Derivatives of sums, products and quotients

Leyla bakes shekerbura, the traditional Azerbaijani Novruz pastry, and sells it by the box. Her weekly revenue is the price of a box times the number of boxes sold: R = p · q. As Novruz approaches, both grow: the price rises by 0.5 manat a week and sales by 5 boxes a week. How fast does her revenue grow? Simply multiplying the two rates (0.5 · 5 = 2.5) is the first idea that comes to mind — and it is wrong. Sums, products and quotients have their own differentiation rules; in this lesson you will learn them and solve Leyla’s problem too.

In the previous lesson you learned the table of derivatives: we know the derivatives of basic functions such as xⁿ, √x, eˣ, ln x, sin x and cos x. In problems, however, functions are built from these “bricks” by adding, subtracting, multiplying and dividing: 3x² − 5x, x² · sin x, ln x / x. A function of a function (such as sin 2x or (3x − 2)⁵) is the topic of the next lesson, the chain rule.

Sums, differences and constant factors

Why is the derivative of a sum the sum of the derivatives? Let f = u + v. When the argument changes by Δx, u changes by Δu and v by Δv, so the sum changes by Δf = Δu + Δv. Divide both sides by Δx: Δf / Δx = Δu / Δx + Δv / Δx. As Δx → 0 the ratios on the right approach u′ and v′, hence f′ = u′ + v′. A constant factor works the same way: the increment of C · u is C · Δu, so C comes out in front of the derivative.

The product rule

The product rule is easy to see on a rectangle. A rectangle with sides u and v has area S = u · v. When the argument grows by Δx, the sides grow by Δu and Δv, and three pieces are added to the area: a strip Δu · v, a strip u · Δv and a small corner Δu · Δv. So ΔS = Δu · v + u · Δv + Δu · Δv.

Now divide by Δx: ΔS / Δx = (Δu / Δx) · v + u · (Δv / Δx) + (Δu / Δx) · Δv. As Δx → 0 the first two terms approach u′ · v and u · v′. In the third, Δu / Δx → u′ but Δv → 0, so it tends to zero. The small corner vanishes, and what is left is the product rule.

Why is (u · v)′ ≠ u′ · v′? The simplest check: x² = x · x. The correct answer is (x²)′ = 2x, while “multiplying the derivatives” would give 1 · 1 = 1. The picture shows the reason too: u′ · v′ appears only in the small corner (Δu · Δv / Δx ≈ u′ · v′ · Δx), and exactly this piece vanishes as Δx → 0. The rule comes from the two strips that remain.

For three factors, apply the rule twice, first treating u · v as one factor: (u · v · w)′ = (u · v)′ · w + (u · v) · w′. Expanding the first bracket as well gives the formula below.

The quotient rule

The quotient rule follows from the product rule. Let q = u / v; then u = q · v. By the product rule, u′ = q′ · v + q · v′. Hence q′ = (u′ − q · v′) / v = (u′ − (u / v) · v′) / v. Multiplying the numerator and the denominator by v gives the quotient rule.

When the numerator is a constant (u = C), its derivative is zero and the first term of the quotient rule disappears: (C / v)′ = (0 · v − C · v′) / v². For such fractions the short formula is handy.

Simplify first, then differentiate

The rules always give the right answer, but not always by the shortest route. Before differentiating, look at the expression for a second — perhaps it can be turned into a sum that needs neither the product nor the quotient rule:

• Expand the brackets if the product consists of two small polynomials: (x + 2)(x − 5) = x² − 3x − 10, whose derivative is 2x − 3.
• Split the fraction term by term if the denominator is a single term: (x² + 1)/x = x + 1/x.
• Write roots and 1/xⁿ as powers: √x = x^(1/2), 1/x³ = x⁻³, 1/√x = x^(−1/2).

Whenever you differentiate an expression, work in this order:

1. 1
   Recognise the structure

   If you were to evaluate the expression, what would the last operation be? Addition or subtraction — the sum rule; multiplication — the product rule; division — the quotient rule. Take constant factors out at once.

2. 2
   Look for a simplification

   Expand brackets, split a fraction with a one-term denominator, write roots as powers. This often makes the product and quotient rules unnecessary.

3. 3
   Write u, v, u′, v′ separately

   If a product or quotient remains, write down four expressions: u, v and their derivatives u′, v′ from the table.

4. 4
   Substitute into the formula

   Put every expression in brackets; in a quotient the numerator starts with u′ · v and the denominator is v².

5. 5
   Simplify and check

   Expand, collect like terms, take out the common factor. Check the result in a second way or at a simple point.

The rules of this lesson cover only adding, subtracting, multiplying and dividing. In functions such as sin 2x, e^(3x) or (3x − 2)⁵ one function sits inside another: you could expand (3x − 2)⁵ and differentiate term by term, but that takes very long. In the next lesson you will learn the chain rule, which differentiates such functions in one step.