Product Rule and Quotient Rule

A rectangle u wide and v tall. Move x on by h and it gains a strip down the right side, a strip along the top and a small corner. Shrink h, divide each piece by h, and see which of the three survive.

The product rule	$(uv)^{'} = u^{'} v + u v^{'}$
The quotient rule	$(\frac{u}{v})^{'} = \frac{u ^{'} v - u v ^{'}}{v ^{2}}, v \neq = 0$

Worked example — Product Rule and Quotient Rule

Area at x	6
Strip on the right	3
Strip on top	4
Corner	2
Area gained, all three	9
u′ times v	3
u times v′	4
The product rule, (uv)′	7

These are the values the calculator shows when the page opens. Change any input above to recompute them.

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Lesson

The theory — Product Rule and Quotient Rule

If u and v are differentiable at x, the derivative of their product is u′v + uv′, and of their quotient (u′v − uv′)/v² wherever v ≠ 0. Both come out of one picture: a rectangle u wide and v tall that grows as x grows.

What each symbol means

u: the rectangle's width, a differentiable function of x
v: its height, another
h: the step in x; Δu = u(x + h) − u(x), and Δv likewise

Where the formula comes from

The rectangle grows. Move x on by h. The width becomes u + Δu and the height v + Δv, and the new area minus the old splits into three pieces: a strip down the right, a strip along the top, and the corner. They are the blue, amber and crimson pieces in the picture.
$(u + Δ u) (v + Δ v) - uv = Δ u v + u Δ v + Δ u Δ v$
Divide by h and let h shrink. Δu/h tends to u′ and Δv/h to v′, so the two strips give u′v + uv′. The corner is (Δu/h)·Δv, a rate times a change, and the change Δv goes to 0 because a function differentiable at x is continuous there. So the corner's line in the second panel falls to zero while the strips level off, and the tempting answer u′v′ is the one piece that disappears.
$\frac{Δ A}{h} = \frac{Δ u}{h} v + u \frac{Δ v}{h} + \to u^{'} \frac{Δ u}{h} \to 0 Δ v ⟶ u^{'} v + u v^{'}$
The quotient from the product. Call the quotient q = u/v, so that u = q·v. Differentiate that product with the rule just found and solve for q′. The minus sign is the top strip qv′, moved to the other side of the equation.
$u = q v \Rightarrow u^{'} = q^{'} v + q v^{'} \Rightarrow q^{'} = \frac{u ^{'} - q v ^{'}}{v} = \frac{u ^{'} v - u v ^{'}}{v ^{2}}$

Assumes: The picture needs both sides positive. The rules do not: they hold for any u and v differentiable at x, with v(x) ≠ 0 for the quotient.

Problems solved in full

Two polynomials, and the tangent at x = 2 4 steps
Differentiate f(x) = (x² + 1)(3x − 2), check the result by multiplying out first, and find the tangent line at x = 2.

Open this problem in the tool
1. Name the two factors and differentiate each. Nothing here needs more than the power rule:
  
  $u = x^{2} + 1, u^{'} = 2 x; v = 3 x - 2, v^{'} = 3$
2. Assemble u′v + uv′ and collect like terms:
  
  $f^{'} (x) = 2 x (3 x - 2) + 3 (x^{2} + 1) = 9 x^{2} - 4 x + 3$
3. Multiply out first and differentiate term by term. The same polynomial comes back, which is the honest check on the rule:
  
  $f (x) = 3 x^{3} - 2 x^{2} + 3 x - 2 \Rightarrow f^{'} (x) = 9 x^{2} - 4 x + 3$
4. At x = 2 the curve stands at 20 and climbs at 31, so the tangent passes through (2, 20) with slope 31:
  
  $f (2) = 5 \cdot 4 = 20, f^{'} (2) = 36 - 8 + 3 = 31, y - 20 = 31 (x - 2)$
Answer

f′(x) = 9x² − 4x + 3, and the tangent at x = 2 is y = 31x − 42. The page prints the rule as 31 at x = 2: 16 from the strip on the right, u′v = 4 × 4, and 15 from the strip on top, uv′ = 5 × 3.

$f^{'} (x) = 9 x^{2} - 4 x + 3, y = 31 x - 42$
A power times a sine 3 steps
Differentiate f(x) = x² sin x and find its slope at x = π/2.

Open this problem in the tool
Show solution
1. The factors and their derivatives:
  
  $u = x^{2}, u^{'} = 2 x; v = sin x, v^{'} = cos x$
2. Assemble. The x can come out as a common factor, which is the form to leave it in:
  
  $f^{'} (x) = 2 x sin x + x^{2} cos x = x (2 sin x + x cos x)$
3. At π/2 the sine is 1 and the cosine 0, so the second term vanishes and only u′v is left:
  
  $f^{'} (\frac{π}{2}) = 2 \cdot \frac{π}{2} \cdot 1 + (\frac{π}{2})^{2} \cdot 0 = π$
Answer

f′(x) = 2x sin x + x² cos x, and f′(π/2) = π ≈ 3.1416. On the page the top strip uv′ reads 0 there. The height sin x has stopped growing at its peak, and the whole rate comes from the width.

$f^{'} (x) = 2 x sin x + x^{2} cos x, f^{'} (\frac{π}{2}) = π \approx 3.1416$
Where x·eˣ is flat 4 steps
Find where the graph of f(x) = x eˣ has a horizontal tangent, and decide whether it is a maximum or a minimum.

Open this problem in the tool
Show solution
1. The factors and their derivatives. eˣ is its own derivative:
  
  $u = x, u^{'} = 1; v = e^{x}, v^{'} = e^{x}$
2. Assemble, then take out the common factor eˣ. An exam wants it factorised, because a product is zero only where one of its factors is:
  
  $f^{'} (x) = e^{x} + x e^{x} = e^{x} (1 + x)$
3. eˣ is never zero, so the slope vanishes only where the bracket does:
  
  $e^{x} > 0 \Rightarrow f^{'} (x) = 0 ⟺ x = - 1, f (- 1) = - e^{- 1} \approx - 0.3679$
4. Test either side. The slope is negative at −2 and positive at 0, so the curve goes down and then up:
  
  $f^{'} (- 2) = - e^{- 2} < 0, f^{'} (0) = 1 > 0$
Answer

A minimum at (−1, −1/e) ≈ (−1, −0.3679). At x = −1 the page prints the rule as 0 from two strips that cancel exactly, 0.3679 and −0.3679. It cannot draw the rectangle there, because one side, u = −1, is negative; the rule does not care.

$f_{m i n} = f (- 1) = - \frac{1}{e} \approx - 0.3679$
A power times a logarithm 3 steps
Differentiate f(x) = x³ ln x for x > 0, and find where it stops falling.

Open this problem in the tool
Show solution
1. The factors and their derivatives:
  
  $u = x^{3}, u^{'} = 3 x^{2}; v = ln x, v^{'} = \frac{1}{x}$
2. Assemble. The second term simplifies, x³ · 1/x = x², and then x² is common to both:
  
  $f^{'} (x) = 3 x^{2} ln x + x^{3} \cdot \frac{1}{x} = 3 x^{2} ln x + x^{2} = x^{2} (3 ln x + 1)$
3. For x > 0, x² is positive, so only the bracket can be zero:
  
  $3 ln x + 1 = 0 \Rightarrow ln x = - \frac{1}{3} \Rightarrow x = e^{- 1/3} \approx 0.7165$
Answer

f′(x) = x²(3 ln x + 1), zero at x = e^(−1/3) ≈ 0.7165. Left of it the bracket is negative and f falls; right of it f rises. At x = 1 the page prints u′v = 0, because ln 1 = 0, and the whole rate of 1 comes from the second strip. Slide x from 0.71 to 0.72 and watch the rule change sign, from −0.0138 to 0.0075.

$f^{'} (x) = x^{2} (3 ln x + 1), f^{'} (e^{- 1/3}) = 0, e^{- 1/3} \approx 0.7165$
A square root times a polynomial 4 steps
Differentiate f(x) = √x (x² − 4) and write the answer as a single fraction.

Open this problem in the tool
Show solution
1. Write √x as a power first; then the power rule applies as usual:
  
  $u = x^{1/2}, u^{'} = \frac{1}{2} x^{- 1/2} = \frac{1}{2 x}; v = x^{2} - 4, v^{'} = 2 x$
2. Assemble:
  
  $f^{'} (x) = \frac{x ^{2} - 4}{2 x} + 2 x x$
3. Put both terms over 2√x. Since √x · √x = x, the second term is 4x²/(2√x):
  
  $2 x x = \frac{4 x ^{2}}{2 x} \Rightarrow f^{'} (x) = \frac{x ^{2} - 4 + 4 x ^{2}}{2 x} = \frac{5 x ^{2} - 4}{2 x}$
4. A check at x = 4, where √x = 2:
  
  $f^{'} (4) = \frac{80 - 4}{2 \cdot 2} = \frac{76}{4} = 19$
Answer

f′(x) = (5x² − 4)/(2√x). At x = 4 the page prints 19: 3 from the strip on the right and 16 from the strip on top.

$f^{'} (x) = \frac{5 x ^{2} - 4}{2 x}$
eˣ sin x, differentiated twice 3 steps
Find f′(x) and f″(x) for f(x) = eˣ sin x.

Open this problem in the tool
Show solution
1. The factors and their derivatives:
  
  $u = e^{x}, u^{'} = e^{x}; v = sin x, v^{'} = cos x$
2. Assemble and take out eˣ:
  
  $f^{'} (x) = e^{x} sin x + e^{x} cos x = e^{x} (sin x + cos x)$
3. f′ is itself a product, eˣ times (sin x + cos x), so the rule runs a second time. The sines cancel:
  
  $f^{''} (x) = e^{x} (sin x + cos x) + e^{x} (cos x - sin x) = 2 e^{x} cos x$
Answer

f′(x) = eˣ(sin x + cos x) and f″(x) = 2eˣ cos x. At x = 0 the page prints f′ = 1, all of it from the top strip, e⁰ cos 0. The second derivative is one step past the page, but you can make it compute it: type u = e^x and v = sin(x) + cos(x), and the rule it prints at 0 is f″(0) = 2.

$f^{'} (x) = e^{x} (sin x + cos x), f^{''} (x) = 2 e^{x} cos x$
A box whose three sides all change 4 steps
Give the rectangle a depth w = 3 − x, so the box has sides x + 1, 2x + 1 and 3 − x. Find the rate at which its volume changes at x = 1, and then ΔV/h exactly for a step h. Press Worked: at x = 1 the page prints (uv)′ = 7, and ΔA/h = 8 at h = ½.

Open this problem in the tool
Show solution
1. Treat the box as the rectangle uv times the depth w and use the product rule twice. The page has already done the inner one:
  
  $(uv w)^{'} = (uv)^{'} w + uv w^{'} = 7 \cdot 2 + 6 \cdot (- 1) = 8$
2. Check it in percent. The sides change at 1 in 2, 2 in 3 and −1 in 2 of themselves, and for a product the rates add:
  
  $\frac{V ^{'}}{V} = \frac{1}{2} + \frac{2}{3} - \frac{1}{2} = \frac{2}{3}, V^{'} = 12 \cdot \frac{2}{3} = 8$
3. Now the exact step. Straight sides give Δu = h, Δv = 2h and Δw = −h, and multiplying out leaves three slabs, three edges and one corner cube:
  
  $Δ V = (2 + h) (3 + 2 h) (2 - h) - 12 = 8 h - 3 h^{2} - 2 h^{3}$
4. Divide by h. The slabs give the derivative; the edges and the cube carry the powers of h that vanish:
  
  $\frac{Δ V}{h} = 8 - 3 h - 2 h^{2}, h = 1 : 3, h = \frac{1}{2} : 6$
Answer

V′(1) = 8, and ΔV/h = 8 − 3h − 2h² exactly. The three slabs u′vw, uv′w and uvw′ (6, 8 and −6) survive the limit; the three edges, −3 in all, come with h, and the cube u′v′w′ = −2 with h². A whole unit step changes the volume by 3, from 12 to 15, which is nowhere near 8: at h = 1 the edges and the cube are as large as anything else.

$V^{'} (1) = 8, \frac{Δ V}{h} = 8 - 3 h - 2 h^{2}$

The wrong rule is the corner 🖖

Press The corner and look at a whole step. The rectangle gains a strip down the right, a strip along the top and a small corner, and that corner is exactly u′v′. Now halve h, and halve it again. The two strips, divided by h, hold their size; the corner shrinks with h and is gone in the limit. So the guess (uv)′ = u′v′ keeps the one piece the limit throws away, and throws away the two it keeps. Leibniz made that guess in a manuscript of 11 November 1675, and a couple of pages later refuted it himself with x times x.

In percent, rates add for a product and subtract for a quotient 🖖

Press Percent. One side grows by 3% of itself per unit of x, the other by 2%, and the area grows by 5%. Divide the product rule by uv and it says this at every x, for every pair: relative rates add. They add when one of them is negative too, which is why raising prices 3% while sales fall 2% lifts revenue by about 1%. Press Ratio and the same two sides give a quotient that grows by 1%, because for a quotient the rates subtract. The minus sign in the quotient rule is that subtraction.

Two strips that cancel in the product add in the quotient 🖖

Press Balance: the sides are sin x and cos x, at x = π/4. The width is growing exactly as fast as the height is shrinking, so the two strips are equal and opposite and the area stops changing. That is the largest this rectangle ever gets. Press tan x and the same two strips go into the quotient rule, where the minus sign turns the cancellation into a sum. A shrinking v makes a quotient larger, so the strip the product lost is one the quotient gains.

Learning path

Differentiation: how rates combine

References (2)

Leibniz's guess that d(xy) is dx·dy, its apparent agreement with one example, and its refutation with x times x (“becomes 2βx, which is quite a different thing from” the product of the increments), in a manuscript dated 11 November 1675; and the later derivation d(xy) = (x + dx)(y + dy) − xy = x dy + y dx + dx dy with dx dy omitted as infinitely small: J. M. Child, The Early Mathematical Manuscripts of Leibniz, translated from the Latin texts published by Carl Immanuel Gerhardt, with critical and historical notes. Chicago and London: Open Court, 1920. Pp. 100–102 (manuscript dated 11 November 1675) and p. 143 (Elementa calculi novi, undated).
The product rule (Theorem 3.5) and quotient rule (Theorem 3.6) as the page states them; the warning that the derivative of a product is not the product of the derivatives, with x² as the example (2x, not 1·1); the proof of the product rule from the limit definition, which uses the continuity of the second factor (the lesson's `assumes`): G. Strang and E. Herman, Calculus Volume 1, OpenStax (2016), section 3.3, “Differentiation Rules”.

By the 42Math team · Reviewed against published sources. Corrections welcome.