Single-Variable-Calculus · Unit 3 · Video 6 · Interactive Practice

Finishing the Geometric Proof: One Rotation, Then the Product and Quotient Rules

IKey Formulas

Formula	Name	Where it comes from
$\angle QPR = \theta + \tfrac{\Delta\theta}{2} \approx \theta$	The angle at $P$	Turn the angle $\theta$ at $O$ through $90\degree$
$\Delta y \approx \Delta\theta\cos\theta \;\Longrightarrow\; \lim\limits_{\Delta\theta \to 0}\dfrac{\Delta y}{\Delta\theta} = \cos\theta$	Derivative of sine, geometrically	Adjacent $=$ hypotenuse $\times \cos$ in $\triangle PQR$
$(uv)' = u'v + uv'$	Product rule	The two strips of a growing rectangle
$\left(\dfrac{u}{v}\right)' = \dfrac{u'v - uv'}{v^2}, \quad v \neq 0$	Quotient rule	The same two terms, joined by a minus

Key Insight: Both new rules change one function at a time. In the rectangle picture the strips $u\,\Delta v$ and $v\,\Delta u$ become the two terms of $(uv)'$ , while the corner $\Delta u\,\Delta v$ dies faster than $\Delta x$ — which is why the product $u'v'$ never appears in either rule.

IIVisualization 1 — One Rotation Names the Angle at $P$

Turning the angle at $O$ through a right angle carries it exactly onto the angle at $P$ .

Angle θ 40° Δθ 20° Rotation 0°

💡 Small-change triangles like $PQR$ return in multivariable calculus, and cutting a curve into such triangles is how arc length is built.

IIIVisualization 2 — The Difference Quotient Becomes $\cos\theta$

Shrinking $\Delta\theta$ drives the difference quotient of $\sin\theta$ onto the graph of $\cos\theta$ .

Δθ (rad) 0.500

IVVisualization 3 — Two Strips and a Vanishing Corner

Growing $u$ and $v$ one at a time adds two strips; the corner is the piece the product rule discards.

Δu 0.60 Δv 0.40

💡 The quotient rule is proved the same way, one function at a time — and it carries the extra condition $v \neq 0$ , since $v^2$ sits in the denominator.

VQuiz Questions

Problem 1 · Reading the Right Triangle

Given: in $\triangle PQR$ the right angle is at $R$ , the hypotenuse $PQ$ has length $\approx \Delta\theta$ , and the angle at $P$ is $\approx \theta$ . The vertical side $PR$ , of length $\Delta y$ , is the side next to that angle. Which relation follows?

✅ Correct! Adjacent

=

hypotenuse

\times \cos(\text{angle})

, and dividing by

\Delta\theta

gives

\Delta y/\Delta\theta \approx \cos\theta

❌ That is the other leg.

\sin\theta

multiplies the hypotenuse to give the side opposite the angle at

P

— the horizontal side

RQ

, not

PR

❌ The cosine multiplies, it does not divide.

\cos\theta = \dfrac{\text{adjacent}}{\text{hypotenuse}} = \dfrac{\Delta y}{\Delta\theta}

, so

\Delta y = \Delta\theta\cos\theta

❌ The hypotenuse is missing.

\cos\theta

alone is a ratio, not a length; it has to be scaled by the hypotenuse

\Delta\theta

❌ Not quite. Write the definition of cosine in this triangle first:

\cos(\angle QPR) = PR/PQ

Show solution

In a right triangle the cosine of an angle is the adjacent side over the hypotenuse. At $P$ :

\cos(\angle QPR) = \frac{PR}{PQ} \approx \frac{\Delta y}{\Delta\theta}

With $\angle QPR \approx \theta$ this rearranges to

\Delta y \approx \Delta\theta\cos\theta

Dividing by $\Delta\theta$ turns the geometry into the difference quotient of the sine, $\dfrac{\Delta y}{\Delta\theta} \approx \cos\theta$ , and letting $\Delta\theta \to 0$ makes both approximations exact:

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

Problem 2 · Product Rule, First Use

Given: $f(x) = x^2 \sin x$ — differentiate it.

✅ Correct! Each factor is differentiated once, in its own term, while the other factor is left alone.

❌ That is $u'v'$ . The derivative of a product is never the product of the derivatives — the vanishing corner

\Delta u\,\Delta v

is exactly the piece that would have produced such a term.

❌ The minus belongs to the quotient rule. The product rule joins its two terms with a plus: both strips add area to the rectangle.

❌ The derivatives are on the wrong factors. In each term exactly one factor wears the prime:

u'v

then

uv'

❌ Not quite. Take

u = x^2

and

v = \sin x

, so

u' = 2x

and

v' = \cos x

, and assemble

u'v + uv'

Show solution

Set $u = x^2$ and $v = \sin x$ , so $u' = 2x$ and $v' = \cos x$ . The product rule gives

(uv)' = u'v + uv' = (2x)(\sin x) + (x^2)(\cos x)

f'(x) = 2x\sin x + x^2\cos x

Check the shape of the answer: each term contains one primed factor and one unprimed factor. A term with two primes ( $2x\cos x$ ) or with no prime ( $x^2\sin x$ ) cannot belong to the rule.

Problem 3 · Both Rules at a Single Point

Given: $u(2) = 3$ , $u'(2) = -1$ , $v(2) = 4$ , $v'(2) = 5$ — evaluate both rules at $x = 2$ .

What is $(uv)'(2)$ ?

What is $\left(\dfrac{u}{v}\right)'(2)$ ?

✅ Correct! The same two products,

u'v = -4

and

uv' = 15

, are added for the product rule and subtracted (then divided by

v^2 = 16

) for the quotient rule.

❌ That is $u'(2)\,v'(2)$ . The product of the derivatives is never a term of either rule.

❌ A sign was dropped.

u'(2) = -1

, so

u'v = (-1)(4) = -4

, not

+4

❌ The numerator is the wrong way round. The derivative of the top function comes first:

u'v - uv'

, not

uv' - u'v

❌ That is the product rule's numerator. The quotient rule joins the same two terms with a minus sign.

❌ Check the denominator. It is

v^2 = 4^2 = 16

, not

v = 4

❌ Check the product rule.

(uv)' = u'v + uv'

, so evaluate

(-1)(4) + (3)(5)

❌ Check the quotient rule.

\left(\frac{u}{v}\right)' = \frac{u'v - uv'}{v^2}

, so evaluate

\frac{(-1)(4) - (3)(5)}{4^2}

Show solution

Both rules are built from the same two products:

u'(2)\,v(2) = (-1)(4) = -4, \qquad u(2)\,v'(2) = (3)(5) = 15

Product rule:

(uv)'(2) = u'v + uv' = -4 + 15 = 11

Quotient rule (the top function is differentiated first, and the denominator is squared):

\left(\frac{u}{v}\right)'(2) = \frac{u'v - uv'}{v^2} = \frac{-4 - 15}{4^2} = \frac{-19}{16}

Since $v(2) = 4 \neq 0$ , the quotient rule applies at this point.

Problem 4 · The Same Triangle Gives the Derivative of Cosine

Given: the same triangle, now read horizontally. The side $RQ$ is opposite the angle $\theta$ at $P$ , and in the first quadrant $Q$ lies to the left of $P$ , so $\Delta x = \cos(\theta + \Delta\theta) - \cos\theta$ is negative. Find the derivative of cosine.

How long is the horizontal side $RQ$ ?

So what is $\lim\limits_{\Delta\theta \to 0} \dfrac{\Delta x}{\Delta\theta}$ ?

✅ Correct! One triangle yields both trigonometric derivatives: its vertical leg gives

\cos x

, its horizontal leg gives

-\sin x

❌ That is the vertical leg.

\cos\theta

scales the side adjacent to the angle at

P

, which is

PR = \Delta y

. The opposite side takes

\sin\theta

❌ The length is right, the sign is not.

RQ

is a length, but

\Delta x

is a displacement to the left, so

\Delta x = -\Delta\theta\sin\theta

❌ Check which leg you are measuring. Opposite

=

hypotenuse

\times \sin(\text{angle})

, with hypotenuse

\approx \Delta\theta

and angle

\approx \theta

❌ Check the sign. Moving counterclockwise in the first quadrant pushes

Q

to the left of

P

, so

\Delta x < 0

while

\Delta\theta > 0

Show solution

In $\triangle PQR$ the angle at $P$ is $\approx \theta$ and the hypotenuse $PQ$ is $\approx \Delta\theta$ . The horizontal side $RQ$ is opposite that angle, so

RQ \approx \Delta\theta \sin\theta

Q sits to the left of $P$ , so the signed horizontal change is $\Delta x \approx -\Delta\theta\sin\theta$ . Dividing by $\Delta\theta$ and letting $\Delta\theta \to 0$ :

\lim_{\Delta\theta \to 0}\frac{\Delta x}{\Delta\theta} = -\sin\theta \qquad\Longrightarrow\qquad \frac{d}{dx}\cos x = -\sin x

This matches the formula proved earlier from the two limits at zero, so the geometric picture reproduces both trigonometric derivatives at a general angle.

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Finishing the Geometric Proof: One Rotation, Then the Product and Quotient Rules

IKey Formulas

IIVisualization 1 — One Rotation Names the Angle at PPP

IIIVisualization 2 — The Difference Quotient Becomes cos⁡θ\cos\thetacosθ

IVVisualization 3 — Two Strips and a Vanishing Corner

VQuiz Questions

IIVisualization 1 — One Rotation Names the Angle at $P$

IIIVisualization 2 — The Difference Quotient Becomes $\cos\theta$