Derivatives and rules (product, quotient, chain): Worked examples

Prerequisites you should have skimmed: Limits and Continuity (what $lim_{h \to 0}$ means) and Power Rule and Basic Derivatives (so $\frac{d}{d x} x^{n} = n x^{n - 1}$ is already yours).

The scenario matrix

Every example below is a tick in one of these boxes. Read the table first — it is the map. ("ML" below is short for machine learning — the field these derivatives power.)

#	Cell (the scenario)	Which rule stress-tested	Cleared by
C1	Product, evaluated at a specific point (plug in a number)	Product	Ex 1
C2	Product of THREE factors (rule used twice)	Product ×2	Ex 2
C3	Quotient at a normal point + sign of the slope	Quotient	Ex 3
C4	Quotient where the derivative is ZERO (turning point)	Quotient	Ex 4
C5a	Denominator → 0, removable hole (the $\frac{0}{0}$ that cancels)	Quotient limit	Ex 5
C5b	Denominator → 0, genuine blow-up (vertical asymptote, $\frac{nonzero}{0}$ )	Quotient limit	Ex 5b
C6	Chain, negative-inside & positive-inside (sign of inner)	Chain	Ex 6
C7	Chain nested THREE deep (three gears)	Chain ×2	Ex 7
C8	All three rules in one expression (machine-learning: sigmoid × input)	Product+Quotient+Chain	Ex 8
C9	Real-world word problem with units	Product	Ex 9
C10	Exam twist: implicit differentiation (curve not solved for $y$ )	Chain + algebra	Ex 10

Recall Quick reminders (open if a rule is fuzzy)

Product ::: $(uv)^{'} = u^{'} v + u v^{'}$ Quotient ::: $(\frac{u}{v})^{'} = \frac{u ^{'} v - u v ^{'}}{v ^{2}}$ Chain ::: $(f (g (x)))^{'} = f^{'} (g (x)) \cdot g^{'} (x)$

Ex 1 — Product at a point · cell C1

Worked example Statement

Let $f (x) = x^{3} cos x$ . Find $f^{'} (x)$ , then evaluate $f^{'} (0)$ .

Forecast: guess the sign of $f^{'} (0)$ before reading on. Is the slope at $x = 0$ positive, negative, or zero?

Step 1 — name the two factors. $u = x^{3}, v = cos x$ . Why this step? The function is a product of two things I can each differentiate — that is the exact trigger for the product rule.

Step 2 — differentiate each piece. $u^{'} = 3 x^{2}$ (power rule), $v^{'} = - sin x$ . Why this step? The product rule needs both derivatives before I can assemble it.

Step 3 — assemble $u^{'} v + u v^{'}$ . $f^{'} (x) = 3 x^{2} cos x + x^{3} (- sin x) = 3 x^{2} cos x - x^{3} sin x$ Why this step? Two rectangle strips: one from $x^{3}$ growing, one from $cos x$ changing.

Step 4 — plug in $x = 0$ . $f^{'} (0) = 3 (0)^{2} cos 0 - (0)^{3} sin 0 = 0 - 0 = 0$ Why this step? The question asked for the slope at a specific point, so after finding the general $f^{'} (x)$ I substitute that point in — a general formula is only useful once evaluated.

Verify: Near $x = 0$ , $x^{3}$ is flat and tiny (its slope $3 x^{2}$ is $0$ there) and it multiplies a bounded $cos$ , so the whole product barely moves — slope $0$ is exactly right. Forecast check: the answer was zero.

Ex 2 — Product of three factors · cell C2

Worked example Statement

Differentiate $g (x) = x e^{x} sin x$ .

Forecast: how many terms will the answer have — 2, 3, or 4? Guess before proceeding.

Step 1 — bracket it as (two things) × (one thing). Set $P = x e^{x}$ (a product) and treat $g = P \cdot sin x$ . Why this step? The product rule is built for two factors. So I nest: apply it once for the outer split, then again inside $P$ .

Step 2 — differentiate the inner product $P = x e^{x}$ . $P^{'} = (1) e^{x} + x e^{x} = e^{x} (1 + x)$ . Why this step? $P$ itself is a product, so it earns its own product rule.

Step 3 — outer product rule on $P \cdot sin x$ . $g^{'} = P^{'} sin x + P (sin x)^{'} = e^{x} (1 + x) sin x + x e^{x} cos x$ Why this step? Now that $P^{'}$ is known, $g = P \cdot sin x$ is just one product of two known pieces, so the standard two-strip product rule finishes it.

Step 4 — tidy. $g^{'} (x) = e^{x} [(1 + x) sin x + x cos x]$ Why this step? Factoring the common $e^{x}$ makes the structure visible and easier to check; not required but clean.

Verify: The general pattern for three factors $ab c$ is $a^{'} b c + a b^{'} c + ab c^{'}$ — three terms. Expanding our answer: $e^{x} sin x + x e^{x} sin x + x e^{x} cos x$ — exactly the three terms ( $a^{'} = 1$ , $b^{'} = e^{x}$ , $c^{'} = cos x$ ). Forecast: 3 terms.

Ex 3 — Quotient at a normal point · cell C3

Worked example Statement

$h (x) = \frac{2 x + 1}{x - 3}$ . Find $h^{'} (x)$ and its value at $x = 1$ .

Forecast: at $x = 1$ the denominator is $1 - 3 = - 2$ . Will the slope be positive or negative?

Step 1 — label top and bottom. $u = 2 x + 1, u^{'} = 2$ ; $v = x - 3, v^{'} = 1$ . Why this step? "Lo D-Hi minus Hi D-Lo" needs each part named.

Step 2 — apply the quotient rule. $h^{'} (x) = \frac{u ^{'} v - u v ^{'}}{v ^{2}} = \frac{2 ( x - 3 ) - ( 2 x + 1 ) ( 1 )}{( x - 3 ) ^{2}}$ Why this step? The minus sign encodes that a growing denominator shrinks the fraction.

Step 3 — simplify the numerator. $2 x - 6 - 2 x - 1 = - 7 \Rightarrow h^{'} (x) = \frac{- 7}{( x - 3 ) ^{2}}$ Why this step? The $x$ terms cancel — a hint the numerator is constant.

Step 4 — evaluate at $x = 1$ . $h^{'} (1) = \frac{- 7}{( 1 - 3 ) ^{2}} = \frac{- 7}{4} = - 1.75$ Why this step? We were asked for the slope at $x = 1$ , so the general derivative must be evaluated at that point to produce the single number the question wants.

Verify: The denominator $(x - 3)^{2}$ is always positive, and the numerator is the negative constant $- 7$ , so this function is decreasing everywhere it is defined. Forecast: negative, confirmed at $- 1.75$ .

Ex 4 — Quotient with a zero derivative (turning point) · cell C4 📈

Figure — Derivatives and rules (product, quotient, chain)

Worked example Statement

$f (x) = \frac{x}{x ^{2} + 1}$ (the sigmoid-ish bump from the parent note). Find where $f^{'} (x) = 0$ and classify those points.

Forecast: look at the figure — two special $x$ -values where the tangent goes flat. Guess them.

Step 1 — quotient rule. $u = x, u^{'} = 1$ ; $v = x^{2} + 1, v^{'} = 2 x$ . $f^{'} (x) = \frac{( 1 ) ( x ^{2} + 1 ) - ( x ) ( 2 x )}{( x ^{2} + 1 ) ^{2}} = \frac{1 - x ^{2}}{( x ^{2} + 1 ) ^{2}}$ Why this step? This matches the parent's example; here we go further and use it.

Step 2 — set the derivative to zero. A fraction is zero only when its numerator is zero (and denominator isn't): $1 - x^{2} = 0 \Rightarrow x = \pm 1$ Why this step? The flat tangent (amber dashes in the figure) is where slope $= 0$ ; solving the numerator finds exactly those $x$ -values.

Step 3 — classify each. At $x = - 1$ : just left, $1 - x^{2} < 0$ (slope $-$ ); just right, $1 - x^{2} > 0$ (slope $+$ ) → valley (min). At $x = + 1$ : slope goes $+ \to -$ → peak (max). Why this step? The sign of $f^{'}$ on each side tells you whether the curve turns up or down there.

Verify: $f (1) = \frac{1}{2}$ , $f (- 1) = - \frac{1}{2}$ — the amber dots in the figure sit at those heights. Forecast: turning points at $x = \pm 1$ .

Ex 5 — Denominator → 0, the removable hole · cell C5a

Worked example Statement

$q (x) = \frac{x ^{2} - 4}{x - 2}$ . Naïvely the quotient rule applies — but something special happens at $x = 2$ . Investigate.

Forecast: is $x = 2$ a genuine "blow-up to infinity", or a removable hole?

Step 1 — check the point directly. At $x = 2$ the bottom is $0$ and the top is $2^{2} - 4 = 0$ . So it is $\frac{0}{0}$ — indeterminate, not automatically infinite. Why this step? The quotient rule assumes $v \neq = 0$ . When $v = 0$ we must first ask what kind of zero this is (see Limits and Continuity).

Step 2 — factor and cancel. $q (x) = \frac{( x - 2 ) ( x + 2 )}{x - 2} = x + 2 (x \neq = 2)$ Why this step? The top shares the factor $(x - 2)$ , so the singularity cancels — it is a removable hole, not a pole.

Step 3 — differentiate the simplified form. For $x \neq = 2$ , $q (x) = x + 2$ , so $q^{'} (x) = 1$ everywhere it is defined. Why this step? Once simplified there is no division left; the derivative is trivially the slope of a line.

Step 4 — the limit at the hole. $lim_{x \to 2} q (x) = 2 + 2 = 4$ so $q$ behaves like the line $y = x + 2$ with a single missing point at $(2, 4)$ . Why this step? At $x = 2$ the function is not defined, so we cannot differentiate at the point — instead we ask what value it approaches. The limit tells us the hole could be "plugged" at height $4$ , which is what makes it removable rather than a true break.

Verify: Plug $x = 3$ into the original: $\frac{9 - 4}{3 - 2} = 5$ , and $x + 2 = 5$ ✓. The two forms agree away from the hole; the slope is a flat $1$ . Forecast: removable hole, not a blow-up.

Ex 5b — Denominator → 0, a genuine blow-up · cell C5b

Worked example Statement

$r (x) = \frac{x + 1}{x - 2}$ . What happens near $x = 2$ ? Contrast this with Ex 5.

Forecast: at $x = 2$ the top is $2 + 1 = 3$ (nonzero) and the bottom is $0$ . Hole, or infinite spike?

Step 1 — check the point directly. At $x = 2$ : top $= 3 \neq = 0$ , bottom $= 0$ . This is $\frac{3}{0}$ — a nonzero over zero, which does not cancel. Why this step? We repeat the Ex 5 diagnosis, but the outcome differs precisely because the numerator is not zero — nothing shares the $(x - 2)$ factor to cancel.

Step 2 — test the two sides of the singularity. Just above $x = 2$ (say $x = 2.01$ ): bottom is a tiny positive number, so $r \to + \infty$ . Just below $x = 2$ (say $x = 1.99$ ): bottom is a tiny negative number, so $r \to - \infty$ . Why this step? A "nonzero / tiny" ratio explodes; its sign flips with the sign of the tiny denominator, so we must check both sides to see the full behaviour.

Step 3 — conclude: vertical asymptote. $lim_{x \to 2^{+}} r (x) = + \infty, lim_{x \to 2^{-}} r (x) = - \infty$ The line $x = 2$ is a vertical asymptote — the curve shoots off in opposite directions on each side, and $r$ can never be "repaired" at $x = 2$ . Why this step? Unlike Ex 5, no finite limit exists, so there is no value to fill the gap — this is the genuine "where the rule breaks" case the matrix promised.

Step 4 — differentiate away from the asymptote. $u = x + 1, u^{'} = 1$ ; $v = x - 2, v^{'} = 1$ : $r^{'} (x) = \frac{( 1 ) ( x - 2 ) - ( x + 1 ) ( 1 )}{( x - 2 ) ^{2}} = \frac{- 3}{( x - 2 ) ^{2}}$ Why this step? The derivative exists for every $x \neq = 2$ ; the numerator $- 3$ being negative tells us $r$ is decreasing on both branches, but the derivative itself also blows up as $x \to 2$ (the tangents become vertical).

Verify: At $x = 3$ : $r = \frac{4}{1} = 4$ ; at $x = 1$ : $r = \frac{2}{- 1} = - 2$ — values on opposite branches, huge in magnitude near $x = 2$ , confirming the asymptote. And $r^{'} (3) = \frac{- 3}{1} = - 3 < 0$ ✓ decreasing. Forecast: infinite blow-up, not a hole.

Ex 6 — Chain rule, sign of the inside matters · cell C6

Worked example Statement

$y = 9 - x^{2}$ (the upper half of a circle of radius 3). Find $y^{'}$ at $x = + 2$ and at $x = - 2$ .

Forecast: on the right half of a hill the slope goes down, on the left half it goes up. Guess the two signs.

Step 1 — spot the nesting. Outer $f (u) = u = u^{1/2}$ , inner $u = 9 - x^{2}$ . Why this step? A function inside a square root is a composition — the chain rule's home turf.

Step 2 — differentiate each layer. $f^{'} (u) = \frac{1}{2} u^{- 1/2} = \frac{1}{2 u}$ ; inner $u^{'} = - 2 x$ . Why this step? Two gears: the root gear and the inside gear. I need both ratios.

Step 3 — multiply (chain). $y^{'} = \frac{1}{2 9 - x ^{2}} \cdot (- 2 x) = \frac{- x}{9 - x ^{2}}$ Why this step? Forgetting $\cdot (- 2 x)$ would drop the sign information entirely.

Step 4 — plug both signs of $x$ . $y^{'} (2) = \frac{- 2}{9 - 4} = \frac{- 2}{5} \approx - 0.894$ $y^{'} (- 2) = \frac{- ( - 2 )}{9 - 4} = \frac{2}{5} \approx + 0.894$ Why this step? The whole point of this cell is to see the sign flip: the $- x$ in the numerator means a positive $x$ gives a negative slope and vice-versa, so we must test both signs to confirm the formula behaves symmetrically.

Verify: On a dome, the right side slopes down (negative ✓) and the left side slopes up (positive ✓), mirror-symmetric — exactly what the $- x$ in the numerator guarantees. Forecast confirmed.

Ex 7 — Triple-nested chain (three gears) · cell C7

Worked example Statement

$y = sin ((3 x + 1)^{2})$ . Differentiate.

Forecast: how many factors will the chain produce — 2 or 3?

Step 1 — peel the layers, outside-in. Layer 1 (outer): $sin (\cdot)$ . Layer 2 (middle): $(\cdot)^{2}$ . Layer 3 (inner): $3 x + 1$ . Why this step? Each ( is a gear. Three shells → three gear ratios to multiply.

Step 2 — differentiate each shell, keeping the inside untouched.

Outer: $\frac{d}{d ( mid )} sin (mid) = cos ((3 x + 1)^{2})$
Middle: $\frac{d}{d ( inner )} (inner)^{2} = 2 (3 x + 1)$
Inner: $\frac{d}{d x} (3 x + 1) = 3$

Why this step? The chain rule says: differentiate outer at the inner, and carry the inners along untouched, layer by layer.

Step 3 — multiply all three. $y^{'} = cos ((3 x + 1)^{2}) \cdot 2 (3 x + 1) \cdot 3 = 6 (3 x + 1) cos ((3 x + 1)^{2})$ Why this step? Rates multiply (the gear-ratio insight from the parent): the total speed-up from $x$ all the way out is the product of every shell's individual rate, so we simply multiply the three factors and combine the constants $2 \cdot 3 = 6$ .

Verify: At $x = 0$ : inner $= 1$ , so $y^{'} = 6 (1) cos (1) \approx 6 (0.5403) \approx 3.242$ . The count of factors is 3 — one per shell. Forecast: 3 factors.

Ex 8 — All three rules together (machine learning) · cell C8

Worked example Statement

The building block $L (x) = x \cdot σ (x)$ where $σ (x) = \frac{1}{1 + e ^{- x}}$ is the SiLU / swish activation. Find $L^{'} (x)$ .

Forecast: the answer should contain $σ$ and $σ (1 - σ)$ . Guess how they combine.

Step 1 — outer structure is a product. $u = x, u^{'} = 1$ ; $v = σ (x)$ , and from the parent note $v^{'} = σ (x) (1 - σ (x))$ (that itself used chain + quotient). Why this step? $L$ is input × activation — a product, so the product rule frames it.

Step 2 — apply the product rule. $L^{'} (x) = (1) σ (x) + x \cdot σ (x) (1 - σ (x))$ Why this step? Two strips: one from $x$ growing, one from the activation changing.

Step 3 — factor $σ$ . $L^{'} (x) = σ (x) [1 + x (1 - σ (x))]$ Why this step? Pulling the common $σ (x)$ out gives the compact form a library like PyTorch actually stores; it feeds straight into Backpropagation.

Step 4 — sanity numbers at $x = 0$ . $σ (0) = \frac{1}{2}$ , so $L^{'} (0) = \frac{1}{2} [1 + 0] = \frac{1}{2} = 0.5$ . Why this step? We pick $x = 0$ because $σ (0) = \frac{1}{2}$ is a value we know exactly and the $x (1 - σ)$ term vanishes — the simplest possible input to catch an algebra slip before trusting the general formula.

Verify: At $x = 0$ , $L (x) = x σ (x) \approx x \cdot \frac{1}{2}$ near the origin, so the slope should be about $\frac{1}{2}$ — matches. Forecast: product rule glued $σ$ to $σ (1 - σ)$ , confirmed.

Ex 9 — Real-world word problem with units · cell C9

Worked example Statement

A rectangular solar panel is expanding as it heats. Its width is $w (t) = 2 + 0.1 t$ metres and its height is $h (t) = 1 + 0.05 t$ metres, where $t$ is in seconds. How fast is the area growing at $t = 10 s$ ?

Forecast: both sides grow, so the area rate should be bigger than either side's rate alone. Guess above or below $0.5 m^{2} / s$ .

Step 1 — area is a product. $A (t) = w (t) h (t)$ . So $A^{'} (t) = w^{'} h + w h^{'}$ — the product rule is the "two growing strips" from the parent. Why this step? Area of a rectangle whose both sides move gains a strip on each side.

Step 2 — the rates. $w^{'} = 0.1 m/s$ , $h^{'} = 0.05 m/s$ (slopes of the linear widths). Why this step? The product rule needs each side's rate of change; since $w$ and $h$ are straight lines in $t$ , their derivatives are just the constant slopes $0.1$ and $0.05$ .

Step 3 — evaluate the sizes at $t = 10$ . $w (10) = 2 + 1 = 3 m$ , $h (10) = 1 + 0.5 = 1.5 m$ . Why this step? The strips' lengths depend on the current side lengths.

Step 4 — assemble with units. $A^{'} (10) = w^{'} h (0.1) (1.5) + w h^{'} (3) (0.05) = 0.15 + 0.15 = 0.30 \frac{m ^{2}}{s}$ Why this step? Plugging the four numbers into $w^{'} h + w h^{'}$ combines both growing strips into the single area-rate the question asked for; the question wants the rate at $t = 10$ , so we use the sizes computed in Step 3.

Verify units: $(m/s) (m) = m^{2} / s$ ✓ — a rate of area, correct. Value $0.30 m^{2} / s$ is below $0.5$ ; the forecast "both sides grow → bigger than one side" holds since each strip is $0.15$ , together $0.30$ .

Ex 10 — Exam twist: implicit differentiation · cell C10

Worked example Statement

A point moves on the circle $x^{2} + y^{2} = 25$ . The curve is not solved for $y$ — you cannot write a single clean $y = f (x)$ for the whole circle. Find the slope $\frac{d y}{d x}$ of the tangent at the point $(3, 4)$ .

Forecast: at $(3, 4)$ we are on the upper-right of the circle. Will the tangent slope be positive or negative? Guess before solving.

Step 1 — differentiate BOTH sides with respect to $x$ , treating $y$ as a function of $x$ . $\frac{d}{d x} (x^{2}) + \frac{d}{d x} (y^{2}) = \frac{d}{d x} (25)$ Why this step? We can't isolate $y$ first, so we differentiate the equation as it stands. Because $y$ secretly depends on $x$ , every $y$ -term needs the chain rule — this is exactly what "implicit differentiation" means.

Step 2 — apply the chain rule to the $y^{2}$ term. $\frac{d}{d x} (x^{2}) = 2 x$ , and $\frac{d}{d x} (y^{2}) = 2 y \cdot \frac{d y}{d x}$ (outer $u^{2}$ gives $2 y$ , times inner rate $\frac{d y}{d x}$ ). The constant $25$ has derivative $0$ : $2 x + 2 y \frac{d y}{d x} = 0$ Why this step? The extra $\frac{d y}{d x}$ factor is the whole point — forgetting it is the classic implicit-differentiation error.

Step 3 — solve algebraically for $\frac{d y}{d x}$ . $2 y \frac{d y}{d x} = - 2 x \Rightarrow \frac{d y}{d x} = - \frac{x}{y}$ Why this step? We want the slope alone, so we isolate $\frac{d y}{d x}$ by moving the $2 x$ across and dividing by $2 y$ . The answer is a formula in both $x$ and $y$ — normal for implicit curves.

Step 4 — plug in the point $(3, 4)$ . $\frac{d y}{d x}_{(3, 4)} = - \frac{3}{4} = - 0.75$ Why this step? The slope formula depends on where you are on the circle, so we substitute the specific point $(3, 4)$ to get the single tangent slope the question asks for.

Verify: The upper half of the circle solved explicitly is $y = 25 - x^{2}$ ; by the Ex 6 method $y^{'} = \frac{- x}{25 - x ^{2}}$ , and at $x = 3$ that is $\frac{- 3}{16} = \frac{- 3}{4} = - 0.75$ ✓ — the implicit and explicit routes agree. A radius to $(3, 4)$ has slope $\frac{4}{3}$ , and the tangent (perpendicular) has slope $- \frac{3}{4}$ , the negative reciprocal ✓. Forecast: negative, confirmed.

Coverage check

Recall Did we hit every matrix cell?

C1 (product@point) ::: Ex 1 C2 (three-factor product) ::: Ex 2 C3 (quotient sign) ::: Ex 3 C4 (quotient zero-slope) ::: Ex 4 C5a (denominator→0 removable hole) ::: Ex 5 C5b (denominator→0 genuine blow-up) ::: Ex 5b C6 (chain, inside sign) ::: Ex 6 C7 (triple nested chain) ::: Ex 7 C8 (all three, machine learning) ::: Ex 8 C9 (word problem + units) ::: Ex 9 C10 (exam twist, implicit differentiation) ::: Ex 10

Onward: these hand-derivatives are exactly the numbers that flow through Gradient and Partial Derivatives and get chained in Backpropagation to power Gradient Descent.

Connections

Derivatives and rules (product, quotient, chain) (Hinglish) — parent topic
Limits and Continuity — the $\frac{0}{0}$ and $\frac{nonzero}{0}$ cases (Ex 5, Ex 5b)
Power Rule and Basic Derivatives — used in every example
Activation Functions — Ex 8 SiLU/swish
Backpropagation, Gradient Descent, Gradient and Partial Derivatives — where these derivatives go