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'

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 $g(x + \Delta x) \to g(x)$ ):

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

$\blacksquare$

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
Ex. 52.2Application
Calculate the derivative of $f(x) = 3x^2 - 4x + 1$ .
Solve online
Ex. 52.3ApplicationAnswer key
Calculate $(\sqrt{x})'$ . Hint: write as $x^{1/2}$ and apply R2.
Solve online
Ex. 52.4Application
Calculate $\left(\dfrac{1}{x^2}\right)'$ .
Solve online
Ex. 52.5ApplicationAnswer key
Calculate $f'(x)$ for $f(x) = 4x^5 - 3x^3 + 7x - 2$ .
Solve online
Ex. 52.6Application
Calculate $\left(-\dfrac{1}{x^3}\right)'$ .
Solve online
Ex. 52.7Application
Calculate $f'(x)$ for $f(x) = x^2\sqrt{x} - x\sqrt{x}$ .
Solve online
Ex. 52.8Application
Calculate $f'(x)$ for $f(x) = x^3 + 2x$ .
Solve online
Ex. 52.9Application
Calculate $\left(\dfrac{1}{\sqrt{x}}\right)'$ .
Solve online
Ex. 52.10ApplicationAnswer key
Calculate $g'(x)$ for $g(x) = 3x^4 - 3x^2$ .
Solve online
Ex. 52.11Application
Calculate $(\sin x \cdot \cos x)'$ .
Solve online
Ex. 52.12ApplicationAnswer key
Calculate $(x e^x)'$ .
Solve online
Ex. 52.13Application
Calculate $(x \ln x)'$ .
Solve online
Ex. 52.14Application
Calculate $(x^2 \sin x)'$ .
Solve online
Ex. 52.15ApplicationAnswer key
Calculate $(e^x \cos x)'$ .
Solve online
Ex. 52.16Application
Calculate $(x^2 e^x)'$ .
Solve online
Ex. 52.17ApplicationAnswer key
Calculate $(\tan x \cdot e^x)'$ .
Solve online
Ex. 52.18Application
Calculate $(x \sin x)'$ .
Solve online
Ex. 52.19Application
Calculate $(x^3 \ln x)'$ .
Solve online
Ex. 52.20Understanding
Generalization of the product rule. If $f$ , $g$ , $h$ are differentiable functions, what is $(fgh)'$ ?
Solve online
Ex. 52.21Application
Calculate $\left(\dfrac{\sin x}{x}\right)'$ for $x \neq 0$ .
Solve online
Ex. 52.22Application
Calculate $\left(\dfrac{e^x}{x}\right)'$ for $x \neq 0$ .
Solve online
Ex. 52.23Application
Calculate $\left(\dfrac{1}{x^2 + 1}\right)'$ .
Solve online
Ex. 52.24Application
Calculate $k'(x)$ for $k(x) = \dfrac{x^2 + 1}{x - 3}$ , $x \neq 3$ .
Solve online
Ex. 52.25Application
Calculate $\left(\dfrac{3x^2 - x}{x^2 - 1}\right)'$ for $x \neq \pm 1$ .
Solve online
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
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
Ex. 52.28Application
Calculate $\left(\dfrac{x}{x^2 + 1}\right)'$ .
Solve online
Ex. 52.29ModelingAnswer key
Find the equation of the tangent line to $f(x) = x^2 + 3x$ at the point $x = 1$ .
Solve online
Ex. 52.30ModelingAnswer key
At what points does the graph of $f(x) = x^3 - 3x$ have a horizontal tangent line?
Solve online
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
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
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
Ex. 52.34Modeling
Find the tangent line to $y = \dfrac{x}{x^2 + 1}$ at $x = 1$ .
Solve online
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
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.
Solve online
Ex. 52.37Understanding
Identify which differentiation rule applies to $h(x) = \dfrac{e^x}{x^2}$ , apply it, and simplify $h'(x)$ .
Solve online
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
Ex. 52.39Challenge
Challenge: product of three functions. Prove that $(fgh)' = f'gh + fg'h + fgh'$ , by applying the product rule twice.
Solve online
Ex. 52.40Proof
Proof. Prove the product rule $(fg)' = f'g + fg'$ from the definition of the derivative by limit.
Solve online

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.