f′Field Guide
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The Product and Quotient Rules

The derivative of a product is not the product of the derivatives.

0 / 10 problems mastered
Product Rule
(fg)′=f g′+g f′(fg)' = f\,g' + g\,f'

“first · d(second) + second · d(first)”

Quotient Rule
(fg)′=g f′−f g′g2\left(\frac fg\right)' = \frac{g\,f' - f\,g'}{g^2}

“low d-high minus high d-low, over low squared”. The order matters because of the minus.

The product rule is an area picture

Lab 4 · grow the rectangle
u = x (width), v = x² (height)–
u·Δv (top strip)–
v·Δu (side strip)–
Δu·Δv (corner)–
Δ(uv) / Δx–
u v′ + v u′ = x·2x + x²·1–

The area uv grows by two strips plus a tiny corner: Δ(uv) = u·Δv + v·Δu + Δu·Δv. Divide by Δx and let Δx → 0. The corner is a product of two small numbers, so it disappears, leaving (uv)′ = u v′ + v u′. Drag Δx toward 0 and watch the two bottom numbers meet.

Worked examples

Example · product rule with eˣ

Differentiate f(x)=xexf(x)=xe^x, then find f′′f''.

  1. First =x=x, second =ex=e^x. f′=x⋅ex+ex⋅1=(x+1)exf'=x\cdot e^x+e^x\cdot1=(x+1)e^x.
  2. Again: f′′=(x+1)ex+ex⋅1=(x+2)exf''=(x+1)e^x+e^x\cdot1=(x+2)e^x.
  3. Pattern: the n-th derivative is (x+n)ex(x+n)e^x.
Example · quotient rule

Differentiate y=x2+x−2x3+6y=\dfrac{x^2+x-2}{x^3+6}.

  1. High =x2+x−2=x^2+x-2, d-high =2x+1=2x+1. Low =x3+6=x^3+6, d-low =3x2=3x^2.
  2. y′=(x3+6)(2x+1)−(x2+x−2)(3x2)(x3+6)2y'=\dfrac{(x^3+6)(2x+1)-(x^2+x-2)(3x^2)}{(x^3+6)^2}
  3. Expand: (2x4+x3+12x+6)−(3x4+3x3−6x2)(2x^4+x^3+12x+6)-(3x^4+3x^3-6x^2)
  4. y′=−x4−2x3+6x2+12x+6(x3+6)2y'=\dfrac{-x^4-2x^3+6x^2+12x+6}{(x^3+6)^2}
Example · from a table (very common exam question)

Given f(5)=1, f′(5)=6, g(5)=−3, g′(5)=2f(5)=1,\ f'(5)=6,\ g(5)=-3,\ g'(5)=2. Find (fg)′(5)(fg)'(5), (f/g)′(5)(f/g)'(5), (g/f)′(5)(g/f)'(5).

  1. (fg)′(5)=f(5)g′(5)+g(5)f′(5)=1(2)+(−3)(6)=−16(fg)'(5)=f(5)g'(5)+g(5)f'(5)=1(2)+(-3)(6)=-16.
  2. (f/g)′(5)=gf′−fg′g2=(−3)(6)−(1)(2)9=−209(f/g)'(5)=\dfrac{g f'-f g'}{g^2}=\dfrac{(-3)(6)-(1)(2)}{9}=-\dfrac{20}{9}.
  3. (g/f)′(5)=fg′−gf′f2=(1)(2)−(−3)(6)1=20(g/f)'(5)=\dfrac{f g'-g f'}{f^2}=\dfrac{(1)(2)-(-3)(6)}{1}=20.

Practice

§3.2 #1
Differentiate f(x)=(1+2x2)(x−x2)f(x)=(1+2x^2)(x-x^2) two ways: Product Rule, and multiplying out first. Do they agree?
§3.2 #5
Differentiate y=x3exy=x^3e^x.
§3.2 #9
Differentiate y=xexy=\dfrac{x}{e^x}.
§3.2 #11
Differentiate g(t)=3−2t5t+1g(t)=\dfrac{3-2t}{5t+1}.
§3.2 #13
Differentiate f(t)=5tt3−t−1f(t)=\dfrac{5t}{t^3-t-1}.
§3.2 #27
Differentiate f(x)=x2exx2+exf(x)=\dfrac{x^2e^x}{x^2+e^x}.
§3.2 #31
Find f′f' and f′′f'' for f(x)=x2exf(x)=x^2e^x.
§3.2 #35
Find the tangent line to y=x21+xy=\dfrac{x^2}{1+x} at (1,12)\left(1,\tfrac12\right).
§3.2 #39
Find the tangent line to the witch of Agnesi y=11+x2y=\dfrac{1}{1+x^2} at (−1,12)\left(-1,\tfrac12\right).
§3.2 · table style
If h(x)=f(x)xh(x)=\dfrac{f(x)}{x}, f(2)=−3f(2)=-3, f′(2)=4f'(2)=4, find h′(2)h'(2).

Traps

(fg)′ = f′g′

False. Check with x⋅x=x2x\cdot x=x^2: f′g′=1f'g'=1, but (x2)′=2x(x^2)'=2x.

Flipping the quotient numerator

It's gf′−fg′g f' - f g', bottom first. Swapping them flips the sign of your answer.

Using quotient rule when you don't need to

x3+1x\dfrac{x^3+1}{x} or sin⁡x5\dfrac{\sin x}{5}: simplify or pull out the constant instead.