A. Rate of Change · Derivatives and Differentiation RulesThe Derivative Function: Tables and Graphs of Tangent Slopes
A2.2 A2.1About 70 minutes
Learning goals- I can build a table of instantaneous rates of change for a polynomial and graph the pairs (x, slope).
- I can recognize the graph of those pairs as the derivative function f′.
- I can describe how the graphs of f and f′ are connected.
- I can see that the derivative of a polynomial of degree n has degree n − 1.
Words to know
derivative
The function f′ whose value at x is the instantaneous rate of change (tangent slope) of f at x.
f′(x)
Notation for the derivative of f, read "f prime of x".
slope graph
A graph of the points (x, tangent slope at x). It is the graph of the derivative.
centred difference
The estimate [f(a + h) − f(a − h)] ÷ (2h) for the tangent slope at x = a, using a small h.
From one slope to a whole function
In the last lesson you found tangent slopes at a handful of points and used their signs. Now we keep the actual numbers. If you find the tangent slope at every x-value and plot the pairs (x, slope), you get the graph of a brand-new function. That function is the derivative.
Remember from MHF4U: in the lesson Reading Rates of Change from Graphs (unit Rates of Change) you sketched rough slope graphs. Today you make them precise and find their equations.
The derivative function
The derivative of f is the function f′ (read "f prime") whose value at each x is the slope of the tangent to y = f(x) at that x:
f′(x) = instantaneous rate of change of f at x = slope of the tangent at (x, f(x))
If f(x) = x², the tangents at x = −3, −2, −1, 0, 1, 2, 3 have slopes −6, −4, −2, 0, 2, 4, 6 (you can check each one with the difference quotient). Every slope is double its x-value. So f′(x) = 2x.
Making the table with technology (or a calculator)
To build a table of tangent slopes for any function, use the centred estimate from MHF4U with a tiny h:
f′(a) ≈ [f(a + 0.001) − f(a − 0.001)] ÷ 0.002
In a graphing tool such as Desmos or GeoGebra, you can type the function f(x), then type g(x) = (f(x + 0.001) − f(x − 0.001))/0.002. The graph of g is (very nearly) the graph of f′. With a scientific calculator, you can do the same thing at five or six x-values and look for a pattern.
The table below gives the slopes for three polynomials. The values are rounded to the nearest whole number, so the tiny errors from using h = 0.001 disappear.
| x | −2 | −1 | 0 | 1 | 2 | 3 |
|---|
| slope of f(x) = 2x + 1 | 2 | 2 | 2 | 2 | 2 | 2 |
| slope of f(x) = x² | −4 | −2 | 0 | 2 | 4 | 6 |
| slope of f(x) = x³ | 12 | 3 | 0 | 3 | 12 | 27 |
| slope of f(x) = x³ − 3x | 9 | 0 | −3 | 0 | 9 | 24 |
Reading the patterns
- Linear: f(x) = 2x + 1 has slope 2 everywhere. Its derivative is the constant function f′(x) = 2.
- Quadratic: the slopes of x² go up by 2 each step, a linear pattern: f′(x) = 2x.
- Cubic: the slopes of x³ are 12, 3, 0, 3, 12, 27, which is 3 × (4, 1, 0, 1, 4, 9), three times the squares of the x-values: f′(x) = 3x². Check: 3(2²) = 12 and 3(3²) = 27. ✓
- x³ − 3x: each slope is 3 less than the slope of x³, so f′(x) = 3x² − 3.
In general, the derivative of a polynomial of degree n is a polynomial of degree n − 1.
Reading the two graphs together
Read f′ like this: where f′ is above the x-axis, f is increasing; where f′ is below, f is decreasing; where f′ crosses the x-axis, f has a turning point. The height of the graph of f′ is the steepness of the graph of f.
Worked exampleExample 1: A derivative from a table
The table gives tangent slopes of f(x) = x² − 4x. Find an equation for f′(x).
x: 0, 1, 2, 3, 4
slope: −4, −2, 0, 2, 4
- The slopes go up by 2 each time x goes up by 1, so the points (x, slope) lie on a line with slope 2.
- When x = 0 the slope is −4, so the line's y-intercept is −4.
- f′(x) = 2x − 4. Check at x = 3: 2(3) − 4 = 2 ✓.
f′(x) = 2x − 4. The zero of f′ at x = 2 matches the vertex of the parabola y = x² − 4x, (2, −4).
Worked exampleExample 2: A cubic pattern
Slopes of the tangent to f(x) = 2x³ are 24, 6, 0, 6, 24 at x = −2, −1, 0, 1, 2. Find f′(x).
- The slopes are symmetric and 0 at x = 0, like a parabola with vertex at the origin.
- Try f′(x) = kx². At x = 1: k(1) = 6, so k = 6.
- Check at x = 2: 6(4) = 24 ✓.
f′(x) = 6x², which is 2 × 3x²: doubling the function doubled every slope.
Worked exampleExample 3: Comparing graphs
For f(x) = x³ − 3x with f′(x) = 3x² − 3, explain how each feature of f′ shows up on the graph of f.
- f′(x) = 0 when 3x² = 3, that is, x = ±1. These are where f has horizontal tangents: the local max (−1, 2) and local min (1, −2).
- f′ is positive for x < −1 and x > 1, the intervals where f increases.
- f′ is negative for −1 < x < 1, where f decreases.
- f′ has its lowest value −3 at x = 0. That is where f is falling most steeply: its tangent at the origin has slope −3.
Zeros of f′ ↔ turning points of f; sign of f′ ↔ increasing/decreasing; the minimum of f′ ↔ the steepest downhill point of f.
Just for funCalculus has a family resemblance: the derivative of a cubic is a quadratic, the derivative of a quadratic is a line, and the derivative of a line is a constant. Every generation is one degree less dramatic.
Hands-on activityMake a slope graph by hand
You need: graph paper (or plain paper with a ruler-drawn grid), a pencil, a ruler and a scientific calculator.
- Draw y = x² − 2x carefully from x = −1 to x = 3 (plot x = −1, 0, 1, 2, 3 and join smoothly).
- At x = −1, 0, 1, 2 and 3, place your ruler as a tangent and estimate its slope by counting rise over run on the grid.
- Check each estimate with the calculator: [f(a + 0.001) − f(a − 0.001)] ÷ 0.002.
- On a second grid directly below, plot the points (x, slope). Join them.
- Find the equation of the slope graph.
What to look for: the slopes should be −4, −2, 0, 2, 4. They lie on the line y = 2x − 2, so f′(x) = 2x − 2. The slope graph crosses the x-axis at x = 1, right below the vertex (1, −1) of the parabola.
Your turn
1What does f′(5) = −2 tell you about the graph of f?
The tangent to y = f(x) at x = 5 has slope −2; f is decreasing there.
2The tangent slopes of f(x) = x² + 6x are 0, 2, 4, 6 at x = −3, −2, −1, 0. Find f′(x).
Slopes rise by 2 per unit, and f′(0) = 6: f′(x) = 2x + 6.
3The derivative of a function is f′(x) = 4. What kind of function is f?
A linear function with slope 4, such as f(x) = 4x + b.
4Slopes of g(x) = −x² at x = −2, −1, 0, 1, 2 are 4, 2, 0, −2, −4. Find g′(x).
5Use the centred estimate with h = 0.001 to estimate the slope of f(x) = x³ at x = 1.5. Compare with 3x².
f(1.501) = 3.381754501 and f(1.499) = 3.368254499, difference 0.013500002; divide by 0.002 to get 6.750001, so the slope is about 6.75. 3(1.5)² = 6.75 ✓.
6A polynomial has degree 5. What is the degree of its derivative?
7Where does the graph of f′ cross the x-axis if f(x) = x² − 10x + 1?
Where f has a horizontal tangent: at the vertex, x = 5 (f′(x) = 2x − 10).
More practice: warm-up
2What is the derivative of f(x) = 7x − 3?
f′(x) = 7 (constant slope).
3What is the derivative of f(x) = x²?
4If f′(x) > 0 on an interval, f is …
increasing on that interval.
5If f(x) = 9, a horizontal line, what is f′(x)?
More practice: on your own
1Tangent slopes of f(x) = 3x² are −12, −6, 0, 6, 12 at x = −2, −1, 0, 1, 2. Find f′(x) and explain how it relates to the derivative of x².
The slopes rise by 6 per unit and f′(0) = 0: f′(x) = 6x, which is 3 × 2x. Tripling the function triples every slope.
2The derivative of f(x) = x³ + 2 is 3x². Explain why adding 2 did not change the derivative.
Adding 2 shifts the graph up 2 units. Every tangent shifts with it but keeps the same slope, so f′ is unchanged.
3Slopes of h(x) = x³ − 12x at x = −3, −2, −1, 0, 1, 2, 3 are 15, 0, −9, −12, −9, 0, 15. Find h′(x) by comparing with the slopes of x³.
The slopes of x³ are 27, 12, 3, 0, 3, 12, 27. Each h-slope is 12 less, so h′(x) = 3x² − 12.
4Sketch in words the graph of f′ if f is a parabola opening down with vertex (4, 10).
f increases for x < 4 and decreases for x > 4, so f′ is a line with negative slope that is positive for x < 4, crosses the x-axis at x = 4, and is negative for x > 4.
5The derivative of a quadratic f is f′(x) = 2x − 8. At what x-value is the vertex of f? Does the parabola open up or down?
Vertex where f′ = 0: x = 4. f′ goes from negative to positive, so f decreases then increases: it opens up.
Challenge
1A table of slopes for an unknown cubic f gives f′(−1) = 0, f′(0) = −3 and f′(1) = 0, and f′ is a quadratic. Find f′(x), then find one possible f.
f′ is a quadratic with zeros ±1: f′(x) = k(x² − 1). f′(0) = −k = −3, so k = 3: f′(x) = 3x² − 3. One possible f is x³ − 3x (any x³ − 3x + C works).
2Using the patterns in this lesson, guess the derivative of f(x) = x⁴. Check your guess at x = 1 with the centred estimate (h = 0.001).
Pattern x² → 2x, x³ → 3x² suggests 4x³. At x = 1: [1.001⁴ − 0.999⁴] ÷ 0.002 = [1.004006004 − 0.996005996] ÷ 0.002 = 0.008000008 ÷ 0.002 ≈ 4.000004, very close to 4(1)³ = 4 ✓.
3Two functions f and g have the same derivative everywhere. What can you say about their graphs?
At every x their tangents are parallel, so the gap f(x) − g(x) never changes. One graph is a vertical translation of the other: f(x) = g(x) + C.
Quiz
1. What does the derivative f′(x) give at each value of x?
2. Tangent slopes of a function at x = 0, 1, 2, 3 are 5, 7, 9, 11. Which could be f′(x)?
3. The derivative of a degree-4 polynomial has degree:
4. The graph of f′ is below the x-axis for 1 < x < 6. What is true of f on that interval?
5. Which pair matches a function with its derivative?
6. If f′(x) = 2x − 6, where does y = f(x) have a horizontal tangent?
Big idea (A2.2): the derivative f′ is itself a function. Its value at each x is the slope of the tangent to f at that x. For polynomials, the derivative is a polynomial of one lower degree: linear → constant, quadratic → linear, cubic → quadratic.
How to help: Ask your teen to point at a place on the graph of f where f′ is zero, positive and negative, and to explain using the words "slope" and "tangent". If you have a computer, the free Desmos graphing calculator lets them type f(x) and g(x) = (f(x + 0.001) − f(x − 0.001))/0.002 to see the derivative appear.
Common difficulties:
- Plotting the y-values of f instead of its slopes when building the slope graph.
- Thinking a high point on f means a high point on f′ (a maximum of f is a zero of f′).
- Forgetting that adding a constant does not change any slopes.