Products and quotients change in two ways

Try

A rectangle’s length and width both change. If you count only the extra strip caused by length growth, you miss the area added by width growth. Sketch a rectangle with both edges slightly extended. Label the two long strips and the small corner rectangle.

Predict

Predict the derivative of x²·x³ in two ways: first combine the powers, then try multiplying their derivatives. The results disagree. Determine which calculation is justified by the power rule already established. This counterexample is enough to reject the proposed “multiply derivatives” shortcut.

Experiment

Let the sides be f(x)=x²+1 and g(x)=3x−2. At x=2, their values are 5 and 4 and their rates are 4 and 3. Break the product rate into contributions.

Contribution	Local computation	Value
Change in first factor	f′(2)g(2)=4·4	16
Change in second factor	f(2)g′(2)=5·3	15
Total product rate	16+15	31

Expand the product as 3x³−2x²+3x−2 and differentiate it independently: 9x²−4x+3 gives 31 at 2. Two different derivations agree. This check is stronger than checking the same formula twice, although it does not prove the general product rule.

Understand

For differentiable f and g, the product rule is (fg)′=f′g+fg′. To derive it at a, start with the difference quotient for f(x)g(x). Add and subtract f(a+h)g(a) in its numerator. Group it as f(a+h)[g(a+h)−g(a)]+g(a)[f(a+h)−f(a)]. After division by h, the limit gives f(a)g′(a)+g(a)f′(a). Here f(a+h) approaches f(a) because differentiability implies continuity. No appeal to a drawing is needed for the proof, although the strips explain the two terms.

For a quotient f/g with g nonzero, the rule is (f/g)′=(f′g−fg′)/g². To establish it directly, put f(a+h)/g(a+h)−f(a)/g(a) over a common denominator. Split its numerator as g(a)[f(a+h)−f(a)]−f(a)[g(a+h)−g(a)]. Divide by h and take limits. Continuity makes the denominator approach g(a)², which is nonzero. This proves differentiability as well as the formula, and explains the numerator order and denominator square.

For (x+1)/(x−1), the numerator becomes 1·(x−1)−(x+1)·1=−2, so the derivative is −2/(x−1)², valid for x≠1. The negative derivative indicates decrease on each of the two separate domain intervals. It does not connect values across the missing input using a theorem that assumes one uninterrupted interval.

Simplifying first is often clearer. For (x²−1)/(x−1), the original function equals x+1 away from 1, so its derivative is 1 there. The original function still has no derivative at 1 because it has no value there. A separately defined continuous extension would be a new function with a different domain.

Revenue provides a nongeometric interpretation. If R(q)=q p(q), then R′=p+q p′: quantity changes and the modeled price changes. Ignoring the second term assumes a constant price, a different model. Treat a continuous quantity q as an approximation if actual goods are indivisible.

Apply

For each practice item, name the factors or numerator and denominator before differentiating. Check the original domain. In the proof item, write the added and subtracted term explicitly. In the revenue item, interpret the sign of the price-change contribution rather than reporting only an algebraic result.

Continue

Products and quotients combine functions side by side. The next lesson handles functions inside functions, where the missing contribution is an inner rate rather than a second additive product term.