Single-Variable-Calculus · Unit 4 · Video 1 · Interactive Practice

The Product Rule: Statement, Example, and Proof

IKey Formulas

Formula	Name	What it says
$(uv)' = u'v + uv'$	Product rule	Each factor is varied once while the other is held fixed, and the two contributions are added
$\dfrac{d}{dx}\big(x^n \sin x\big) = n x^{n-1}\sin x + x^n \cos x$	The worked example	$u = x^n$ and $v = \sin x$ , so $u' = nx^{n-1}$ and $v' = \cos x$
$\Delta(uv) = \Delta u \cdot v(x + \Delta x) + u(x)\,\Delta v$	Add-and-subtract identity	Add and subtract $u(x)\,v(x + \Delta x)$ , then factor $u(x)$ out of the last two terms
$\dfrac{\Delta(uv)}{\Delta x} = \dfrac{\Delta u}{\Delta x}\,v(x + \Delta x) + u(x)\,\dfrac{\Delta v}{\Delta x}$	Difference quotient	Letting $\Delta x \to 0$ needs $v(x + \Delta x) \to v(x)$

Key Insight: A product differentiates into a sum, and the false guess $(uv)' = u'v'$ never appears: the only $\Delta u\,\Delta v$ in sight is swallowed by the factor $v(x + \Delta x)$ , which collapses to $v(x)$ in the limit.

IIThe Rule on $x^n \sin x$

The tangent slope is $u'v + uv'$ , never the product $u'v'$ of the two derivatives.

x_0

1.20

IIIWhere the Two Terms Come From

Growing $x$ by $\Delta x$ grows the rectangle $uv$ by an L-shaped region made of exactly two pieces.

Step 1

Step 1 — The change in the product

\Delta(uv) = u(x + \Delta x)\,v(x + \Delta x) - u(x)\,v(x)

Step 2 — Make $\Delta u$ appear

\big[u(x + \Delta x) - u(x)\big]\,v(x + \Delta x)

Step 3 — That bracket subtracted $u(x)v(x + \Delta x)$ , so add it back

+\; u(x)\,v(x + \Delta x)

Step 4 — The original $-\,u(x)v(x)$ is still owed

-\; u(x)\,v(x)

Step 5 — The last two terms share $u(x)$

u(x)\,v(x + \Delta x) - u(x)\,v(x) = u(x)\big[v(x + \Delta x) - v(x)\big]

Step 6 — The add-and-subtract identity

\Delta(uv) = \big[u(x + \Delta x) - u(x)\big]v(x + \Delta x) + u(x)\big[v(x + \Delta x) - v(x)\big]

\Delta(uv) = \Delta u \cdot v(x + \Delta x) + u(x)\,\Delta v

IVThe Limit $\Delta x \to 0$

Divided by $\Delta x$ , each term settles onto its own limit, and the two add to $(uv)'$ .

Δx 1.200

💡 The step $v(x + \Delta x) \to v(x)$ is not bookkeeping — it is the continuity of $v$ , which holds because $v$ is differentiable (Lecture 2). Group the brackets the other way and it is the continuity of $u$ that the same proof needs.

VQuiz Questions

Problem 1 · Applying the Rule

Given: $f(x) = x^4 \sin x$ — find $f'(x)$ .

✅ Correct! With

u = x^4

and

v = \sin x

u'v = 4x^3\sin x

differentiates the first and keeps the second, and

uv' = x^4\cos x

keeps the first and differentiates the second.

❌ That is $u'v'$ . Differentiating both factors at once is the wrong guess the video names: a product differentiates into a sum of two terms, not a single product of derivatives.

❌ Check the sign. The product rule adds its two terms. The minus sign belongs to the quotient rule and to

\frac{d}{dx}\cos x

— neither is in play here.

❌ The factors are crossed. Each term pairs a differentiated factor with the undifferentiated other one, so

x^4

must sit next to

\cos x

, not next to

\sin x

Show solution

Name the pieces, then differentiate each:

u = x^4, \quad u' = 4x^3, \qquad v = \sin x, \quad v' = \cos x

Apply $(uv)' = u'v + uv'$ — derivative of the first times the second as it stands, plus the first as it stands times the derivative of the second:

f'(x) = \underbrace{4x^3 \cdot \sin x}_{u'v} + \underbrace{x^4 \cdot \cos x}_{uv'} = 4x^3\sin x + x^4\cos x

This is the video's $\frac{d}{dx}(x^n\sin x) = nx^{n-1}\sin x + x^n\cos x$ with $n = 4$ .

Problem 2 · The Cosine's Minus Sign

Given: $g(x) = x^3 \cos x$ — find $g'(x)$ .

✅ Correct! The product rule still adds its two terms; the minus arrives inside the second one, from

v' = (\cos x)' = -\sin x

❌ The second factor's derivative is missing its sign.

\frac{d}{dx}\cos x = -\sin x

, so

uv' = x^3 \cdot (-\sin x) = -x^3\sin x

❌ That is $u'v'$ . You multiplied

3x^2

-\sin x

. The rule never multiplies the two derivatives together — it pairs each one with the other factor left alone.

❌ The trig functions are swapped. The undifferentiated

v

\cos x

and the differentiated

v'

-\sin x

, so

\cos x

belongs with

3x^2

and

\sin x

with

x^3

Show solution

Here $u = x^3$ and $v = \cos x$ , so

u' = 3x^2, \qquad v' = -\sin x

and the rule gives

g'(x) = u'v + uv' = 3x^2 \cdot \cos x + x^3 \cdot (-\sin x) = 3x^2\cos x - x^3\sin x

The rule itself is unchanged — it is $v'$ , not the product rule, that supplies the minus sign.

Problem 3 · Inside the Proof

Given: the change $\Delta(uv) = u(x + \Delta x)\,v(x + \Delta x) - u(x)\,v(x)$ , rewritten so that $\Delta u$ and $\Delta v$ both appear — identify the two moves that make the proof work.

Which product is added and subtracted?

Which limit is the one that needs the continuity of $v$ ?

✅ Correct! Adding and subtracting

u(x)v(x + \Delta x)

splits the change into

\Delta u \cdot v(x + \Delta x) + u(x)\,\Delta v

, and only the leftover factor

v(x + \Delta x)

has to be moved to

v(x)

by continuity.

❌ That product is already there.

-\,u(x)v(x)

is the "old value" in the original change, so adding and subtracting it would change nothing.

❌ Wrong pairing. The first bracket is multiplied by

v

at the shifted point,

v(x + \Delta x)

, so the term introduced by expanding it is

u(x)v(x + \Delta x)

❌ No corner term is ever added. In this grouping

\Delta u\,\Delta v

is already inside

\Delta u \cdot v(x + \Delta x)

, because that factor is

v

at the taller, shifted height.

❌ That limit is the definition of a derivative.

\Delta u / \Delta x \to du/dx

holds because

u

is differentiable — no continuity argument is needed to pass to it.

❌ That limit is the definition of a derivative.

\Delta v / \Delta x \to dv/dx

is just

v

being differentiable. The delicate factor is the one with no

\Delta x

underneath it.

❌ That is the left-hand side.

\Delta(uv)/\Delta x \to d(uv)/dx

is the definition of the derivative being computed — it is the conclusion, not the step that needs justifying.

Show solution

The added and subtracted product. To make $\Delta u$ appear, write the first bracket against $v(x + \Delta x)$ :

\big[u(x + \Delta x) - u(x)\big]v(x + \Delta x) = u(x + \Delta x)v(x + \Delta x) - u(x)v(x + \Delta x)

The second piece was not in the original change, so add $u(x)v(x + \Delta x)$ back; the original $-\,u(x)v(x)$ is still owed, so subtract it as well. The last two terms share $u(x)$ :

u(x)v(x + \Delta x) - u(x)v(x) = u(x)\big[v(x + \Delta x) - v(x)\big]

\Longrightarrow \quad \Delta(uv) = \Delta u \cdot v(x + \Delta x) + u(x)\,\Delta v

The delicate limit. Dividing by $\Delta x$ puts one $\Delta x$ under $\Delta u$ and one under $\Delta v$ :

\frac{\Delta(uv)}{\Delta x} = \frac{\Delta u}{\Delta x}\,v(x + \Delta x) + u(x)\,\frac{\Delta v}{\Delta x}

Two of the three limits are definitions of derivatives. The remaining factor $v(x + \Delta x)$ carries no $\Delta x$ underneath, and sending it to $v(x)$ is exactly the continuity of $v$ — available because a differentiable function is continuous.

Problem 4 · The Example at a Point

Given: $f(x) = x^n \sin x$ for a positive integer $n$ — find $f'(\pi)$ .

✅ Correct! At

x = \pi

the first term dies because

\sin\pi = 0

, and the second survives as

\pi^n\cos\pi = -\pi^n

❌ Only the first term vanishes.

f(\pi) = 0

and

\sin\pi = 0

kills

nx^{n-1}\sin x

, but the curve is still moving: the term

x^n\cos x

is nonzero there.

❌ Check the sign of the cosine.

\cos\pi = -1

, not

+1

, so the surviving term is

\pi^n \cdot (-1)

❌ You used $\sin\pi = 1$ . It is

\sin\frac{\pi}{2}

that equals

1

;

\sin\pi = 0

, so the whole first term drops out.

Show solution

Differentiate first, with $u = x^n$ and $v = \sin x$ :

f'(x) = nx^{n-1}\sin x + x^n\cos x

Now substitute $x = \pi$ , where $\sin\pi = 0$ and $\cos\pi = -1$ :

f'(\pi) = n\pi^{n-1}\cdot 0 + \pi^n \cdot (-1) = -\pi^n

Each term must be evaluated separately: the product rule produces a sum, so one term vanishing does not make the derivative vanish. Checking $n = 1$ : $f(x) = x\sin x$ , $f'(x) = \sin x + x\cos x$ , and $f'(\pi) = 0 - \pi = -\pi$ . ✓

Solved: 0 / 4

The Product Rule: Statement, Example, and Proof

IKey Formulas

IIThe Rule on xnsin⁡xx^n \sin xxnsinx

IIIWhere the Two Terms Come From

IVThe Limit Δx→0\Delta x \to 0Δx→0

VQuiz Questions

IIThe Rule on $x^n \sin x$

IVThe Limit $\Delta x \to 0$