When two functions are multiplied and both are changing, the product changes twice over: once because the first factor moves and once because the second does. The product rule adds the two contributions, f′·g + f·g′. For polynomials it is checked by expanding the product and differentiating term by term, and the lab compares the two results exactly.
Core ideas
Three ideas: why there are two terms, how the rule reads, and a quick check that catches the wrong rule.
Picture a rectangle of width f and height g. Its area is f·g. If the width grows, the area gains a strip of the old height; if the height grows, it gains a strip of the old width. The rate of the area is the sum of the two strips' rates, and that is the rule.
The width's rate f′ acts on the height as it stands, g, and the height's rate g′ acts on the width as it stands, f. So the rule is f′·g + f·g′: a sum of two products, each holding one factor still and differentiating the other.
If f has degree 2 and g has degree 1, the product has degree 3 and its derivative has degree 2. The product of the derivatives has degree 1 + 0 = 1. The two cannot agree, whatever the coefficients.
The material
The temptation and its refutation first, then the statement and its proof, then two checks and the case of a square.
A product of two polynomials can always be multiplied out and differentiated term by term. The product rule is for the other case, when the factors are better left as they are, and the way to trust it is to check it against the expansion.
Take f(t) = t² and g(t) = t + 1. Multiplied out, f·g = t³ + t², and the power rule gives the derivative of t³ + t² as 3t² + 2t. The tempting rule is to differentiate each factor and multiply: f′ = 2t and g′ = 1, so the product is 2t·1 = 2t. That is not 3t² + 2t, and it cannot be, since its degree is 1 and the true derivative's is 2.
For polynomials f and g, (f·g)′ = f′·g + f·g′.
The numerator of the quotient of the product is f(t + h)·g(t + h) − f(t)·g(t). Add and subtract f(t)·g(t + h) and regroup: [f(t + h) − f(t)]·g(t + h) + f(t)·[g(t + h) − g(t)].
Divide by h. The quotient of the product is the quotient of f times g(t + h), plus f(t) times the quotient of g. Each quotient is a polynomial in h after the cancelling, and so is g(t + h).
Read off the part free of h: the constant term of the quotient of f is f′(t), the constant term of g(t + h) is g(t), and the constant term of the quotient of g is g′(t). The constant term of a product is the product of the constant terms, which gives f′(t)·g(t) + f(t)·g′(t).
□
For f = t² and g = t + 1: f′·g = 2t·(t + 1) = 2t² + 2t and f·g′ = t²·1 = t². Their sum is 3t² + 2t, which is the derivative of the expanded product t³ + t² found above.
The lab does this on any pair of polynomials, and its rule tile prints the correct sum beside the wrong product so the two can be compared. Its equality tile says whether the expanded product's derivative and the rule's output are the same polynomial.
With f = t³ − t and g = t² + 2 the expansion is t⁵ + t³ − 2t, and its derivative is 5t⁴ + 3t² − 2. The rule reaches the same polynomial from the factors: f′·g = (3t² − 1)(t² + 2) = 3t⁴ + 5t² − 2 and f·g′ = (t³ − t)·2t = 2t⁴ − 2t², which add to 5t⁴ + 3t² − 2. For larger factors the rule is less work than the expansion, and for functions that cannot be multiplied out it is the only route.
Put g = f and the rule says (f²)′ = f′·f + f·f′ = 2·f·f′. Take f = t + 1: the square is t² + 2t + 1, whose derivative is 2t + 2, and the rule gives 2·(t + 1)·1, the same. The derivative of a square is twice the function times its own derivative, not f′ squared and not 2·f′. The lab's second preset squares t²: the true derivative of t⁴ is 4t³ and the wrong rule gives 2t·2t = 4t². At t = 0 and at t = 1 the two agree, so a check at one of those places would pass the wrong rule; the degrees, 3 against 2, say the two polynomials are different whatever any single place says. The next lesson carries the square one step further, to every power.
Interactive
Difference quotients of polynomials, by algebra
Method
Four moves. The second is the one the wrong rule skips.
Find f′ and g′ by the power rule. Write them down before combining anything.
Form f′·g, the derivative of the first with the second as it stands, and f·g′, the first as it stands with the derivative of the second.
The rule is a sum. Collect like powers and write the result in descending order.
The result has degree one less than f·g, namely the two degrees added, minus 1. If you have time, multiply out and differentiate; the two polynomials must be identical.
Take f = t³ − t and g = t² + 2. Each line is exact, and the last two check the answer against the other route.
The two routes agree, and the product of the derivatives is a different polynomial of the wrong degree. The lab's equality tile reads equal for this pair.
For a rehearsal, differentiate t·(t² + 1) by the rule and by expanding, and compare.
You should be able to differentiate each factor, form the two crossed products and add them, check the degree of the result, expand to verify, and say why the product of the derivatives is not the derivative of the product.
Something to change about this lesson? Write it down, tick it off when it is handled, and export the list when you want to act on it. Kept in this browser only: nothing is sent anywhere, and nobody is notified.
No recommendations yet for this lesson.