Lesson · Rates of Change and the Derivative · The derivative as a function

The Derivative as a Function

Each place has its own derivative, so the derivative is itself a function of t. For a polynomial it is found by the power rule: each term c·tⁿ becomes n·c·tⁿ⁻¹, and a constant term disappears. Where the result is positive the graph rises, where it is negative the graph falls, and where it is zero the graph is momentarily level.

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(c·tⁿ)′ = n·c·tⁿ⁻¹ a constant has derivative 0 (p + q)′ = p′ + q′ p′ > 0: rising p′ < 0: falling t³ − 6t² + 9t has p′ = 3t² − 12t + 9

The derivative as a function, by the power rule

Core ideas

What this lesson establishes

Three ideas: the rule, why it is allowed, and what the sign says about the graph.

The derivative is a function of t

Compute f′(a) for every a and you get a new function, f′(t). For t² the constant term of the quotient is 2t, so (t²)′ = 2t. For t³ it is 3t². These are the same constant terms found in the polynomial lesson, now read as formulas.

The power rule: bring the exponent down

The pattern is that t² has derivative 2t, t³ has 3t² and t⁴ has 4t³: the derivative of tⁿ is n·tⁿ⁻¹. The reason is the expansion (t + h)ⁿ = tⁿ + n·tⁿ⁻¹·h + (terms with h²), so the quotient is n·tⁿ⁻¹ plus terms that each contain h. A constant c multiplies through, and a constant term has quotient (c − c)/h = 0.

The sign of p′ is the direction of the graph

A positive derivative at t means the secant slopes there are positive, so the graph is rising. A negative one means it is falling. Where p′ is zero the tangent is horizontal. Solving p′ = 0 and testing the sign on each side of a zero reads the whole shape of the graph.

The material

A rule, a reason, and a graph read from a sign

The power rule with its proof, the sum and constant rules, one cubic read completely, and the three errors the rule invites.

Theorem · The power rule

For a whole number n ≥ 1, the derivative of tⁿ is n·tⁿ⁻¹. For a constant c, the derivative of c is 0.

Expand (t + h)ⁿ by multiplying n copies of t + h together. Taking t from every copy gives tⁿ; taking h from one copy and t from the other n − 1 can be done in n ways, which gives n·tⁿ⁻¹·h; every other choice takes h from at least two copies, so every later term contains h². For n = 3 this is the t³ + 3t²h + 3th² + h³ of the polynomial lesson.

Subtracting tⁿ and dividing by h leaves n·tⁿ⁻¹ plus terms that each still contain h, so the constant term in h is n·tⁿ⁻¹. For a constant c the numerator c − c is 0, so the quotient is (c − c)/h = 0 for every h.

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Two more facts let the rule handle a whole polynomial. The quotient of a sum is the sum of the quotients, because subtraction and division distribute over a sum, so (p + q)′ = p′ + q′. And a constant factor comes through unchanged: (c·p)′ = c·p′. Together they give the rule for c·tⁿ: n·c·tⁿ⁻¹, term by term.

Example · A cubic, term by term

p(t) = t³ − 6t² + 9t. The three terms give 3t², −12t and 9, so p′(t) = 3t² − 12t + 9.

Factor it: 3t² − 12t + 9 = 3(t² − 4t + 3) = 3(t − 1)(t − 3). The zeros are at t = 1 and t = 3.

Reading the graph from the sign

The product 3(t − 1)(t − 3) is positive when both brackets have the same sign and negative when they differ. Test one point in each region: at t = 0 it is 3·(−1)·(−3) = 9, positive; at t = 2 it is 3·1·(−1) = −3, negative; at t = 4 it is 3·3·1 = 9, positive.

p′(t) = 3(t − 1)(t − 3) t < 1 p′ > 0 rising 1 < t < 3 p′ < 0 falling t > 3 p′ > 0 rising

That is what the graph of t³ − 6t² + 9t does: it rises to t = 1, where p = 4, falls to t = 3, where p = 0, and rises again. The derivative has told us the shape without plotting a point. The lab prints the derivative, its rational zeros and the intervals of rising and falling.

Three errors the rule invites

Forgetting the coefficient: the derivative of t³ is 3t², not t².
Treating a constant as if it had a rate: the derivative of 5 is 0, not 1, and the derivative of t + 5 is 1.
Differentiating the factors of a product separately: that is not this rule, and a later lesson shows why not. Expand first.

Interactive

The derivative

p′ and p″ term by term, and their signs

Differentiate a polynomial and read its sign

Type a polynomial in t. The lab applies the power rule term by term, finds the rational zeros of the derivative, and lists where the polynomial rises and falls. The plot shows the polynomial and its derivative together; the derivative crosses zero where the graph turns.

Polynomial

p(t)

Derivatives shown

p′(t)—

p″(t)—

Zeros of p′—

Where p rises and falls—

Inflection points—

A polynomial in t of degree 10 at most: t^3 - 6t^2 + 9t.

Method

Differentiating and reading a polynomial

Five moves. The first two are mechanical, and the rest read the answer.

Differentiate each term

c·tⁿ becomes n·c·tⁿ⁻¹. A constant term goes. Do not skip the coefficient.

Add the results

The derivative of the sum is the sum of the derivatives. Write it in order of descending powers.

Solve p′ = 0

Factor, or use the quadratic formula for a quadratic. The zeros are where the tangent is horizontal.

Test the sign between the zeros

Pick one point in each region and evaluate p′. Positive means rising, negative means falling.

Check against the plot

The graph should turn at each zero where the sign changes. If it does not, recheck the coefficient of each term.

The power rule and the sign

Score 0 / 0

Choose the answer you can justify.

A cubic, from rule to graph

The polynomial is p(t) = t³ − 6t² + 9t. The derivative, the factoring and the signs are each exact.

p(t) = t³ − 6t² + 9t p′(t) = 3t² − 12t + 9 = 3(t − 1)(t − 3) zeros: t = 1, t = 3 t = 0: p′ = 9 rising t = 2: p′ = −3 falling t = 4: p′ = 9 rising p(1) = 4, p(3) = 0

The graph rises to a height of 4 at t = 1, falls to 0 at t = 3, and rises for ever after. Neither height was found by plotting.

For a rehearsal, differentiate t² − 4t. The derivative is 2t − 4, the zero is at 2, and the graph falls before it and rises after, which is a bowl.

Common mistakes

Writing the derivative of tⁿ as tⁿ⁻¹, or the derivative of a constant as 1The exponent comes down as a coefficient: the derivative of t³ is 3t², not t². The polynomial lesson found 3t² as the constant term of the cube's quotient, and t² would give a slope of 1 at t = 1 where the quotients head for 3. A constant is the other slip: every quotient of f(t) = 5 is (5 − 5)/h = 0, so its derivative is 0, not 1.

Reading p′ > 0 as “p is positive”The sign of p′ is the direction of the graph, not its height. For t² − 4t at t = 3, p′ = 2 is positive, so the graph is rising, while p = −3 is negative, so the graph is below the axis. The two signs answer different questions, and the derivative answers only the first.

Dropping the sign check once the zeros are foundA zero of p′ is a place where the tangent is level. It does not say whether the graph rises or falls on either side. The signs have to be tested between the zeros, and the next lesson shows a zero where the sign does not change.

Check your understanding

✓

Finish when you can differentiate a polynomial term by term and say from the sign of the result where the graph rises and falls.

You should be able to apply the power rule to every term, factor the result where it factors, solve for the zeros, test one point in each region, and state the intervals of rising and falling.

Note: Where the derivative is zero is where the graph can turn, and not every zero is a turn. The rest of the course treats those zeros, starting in “Where the Rate Is Zero”, then the rules for products and compositions, the second derivative and the exponential.