $(x \ln x)'$ .
Solve online Active Calculus · §2.1
Show solution
Product rule with $f = x$ and $g = \ln x$ : $(x \ln x)' = 1 \cdot \ln x + x \cdot (1/x) = \ln x + 1$ .
Ex. 52.14Application
Calculate $(x^2 \sin x)'$ .
Solve online Active Calculus · §2.3
Show solution
Product rule with $f = x^2$ and $g = \sin x$ : $(x^2 \sin x)' = 2x \sin x + x^2 \cos x$ .
Ex. 52.15ApplicationAnswer key
Calculate $(e^x \cos x)'$ .
Solve online Active Calculus · §2.3
Show solution
Product rule with $f = e^x$ and $g = \cos x$ : $(e^x \cos x)' = e^x \cos x + e^x(-\sin x) = e^x(\cos x - \sin x)$ .
Ex. 52.16Application
Calculate $(x^2 e^x)'$ .
Solve online OpenStax Calculus Vol. 1 · §3.3
Show solution
Product rule with $f = x^2$ and $g = e^x$ : $(x^2 e^x)' = 2x e^x + x^2 e^x = e^x(2x + x^2) = e^x x(x + 2)$ .
Ex. 52.17ApplicationAnswer key
Calculate $(\tan x \cdot e^x)'$ .
Solve online Active Calculus · §2.3
Show solution
Product rule with $f = \tan x$ and $g = e^x$ : $(\tan x \cdot e^x)' = \sec^2 x \cdot e^x + \tan x \cdot e^x = e^x(\tan x + \sec^2 x)$ .
Ex. 52.18Application
Calculate $(x \sin x)'$ .
Solve online OpenStax Calculus Vol. 1 · §3.3
Show solution
Product rule with $f = x$ and $g = \sin x$ : $(x \sin x)' = 1 \cdot \sin x + x \cdot \cos x = \sin x + x \cos x$ .
Ex. 52.19Application
Calculate $(x^3 \ln x)'$ .
Solve online OpenStax Calculus Vol. 1 · §3.3
Show solution
Product rule with $f = x^3$ and $g = \ln x$ : $(x^3 \ln x)' = 3x^2 \ln x + x^3 \cdot (1/x) = 3x^2 \ln x + x^2 = x^2(3 \ln x + 1)$ .
Ex. 52.20Understanding
Generalization of the product rule. If $f$ , $g$ , $h$ are differentiable functions, what is $(fgh)'$ ?
Select the correct option
$f'gh + fg'h + fgh'$ $f'g'h'$ $f'gh - fg'h + fgh'$ $(fg)'h + f(gh)'$ , without simplifying
Select an option first
Solve online Active Calculus · §2.3
Show solution
Apply the product rule twice: $(fgh)' = ((fg)h)' = (fg)'h + (fg)h' = (f'g + fg')h + fgh' = f'gh + fg'h + fgh'$ . Each term differentiates exactly one of the three factors and keeps the other two.
Show step-by-step (with the why)
1. Associate the product of three factors. Write $(fgh) = (fg) \cdot h$ .
2. Apply R5 to the product of two factors $(fg)$ and $h$ . $((fg) \cdot h)' = (fg)' \cdot h + (fg) \cdot h'$ .
3. Expand $(fg)'$ using the product rule. $(fg)' = f'g + fg'$ .
4. Distribute and simplify. $(f'g + fg') \cdot h + fg \cdot h' = f'gh + fg'h + fgh'$ .
5. General pattern. For $n$ factors, the derivative is a sum of $n$ terms: in each term, exactly one factor is differentiated and the others remain intact.
Mental shortcut: the distractors above test the error of multiplying derivatives ( $f'g'h'$ ) or reversing a sign. The only correct one has exactly 3 terms, each with exactly one "prime".
Ex. 52.21Application
Calculate $\left(\dfrac{\sin x}{x}\right)'$ for $x \neq 0$ .
Solve online Active Calculus · §2.3
Show solution
Quotient rule with $f = \sin x$ and $g = x$ : $\left(\frac{\sin x}{x}\right)' = \frac{\cos x \cdot x - \sin x \cdot 1}{x^2} = \frac{x\cos x - \sin x}{x^2}$ .
Ex. 52.22Application
Calculate $\left(\dfrac{e^x}{x}\right)'$ for $x \neq 0$ .
Solve online Active Calculus · §2.3
Show solution
Quotient rule with $f = e^x$ and $g = x$ : $\left(\frac{e^x}{x}\right)' = \frac{e^x \cdot x - e^x \cdot 1}{x^2} = \frac{e^x(x-1)}{x^2}$ .
Ex. 52.23Application
Calculate $\left(\dfrac{1}{x^2 + 1}\right)'$ .
Solve online OpenStax Calculus Vol. 1 · §3.3
Show solution
Quotient rule with $f = 1$ and $g = x^2 + 1$ : $\left(\frac{1}{x^2+1}\right)' = \frac{0 \cdot (x^2+1) - 1 \cdot 2x}{(x^2+1)^2} = \frac{-2x}{(x^2+1)^2}$ .
Ex. 52.24Application
Calculate $k'(x)$ for $k(x) = \dfrac{x^2 + 1}{x - 3}$ , $x \neq 3$ .
Solve online OpenStax Calculus Vol. 1 · §3.3
Show solution
Quotient rule with $f = x^2 + 1$ and $g = x - 3$ : numerator = $2x(x-3) - (x^2+1) \cdot 1 = 2x^2 - 6x - x^2 - 1 = x^2 - 6x - 1$ . Denominator: $(x-3)^2$ . Thus $k'(x) = \dfrac{x^2 - 6x - 1}{(x-3)^2}$ .
Show step-by-step (with the why)
1. Identify numerator and denominator. $f = x^2 + 1$ , $g = x - 3$ .
2. Calculate the derivatives. $f' = 2x$ and $g' = 1$ .
3. Apply R6. Numerator: $f'g - fg' = 2x(x-3) - (x^2+1) \cdot 1$ .
4. Expand the numerator. $2x^2 - 6x - x^2 - 1 = x^2 - 6x - 1$ .
5. Write the result. $k'(x) = (x^2 - 6x - 1)/(x-3)^2$ .
Tip: the most common error in the quotient rule is to reverse the order of the numerator ( $fg' - f'g$ instead of $f'g - fg'$ ). Memorize: "first differentiate the top, then the bottom, with a minus sign".
Ex. 52.25Application
Calculate $\left(\dfrac{3x^2 - x}{x^2 - 1}\right)'$ for $x \neq \pm 1$ .
Solve online OpenStax Calculus Vol. 1 · §3.3
Show solution
Quotient rule with $f = 3x^2 - x$ and $g = x^2 - 1$ . $f' = 6x - 1$ , $g' = 2x$ . Numerator: $(6x-1)(x^2-1) - (3x^2-x)(2x)$ . Expanding: $6x^3 - 6x - x^2 + 1 - 6x^3 + 2x^2 = x^2 - 6x + 1$ . Denominator: $(x^2-1)^2$ .
Ex. 52.26ApplicationAnswer key
Differentiate $\cot x = \dfrac{\cos x}{\sin x}$ using the quotient rule and show that $(\cot x)' = -\csc^2 x$ .
Solve online OpenStax Calculus Vol. 1 · §3.5
Show solution
Differentiate $\cot x = \cos x / \sin x$ using the quotient rule: numerator = $(-\sin x)(\sin x) - (\cos x)(\cos x) = -(\sin^2 x + \cos^2 x) = -1$ . Denominator: $\sin^2 x$ . Thus $(\cot x)' = -1/\sin^2 x = -\csc^2 x$ .
Ex. 52.27Application
Differentiate $\sec x = \dfrac{1}{\cos x}$ using the quotient rule and show that $(\sec x)' = \sec x \tan x$ .
Solve online OpenStax Calculus Vol. 1 · §3.5
Show solution
Differentiate $\sec x = 1/\cos x$ using the quotient rule with $f = 1$ , $g = \cos x$ : numerator = $0 \cdot \cos x - 1 \cdot (-\sin x) = \sin x$ . Denominator: $\cos^2 x$ . Thus $(\sec x)' = \sin x / \cos^2 x = (1/\cos x)(\sin x/\cos x) = \sec x \tan x$ .
Ex. 52.28Application
Calculate $\left(\dfrac{x}{x^2 + 1}\right)'$ .
Solve online OpenStax Calculus Vol. 1 · §3.3
Show solution
Quotient rule with $f = x$ and $g = x^2 + 1$ : numerator = $1 \cdot (x^2+1) - x \cdot 2x = x^2 + 1 - 2x^2 = 1 - x^2$ . Denominator: $(x^2+1)^2$ . Thus $y' = (1-x^2)/(x^2+1)^2$ .
Ex. 52.29ModelingAnswer key
Find the equation of the tangent line to $f(x) = x^2 + 3x$ at the point $x = 1$ .
Solve online Active Calculus · §2.1
Show solution
Point: $f(1) = 1 + 3 = 4$ . Derivative: $f'(x) = 2x + 3$ . Slope: $f'(1) = 5$ . Tangent: $y - 4 = 5(x - 1)$ , or $y = 5x - 1$ .
Show step-by-step (with the why)
1. Calculate the point of tangency. $f(1) = 1^2 + 3(1) = 4$ . Point: $(1, 4)$ .
2. Calculate the derivative. $f'(x) = 2x + 3$ by rules R2 and R3.
3. Slope at the point. $f'(1) = 2(1) + 3 = 5$ .
4. Equation of the tangent. $y - 4 = 5(x - 1) \Rightarrow y = 5x - 1$ .
5. Verification. At $x = 1$ : $y = 5(1) - 1 = 4$ \checkmark.
Tip: the tangency point step is often forgotten. Without it, you have the slope but not the line.
Ex. 52.30ModelingAnswer key
At what points does the graph of $f(x) = x^3 - 3x$ have a horizontal tangent line?
Solve online Active Calculus · §2.1
Show solution
A horizontal tangent occurs when $f'(x) = 0$ . For $f(x) = x^3 - 3x$ : $f'(x) = 3x^2 - 3 = 0 \Rightarrow x^2 = 1 \Rightarrow x = \pm 1$ . The points are $(1, -2)$ and $(-1, 2)$ .
Ex. 52.31Modeling
An object has position $s(t) = t^3 - 6t^2 + 9t$ meters ( $t$ in seconds). Calculate $v(t)$ and $a(t)$ . Evaluate at $t = 2$ and determine when the object is at rest.
Solve online OpenStax Calculus Vol. 1 · §3.4
Show solution
Velocity: $v(t) = s'(t) = 3t^2 - 12t + 9$ m/s. Acceleration: $a(t) = v'(t) = 6t - 12$ m/s². At $t = 2$ : $v(2) = 12 - 24 + 9 = -3$ m/s (receding); $a(2) = 0$ (velocity inflection). The object stops when $3t^2 - 12t + 9 = 0 \Rightarrow t = 1$ or $t = 3$ .
Show step-by-step (with the why)
1. Identify. Position $s(t) = t^3 - 6t^2 + 9t$ . Velocity = derivative of position; acceleration = derivative of velocity.
2. Differentiate for velocity. $v(t) = s'(t) = 3t^2 - 12t + 9$ .
3. Differentiate again for acceleration. $a(t) = v'(t) = 6t - 12$ .
4. Evaluate at $t = 2$ . $v(2) = 12 - 24 + 9 = -3$ m/s. The negative sign indicates return.
5. When does it stop? $v = 0: 3t^2 - 12t + 9 = 0 \Rightarrow t^2 - 4t + 3 = 0 \Rightarrow t = 1$ or $t = 3$ .
Curiosity: at $t = 2$ , the acceleration is zero — the object is at the inflection point of velocity, transitioning from deceleration to acceleration (or vice versa).
Ex. 52.32Modeling
Cost function: $C(q) = 500 + 50q + 0.1q^2$ (reais). Calculate the marginal cost $C'(q)$ and evaluate at $q = 100$ .
Solve online OpenStax Calculus Vol. 1 · §3.4
Show solution
Marginal cost: $C'(q) = 50 + 0.2q$ . At $q = 100$ : $C'(100) = 50 + 20 = 70$ reais per unit. This means that producing the 101st unit costs approximately R$ 70 more.
Ex. 52.33Modeling
Total revenue: $R(q) = q(200 - q)$ . Calculate the marginal revenue $R'(q)$ and determine the quantity that maximizes revenue.
Solve online OpenStax Calculus Vol. 1 · §3.4
Show solution
Expand: $R(q) = 200q - q^2$ . Marginal revenue: $R'(q) = 200 - 2q$ . Maximum revenue when $R'(q) = 0 \Rightarrow q = 100$ units. Maximum revenue: $R(100) = 100 \cdot 100 = 10000$ reais.
Ex. 52.34Modeling
Find the tangent line to $y = \dfrac{x}{x^2 + 1}$ at $x = 1$ .
Solve online OpenStax Calculus Vol. 1 · §3.3
Show solution
Function: $y = x/(x^2+1)$ . Point: $y(1) = 1/2$ . Derivative (quotient): $y' = (1-x^2)/(x^2+1)^2$ . At $x = 1$ : $y'(1) = 0$ . Equation of the tangent: $y = 1/2$ (horizontal line). The point $x = 1$ is a local maximum of the function.
Ex. 52.35Modeling
For $s(t) = t^3 - 6t^2 + 9t$ , determine: (a) the velocity at $t = 2$ ; (b) when the object is at rest.
Solve online OpenStax Calculus Vol. 1 · §3.4
Show solution
From question 52.31: $v(t) = 3t^2 - 12t + 9$ . At $t = 2$ : $v(2) = 12 - 24 + 9 = -3$ m/s (receding). The object stops when $v = 0 \Rightarrow 3(t^2 - 4t + 3) = 0 \Rightarrow t = 1$ or $t = 3$ .
Ex. 52.36UnderstandingAnswer key
Error identification. A student calculated $(x^2 \cdot x^3)' = 2x \cdot 3x^2 = 6x^3$ . Is this correct or incorrect? Justify and correct if necessary.
Select the correct option
Incorrect — $(fg)' \neq f'g'$ . Correct: $2x \cdot x^3 + x^2 \cdot 3x^2 = 5x^4$ Correct — the calculation is rightIncorrect only in the exponent — it should be $6x^4$ Incorrect — the result should be negative
Select an option first
Solve online Active Calculus · §2.3
Show solution
The student made the classic mistake: $(fg)' \neq f' \cdot g'$ . The correct rule is the product rule: $(x^2 \cdot x^3)' = 2x \cdot x^3 + x^2 \cdot 3x^2 = 2x^4 + 3x^4 = 5x^4$ . Direct alternative: $x^2 \cdot x^3 = x^5$ , therefore $(x^5)' = 5x^4$ . Consistent.
Show step-by-step (with the why)
1. Identify the error. The student calculated $(x^2)' \cdot (x^3)' = 2x \cdot 3x^2 = 6x^3$ . This multiplies the derivatives — an invalid operation.
2. Correct product rule. $(x^2 \cdot x^3)' = 2x \cdot x^3 + x^2 \cdot 3x^2 = 2x^4 + 3x^4 = 5x^4$ .
3. Alternative verification. $x^2 \cdot x^3 = x^5$ . By the power rule: $(x^5)' = 5x^4$ . Consistent.
4. Moral. $(fg)' = f'g'$ would only be valid if the product rule were multiplicative — but it is not. Leibniz requires the two cross terms.
Mental shortcut: the distractors "correct" and "only exponent" test whether the student accepts the error or underestimates it. The correct answer identifies the specific type of error: multiplying derivatives.
Ex. 52.37Understanding
Identify which differentiation rule applies to $h(x) = \dfrac{e^x}{x^2}$ , apply it, and simplify $h'(x)$ .
Solve online Active Calculus · §2.3
Show solution
Identify the structure: $h(x) = e^x / x^2$ is a quotient of $f = e^x$ by $g = x^2$ . Apply R6: $h'(x) = (e^x \cdot x^2 - e^x \cdot 2x) / x^4 = e^x(x^2 - 2x)/x^4 = e^x(x-2)/x^3$ .
Ex. 52.38Understanding
Concept. Why is the derivative of $e^x$ "special"? Explain what $(e^x)' = e^x$ means in geometric and numerical terms.
Solve online Active Calculus · §2.1
Show solution
The identity $(e^x)' = e^x$ means that the growth rate of $e^x$ at a point is equal to the function's value at that point. At $x = 0$ : $e^0 = 1$ and the slope is also 1. At $x = 2$ : $e^2 \approx 7.39$ and the slope is also $\approx 7.39$ . This is the property that defines Euler's number $e \approx 2.718$ — the unique base for which the exponential is its own derivative.
Ex. 52.39Challenge
Challenge: product of three functions. Prove that $(fgh)' = f'gh + fg'h + fgh'$ , by applying the product rule twice.
Solve online Active Calculus · §2.3
Show solution
By associativity: $(fgh)' = ((fg)h)' = (fg)'h + (fg)h' = (f'g + fg')h + fgh' = f'gh + fg'h + fgh'$ . Each of the three terms differentiates exactly one factor.
Show step-by-step (with the why)
1. Group two factors. Write $fgh = (fg) \cdot h$ .
2. Apply R5 to the pair. $((fg)h)' = (fg)'h + (fg)h'$ .
3. Expand $(fg)'$ . $(fg)' = f'g + fg'$ .
4. Distribute and simplify. $(f'g + fg')h + fgh' = f'gh + fg'h + fgh'$ .
5. Pattern: in each of the 3 terms, exactly one of the three factors has been differentiated. For $n$ factors, there are $n$ terms.
Curiosity: this result is the generalized Leibniz rule. For $n$ factors, $(f_1 f_2 \cdots f_n)' = \sum_{k=1}^n f_1 \cdots f_k' \cdots f_n$ .
Ex. 52.40Proof
Proof. Prove the product rule $(fg)' = f'g + fg'$ from the definition of the derivative by limit.
Solve online Active Calculus · §2.3
Show solution
The proof starts from the definition: $(fg)'(x) = \lim_{h \to 0} \frac{f(x+h)g(x+h) - f(x)g(x)}{h}$ . Add and subtract $f(x+h)g(x)$ in the numerator, obtaining two terms: $[f(x+h)-f(x)]g(x+h)/h$ and $f(x)[g(x+h)-g(x)]/h$ . In the limit, the first converges to $f'(x)g(x)$ (using continuity of $g$ : $g(x+h) \to g(x)$ ) and the second to $f(x)g'(x)$ . Therefore $(fg)' = f'g + fg'$ . $\blacksquare$
Show step-by-step (with the why)
1. Write the definition. $(fg)'(x) = \lim_{h \to 0} \frac{f(x+h)g(x+h) - f(x)g(x)}{h}$ .
2. Algebraic trick. Add and subtract $f(x+h)g(x)$ in the numerator. The numerator becomes $[f(x+h)-f(x)]g(x+h) + f(x+h)[g(x+h)-g(x)]$ . (Note: we use $f(x+h)$ , not $f(x)$ , in the second group — this is correct and simplifies the separation.)
3. Separate the limits. The first tends to $f'(x) \cdot \lim_{h \to 0} g(x+h) = f'(x)g(x)$ . The second tends to $\lim_{h\to 0} f(x+h) \cdot g'(x) = f(x)g'(x)$ (by the continuity of $f$ ).
4. Conclusion. $(fg)'(x) = f'(x)g(x) + f(x)g'(x)$ . $\blacksquare$
Note: the trick of adding and subtracting an intermediate term is common in calculus proofs. Memorize the strategy: "introduce and cancel an auxiliary term to separate two independent limits".

Sources

Active Calculus 2.0 — Boelkins · 2024 · §2.1 (Elementary Rules), §2.2 (Sine and Cosine), §2.3 (Product and Quotient). Primary source. CC-BY-NC-SA.
OpenStax Calculus Volume 1 — OpenStax · 2016 · §3.3 (Differentiation Rules), §3.4 (Derivatives as Rates of Change), §3.5 (Derivatives of Trigonometric Functions). CC-BY-NC-SA.
APEX Calculus — Hartman et al. · 2023 · §2.3 (Basic Rules), §2.4 (Product and Quotient). CC-BY-NC.