Product rule — proof: Exercises — Maths deep dive

This page is a self-test ladder for Product rule — proof. Each rung gets harder: first you just recognise the rule, then apply it, then analyse trickier structures, then synthesise with other rules, and finally reach mastery where you prove things.

Every symbol used here was built in the parent note. Quick reminder of the tool we lean on:

Level 1 — Recognition

Here you only need to spot the two factors $u, v$ , find $u^{'}, v^{'}$ , and slot them in. No simplifying tricks.

L1.1 Differentiate $f (x) = x^{3} cos x$ .

Recall Solution L1.1

WHAT: name the factors. Let $u = x^{3}$ and $v = cos x$ . WHY: the function is a product of two things we already know how to differentiate (Power rule for $x^{3}$ , standard fact for $cos x$ ). $u^{'} = 3 x^{2}, v^{'} = - sin x$ Now slot into $(uv)^{'} = u^{'} v + u v^{'}$ : $f^{'} (x) = 3 x^{2} cos x + x^{3} (- sin x) = 3 x^{2} cos x - x^{3} sin x$

L1.2 Differentiate $g (x) = (2 x + 1) e^{x}$ .

Recall Solution L1.2

Let $u = 2 x + 1$ , $v = e^{x}$ . Then $u^{'} = 2$ , $v^{'} = e^{x}$ . $g^{'} (x) = 2 e^{x} + (2 x + 1) e^{x} = e^{x} (2 x + 3)$ The last step just factors out the common $e^{x}$ — optional but tidy.

L1.3 Differentiate $h (x) = x ln x$ for $x > 0$ .

Recall Solution L1.3

Let $u = x$ , $v = ln x$ . Then $u^{'} = 1$ , $v^{'} = \frac{1}{x}$ . $h^{'} (x) = 1 \cdot ln x + x \cdot \frac{1}{x} = ln x + 1$

Level 2 — Application

Now the factors are less obvious, or you must clean up the algebra afterward.

L2.1 Differentiate $f (x) = x^{2} e^{3 x}$ . (You may quote $(e^{3 x})^{'} = 3 e^{3 x}$ from Chain rule.)

Recall Solution L2.1

Let $u = x^{2}$ , $v = e^{3 x}$ . Then $u^{'} = 2 x$ , and $v^{'} = 3 e^{3 x}$ (chain rule on the inside $3 x$ ). $f^{'} (x) = 2 x e^{3 x} + x^{2} \cdot 3 e^{3 x} = e^{3 x} (3 x^{2} + 2 x)$

L2.2 Differentiate $g (x) = x sin x$ for $x > 0$ . Write $x = x^{1/2}$ .

Recall Solution L2.2

Let $u = x^{1/2}$ , $v = sin x$ . By the power rule $u^{'} = \frac{1}{2} x^{- 1/2}$ , and $v^{'} = cos x$ . $g^{'} (x) = \frac{1}{2} x^{- 1/2} sin x + x^{1/2} cos x = \frac{s i n x}{2 x} + x cos x$

L2.3 Find the slope of $y = x^{2} (x - 3)$ at $x = 1$ using the product rule (do not expand first).

Recall Solution L2.3

Let $u = x^{2}$ , $v = x - 3$ . Then $u^{'} = 2 x$ , $v^{'} = 1$ . $y^{'} = 2 x (x - 3) + x^{2} \cdot 1 = 2 x^{2} - 6 x + x^{2} = 3 x^{2} - 6 x$ At $x = 1$ : $y^{'} (1) = 3 (1) - 6 (1) = - 3$ . The slope is $- 3$ .

Level 3 — Analysis

Here you decide the structure: three factors, a product hiding inside a quotient, or reading a rule off a graph.

L3.1 Differentiate $f (x) = x e^{x} sin x$ (three factors).

Recall Solution L3.1

WHAT: the three-factor pattern from Example 3 in the parent — differentiate one factor at a time, keep the others fixed, sum the three results. $(uv w)^{'} = u^{'} v w + u v^{'} w + uv w^{'}$ With $u = x, v = e^{x}, w = sin x$ ( $u^{'} = 1, v^{'} = e^{x}, w^{'} = cos x$ ): $f^{'} (x) = 1 \cdot e^{x} sin x + x e^{x} sin x + x e^{x} cos x$ $= e^{x} (sin x + x sin x + x cos x)$

L3.2 A function is written $F (x) = \frac{x ^{2}}{cos x}$ . Rewrite it as a product $x^{2} (cos x)^{- 1}$ and differentiate using the product rule + chain rule (this is where the Quotient rule secretly comes from).

Recall Solution L3.2

Let $u = x^{2}$ , $v = (cos x)^{- 1}$ . Then $u^{'} = 2 x$ . For $v^{'}$ use the chain rule on $(cos x)^{- 1}$ : outer power $- 1$ , inner $cos x$ with derivative $- sin x$ : $v^{'} = - 1 \cdot (cos x)^{- 2} \cdot (- sin x) = \frac{s i n x}{c o s ^{2} x}$ Product rule: $F^{'} (x) = 2 x (cos x)^{- 1} + x^{2} \cdot \frac{s i n x}{c o s ^{2} x} = \frac{2 x}{c o s x} + \frac{x ^{2} s i n x}{c o s ^{2} x}$ Over a common denominator: $F^{'} (x) = \frac{2 x cos x + x ^{2} sin x}{cos ^{2} x}$ — exactly what the quotient rule gives.

L3.3 (geometric) The picture shows the growing-rectangle model. Width $u (x)$ and height $v (x)$ both increase. At the instant shown, $u = 4, v = 3, u^{'} = 2, v^{'} = 5$ (rates per unit $x$ ). How fast is the area $A = uv$ growing?

Recall Solution L3.3

WHAT IT LOOKS LIKE: in the figure the area grows by a right strip of rate $v u^{'}$ and a top strip of rate $u v^{'}$ ; the corner $Δ u Δ v$ is second-order tiny and does not contribute to the instantaneous rate. $A^{'} = u^{'} v + u v^{'} = (2) (3) + (4) (5) = 6 + 20 = 26$ The area is growing at $26$ square units per unit $x$ .

Level 4 — Synthesis

Combine the product rule with other rules and with problem-solving (tangent lines, roots).

L4.1 Find the equation of the tangent line to $y = x^{2} e^{x}$ at $x = 0$ .

Recall Solution L4.1

Point: $y (0) = 0^{2} \cdot e^{0} = 0$ , so the point is $(0, 0)$ . Slope: with $u = x^{2}, v = e^{x}$ : $y^{'} = 2 x e^{x} + x^{2} e^{x} = e^{x} (2 x + x^{2})$ . At $x = 0$ : $y^{'} (0) = e^{0} (0 + 0) = 0$ . Tangent line: through $(0, 0)$ with slope $0$ is $y = 0$ (the $x$ -axis). The curve touches flat there.

L4.2 For $f (x) = (x^{2} - 1) e^{x}$ , find all $x$ where $f^{'} (x) = 0$ .

Recall Solution L4.2

Let $u = x^{2} - 1, v = e^{x}$ . Then $u^{'} = 2 x$ , $v^{'} = e^{x}$ . $f^{'} (x) = 2 x e^{x} + (x^{2} - 1) e^{x} = e^{x} (x^{2} + 2 x - 1)$ Since $e^{x} > 0$ always, we need $x^{2} + 2 x - 1 = 0$ . By the quadratic formula: $x = \frac{- 2 \pm 4 + 4}{2} = - 1 \pm 2$ So $x = - 1 + 2 \approx 0.4142$ and $x = - 1 - 2 \approx - 2.4142$ .

L4.3 Let $f (x) = u (x) v (x)$ where at $x = 2$ we know $u (2) = 3, u^{'} (2) = - 1, v (2) = 5, v^{'} (2) = 4$ . Find $f^{'} (2)$ — you are not given formulas for $u, v$ , only their values.

Recall Solution L4.3

The product rule needs only the four numbers, not formulas: $f^{'} (2) = u^{'} (2) v (2) + u (2) v^{'} (2) = (- 1) (5) + (3) (4) = - 5 + 12 = 7$

Level 5 — Mastery

Prove and generalise. You are now the one wielding the rule, not the one being tested by it.

L5.1 Use the product rule to prove the power rule $(x^{4})^{'} = 4 x^{3}$ by writing $x^{4} = x \cdot x^{3}$ and assuming $(x^{3})^{'} = 3 x^{2}$ .

Recall Solution L5.1

Let $u = x, v = x^{3}$ , so $u^{'} = 1, v^{'} = 3 x^{2}$ (the assumed fact). $(x^{4})^{'} = (1) (x^{3}) + (x) (3 x^{2}) = x^{3} + 3 x^{3} = 4 x^{3}$ This is the induction step from Example 2: the product rule bootstraps the Power rule one degree at a time.

L5.2 Prove the general three-factor rule $(uv w)^{'} = u^{'} v w + u v^{'} w + uv w^{'}$ starting only from the two-factor product rule.

Recall Solution L5.2

WHAT: treat $uv w$ as a product of two blocks $(uv)$ and $w$ . WHY: the two-factor rule is all we are allowed; group so it applies once, then again to the inner product. $(uv w)^{'} = ((uv) w)^{'} = (uv)^{'} w + (uv) w^{'} (two-factor rule)$ Now expand the inner $(uv)^{'} = u^{'} v + u v^{'}$ : $= (u^{'} v + u v^{'}) w + uv w^{'} = u^{'} v w + u v^{'} w + uv w^{'} ■$

L5.3 Suppose $f (x) = [g (x)]^{2}$ where $g$ is differentiable. Using only the product rule (write $[g]^{2} = g \cdot g$ ), show $f^{'} (x) = 2 g (x) g^{'} (x)$ . This is the Chain rule for a square, derived without invoking it.

Recall Solution L5.3

Write $f = u \cdot v$ with $u = g$ and $v = g$ , so $u^{'} = g^{'}$ and $v^{'} = g^{'}$ . $f^{'} (x) = u^{'} v + u v^{'} = g^{'} \cdot g + g \cdot g^{'} = 2 g (x) g^{'} (x) ■$ Quick sanity check with $g (x) = x^{3}$ : $f = x^{6}$ , and $2 g g^{'} = 2 x^{3} \cdot 3 x^{2} = 6 x^{5} = (x^{6})^{'}$ . ✓

L5.4 (Leibniz preview) For $f = uv$ , the first derivative is $u^{'} v + u v^{'}$ . Differentiate again to find $f^{''} = (uv)^{''}$ in terms of $u, v$ and their derivatives. Notice the binomial-coefficient pattern that leads to the Leibniz rule (nth derivative of a product).

Recall Solution L5.4

Differentiate $f^{'} = u^{'} v + u v^{'}$ term by term, applying the product rule to each term: $(u^{'} v)^{'} = u^{''} v + u^{'} v^{'}, (u v^{'})^{'} = u^{'} v^{'} + u v^{''}$ Add them: $f^{''} = u^{''} v + u^{'} v^{'} + u^{'} v^{'} + u v^{''} = u^{''} v + 2 u^{'} v^{'} + u v^{''}$ The coefficients $1, 2, 1$ are the binomial coefficients $(0 2), (1 2), (2 2)$ — exactly the Leibniz rule pattern $f^{(n)} = k = 0 \sum n (k n) u^{(n - k)} v^{(k)}$ .

Recall Self-check summary (click after finishing)

L1.1 ::: $3 x^{2} cos x - x^{3} sin x$ L1.2 ::: $e^{x} (2 x + 3)$ L1.3 ::: $ln x + 1$ L2.1 ::: $e^{3 x} (3 x^{2} + 2 x)$ L2.2 ::: $\frac{sin x}{2 x} + x cos x$ L2.3 slope at $x = 1$ ::: $- 3$ L3.1 ::: $e^{x} (sin x + x sin x + x cos x)$ L3.3 area rate ::: $26$ L4.1 tangent ::: $y = 0$ L4.2 roots ::: $x = - 1 \pm 2$ L4.3 ::: $7$ L5.4 ::: $u^{''} v + 2 u^{'} v^{'} + u v^{''}$

Connections

Product rule — proof — the parent this page drills.
Power rule — bootstrapped in L5.1.
Chain rule — needed inside factors (L2.1, L3.2) and rederived in L5.3.
Quotient rule — reconstructed from product + chain in L3.2.
Limit definition of the derivative — the engine behind every rule here.
Differentiability implies continuity — the hidden lemma in the parent's proof.
Leibniz rule (nth derivative of a product) — previewed in L5.4.