Lesson · Rates of Change and the Derivative · Average rates

The Difference Quotient as a Polynomial in h

When the function is a polynomial, the difference quotient can be simplified by algebra alone. Expand p(t + h), subtract p(t), and every surviving term has a factor h, so dividing by h leaves a polynomial in h. Its constant term, the part that does not contain h, is what the next lessons call the rate at t. No limit is taken; the cancelling is exact.

Open the interactive lesson Practice

(p(t + h) − p(t))/h, expanded and cancelled every term of the numerator has a factor h what is left is a polynomial in h t²: 2t + h t³: 3t² + 3t·h + h² its h-free part: 2t, 3t²

The quotient of a polynomial, simplified

Core ideas

What this lesson establishes

Three ideas: why the h cancels, what is left, and the order of operations that stops it going wrong.

The numerator always has a factor of h

When you expand p(t + h), the terms that contain no h add up to exactly p(t): that is what it means to put h = 0. Subtracting p(t) removes them, so every term left has at least one h. That is why the division by h always comes out even for a polynomial, with nothing left over.

What remains is a polynomial in h

After the division the result is a sum of terms in h, with coefficients that depend on t. For t² it is 2t + h. For t³ it is 3t² + 3t·h + h². The step has not been made small, and nothing has been dropped: this is an exact rewriting of the same quotient.

Cancel first, then read off the part without h

The part that does not contain h is the constant term. It is 2t for t² and 3t² for t³. It is the value the quotient takes when h is replaced by 0 in the simplified form, which is safe because the division is already done. Doing it in the other order is dividing zero by zero.

The material

Expand, subtract, divide, read

One polynomial worked through completely, then the same moves on a quartic, and the order of operations that matters.

“Average Rate of Change” found that the rate of t² at 1 was 2 + h. That was one place. Do the same computation at a general t and the answer is a formula in t and h together.

p(t) = t² p(t + h) = t² + 2th + h² p(t + h) − p(t) = 2th + h² (p(t + h) − p(t))/h = 2t + h

At t = 1 this is 2 + h, the result of the previous lesson, and at t = 2 it is 4 + h. One expansion has done every place at once. The terms of p(t + h) that have no h are t² alone, which the subtraction removes; every term that survives has an h, and the division by h takes one of them away.

Definition · The quotient as a polynomial in h

For a polynomial p, the expression (p(t + h) − p(t))/h, after the expansion, subtraction and cancellation, is a polynomial in h whose coefficients are polynomials in t.

Its constant term is the coefficient of h to the power 0: the part of the quotient that does not contain h.

The cube

For p(t) = t³ the expansion is (t + h)³ = t³ + 3t²h + 3th² + h³. Subtracting t³ removes the first term, and what is left is 3t²h + 3th² + h³. Each term has a factor h, and dividing by h gives 3t² + 3th + h².

p(t) = t³ p(t + h) − p(t) = 3t²h + 3th² + h³ divide by h: 3t² + 3th + h² the part without h: 3t²

The constant term is 3t². The terms that contain h are 3th and h²; they are the part of the rate that depends on how big the step is, and the constant term is the part that does not.

Example · A quartic with two terms

Let p(t) = t⁴ − 2t². The first term contributes 4t³h + 6t²h² + 4th³ + h⁴ to the numerator, and the second contributes −2·(2th + h²), which is −4th − 2h².

Dividing the whole numerator by h gives 4t³ + 6t²h + 4th² + h³ − 4t − 2h. The lab groups it by powers of h: the constant term is 4t³ − 4t, and the remaining terms all contain h.

Why the order matters

The tempting shortcut is to put h = 0 into the quotient straight away. The numerator becomes p(t) − p(t) = 0 and the denominator becomes 0, so the quotient is 0/0, which has no value. The simplified form is different: after the h has been cancelled out of the numerator, 2t + h is a perfectly good polynomial and has a value at h = 0. The step is allowed only after the cancelling, because the cancelling is what removed the zero from the denominator.

Read this way, the constant term is not a limit that has to be believed. It is a coefficient, found by the same expanding and collecting that Algebra's Polynomials and Factoring does. What it means as a rate, and why it is the right thing to call the rate at t, is the question for the next lesson.

Interactive

The quotient in h

Difference quotients of polynomials, by algebra

Method

Simplifying a quotient

Four moves. The order is the point: the division comes before any step is made small.

Expand p(t + h)

Replace every t by t + h and multiply out. For a power use the binomial pattern: (t + h)³ has coefficients 1, 3, 3, 1.

Subtract p(t)

The terms with no h cancel, since they sum to p(t). Check that every term left has an h; if one does not, the expansion has an error.

Divide each term by h

Do it term by term. Each h becomes 1, each h² becomes h, and so on. The result is a polynomial in h.

Read off the constant term

Collect the terms that contain no h. That coefficient, a polynomial in t, is the part of the rate that does not depend on the step.

Cancelling the h

Score 0 / 0

Choose the answer you can justify.

The cube, term by term

The polynomial is p(t) = t³. The expansion is the binomial pattern, and each line is exact.

p(t + h) = t³ + 3t²h + 3th² + h³ p(t + h) − p(t) = 3t²h + 3th² + h³ divide by h: 3t² + 3th + h² constant term in h: 3t² check at t = 1, h = 1/4: 3 + 3/4 + 1/16 = 61/16

The check at the end agrees with the quotient computed the long way. The output at 5/4 is 125/64, the change from t³ = 1 is 61/64, and dividing by 1/4 gives 61/16.

For a rehearsal, repeat this for p(t) = t⁴, whose expansion has the coefficients 1, 4, 6, 4, 1. The constant term is 4t³. The pattern across t², t³ and t⁴ is that the constant term is the exponent times the power one lower, and the lesson on the derivative as a function uses it.

Common mistakes

Setting h to zero before dividingThe quotient (p(t + h) − p(t))/h is undefined at h = 0: both the numerator and the denominator are zero. The constant term is found by cancelling an h from the numerator first, and then reading the result. For t², putting h = 0 in the quotient gives 0/0, and putting it in 2t + h gives 2t.

Dropping terms because h is smallNothing in this lesson drops a term. 3t² + 3th + h² is exactly equal to the quotient for every h, and 3th and h² are part of it. They are not the constant term, and that is all that distinguishes them.

Forgetting a term of the expansionThe most common slip is (t + h)³ = t³ + h³. The middle terms 3t²h and 3th² are what make the answer 3t² and not 0. Check by putting in numbers: (1 + 1)³ = 8, not 1 + 1 = 2.

Check your understanding

✓

Finish when you can expand a polynomial's difference quotient, cancel the h, and name its constant term.

You should be able to expand p(t + h), subtract p(t), divide by h and write the result as a polynomial in h, say why the division comes out even, and say why the order of the steps is not optional.

Note: The constant term is a number once a place is chosen: 2 at t = 1 for t². Whether that number is the rate at the place, in the sense of a column of average rates heading somewhere, is checked next in “What the Quotients Approach”.