The derivative as a function
The derivative at a point is one slope at one place. If you record the slope at every place, what kind of object do you get, and how do you read it off a graph or a table of measurements?
Where you are 3 in Definition of the derivative
When you reach for it
The derivative at a point gives one number, f'(a), the slope of the graph at one input. Finding a derivative from the definition does the same limit with x left as a letter and gets a formula. This lesson treats that result as a function in its own right, with a domain and a graph of its own.
Reach for it when the slope is needed everywhere at once, and especially when there is no formula. A graph of f is enough to sketch the graph of f'; a graph of f' is enough to sketch f, up to its height; a table of measurements is enough to estimate f' at every row. Day 1 works with graphs and sets out the notation. Day 2 works with tables.
The procedure
Day 1: the derivative is a function
The derivative of f is the function f' whose value at each input x is the slope of the graph of f at (x, f(x)):
Its domain is every x where this limit exists, which can be smaller than the domain of f: |x| is defined at 0 but has no derivative there.
Slopes become heights. To graph f', read the slope of the graph of f at an input x and plot that number as a height above x on a second set of axes. Each feature of f' comes from a feature of f:
| graph of f at or near x | graph of f' |
|---|---|
| level tangent | on the x-axis: f'(x) = 0 |
| rising as x increases | on or above the x-axis |
| falling as x increases | on or below the x-axis |
| steeper | farther from the x-axis |
| straight piece of slope m | horizontal piece at height m |
| corner | a gap: f'(x) does not exist |
“On or above” is needed because a rising graph can level off for an instant: x^3 rises everywhere, yet its derivative is 0 at x = 0. Each row follows from the definition; if f rises near x, every secant slope there is positive, so their limit cannot be negative. Precise statements and converses belong to Increasing, decreasing, constant, positive and negative intervals. A sketch made this way gets signs, zeros and rough sizes right; exact heights need a tangent line through known points or a formula.
Going back from f' to f. Read the table from right to left: where f' > 0 the graph of f rises, where f' < 0 it falls, and the larger |f'|, the steeper. That fixes the shape of f but not its height, because a constant cancels in every difference quotient:
so f and f + C have the same derivative. Conversely, two functions with the same derivative on an interval differ by a constant there; that needs the mean value theorem and is stated here without proof. So f' fixes f up to a vertical shift, and one known value f(a) fixes the shift.
Notation.
| prime notation | Leibniz notation |
|---|---|
| f'(x) | \dfrac{df}{dx}, or \dfrac{dy}{dx} when y = f(x) |
| derivative of a formula | \dfrac{d}{dx}\big[x^3\big] = 3x^2 |
| f'(a) | \left.\dfrac{dy}{dx}\right\vert_{x = a} |
\dfrac{dy}{dx} is read “dee y dee x”. It recalls \dfrac{\Delta y}{\Delta x}, the secant slope whose limit it is, but here it is one symbol, not a fraction. \dfrac{d}{dx} is an instruction, “differentiate with respect to x”. Leibniz notation names the input variable (for a position s at time t the velocity is \dfrac{ds}{dt}) and carries units: units of y per unit of x, read in a setting in The derivative as a rate of change.
Day 2: estimating from a table
When f is known only at listed inputs, no limit can be taken; the best available is a secant slope over the shortest interval the table offers. Rates at an instant from shrinking intervals did this for one rate; here the quotients are applied at every row, turning a table of f into a table of f'. At a row x_i with neighbors x_{i-1} < x_i < x_{i+1}:
At the first row only the forward quotient exists, at the last only the backward one.
Equal spacing. With rows h apart, the centered quotient is the average of the other two and its interval is centered on x_i. For a smooth function it is usually far more accurate: its error shrinks like h^2, against h for a one-sided quotient (stated without proof; Rates at an instant from shrinking intervals saw it happen for a formula).
Uneven spacing. Write h_1 = x_i - x_{i-1}, h_2 = x_{i+1} - x_i, and m_1, m_2 for the backward and forward quotients. Since f(x_{i+1}) - f(x_{i-1}) = h_1 m_1 + h_2 m_2,
a weighted average leaning toward the slope over the longer interval, the one reaching farther from x_i. If f is differentiable on [x_{i-1}, x_{i+1}], the mean value theorem (used here without proof) says each one-sided quotient equals f' somewhere inside its own interval, so if f' is steadily rising or steadily falling, f'(x_i) lies between m_1 and m_2. A single estimate that weights the nearer interval more is
the slope at x_i of the parabola through the three data points (stated here without proof; exercise 21 shows the formula is exact for every quadratic).
Worked examples
Day 1: slopes into heights
The top of Figure 1 shows the graph of a function f. Its tangents at (1, 6) and (3, 2) are horizontal, and its tangent at (2, 4) passes through (3, 1). Sketch the graph of f'. Then check the sketch against the formula behind the graph, f(x) = x^3 - 6x^2 + 9x + 2.
SolutionThe level tangents give the zeros: f'(1) = 0 and f'(3) = 0.
The signs come from rising and falling. Left of x = 1 the graph rises, so f' > 0; between 1 and 3 it falls, so f' < 0; right of 3 it rises, so f' > 0.
The sizes come from steepness. Between 1 and 3 the fall is steepest halfway, at (2, 4), where the tangent has slope
so f' dips to a lowest point near (2, -3). Outside [1, 3] the graph of f gets steeper and steeper, so f' grows without bound both ways. The graph of f' is U-shaped (Figure 1, bottom).
Check with the definition, expanding as in Finding a derivative from the definition:
Dividing by h \ne 0 and letting h \to 0 leaves
an upward parabola with zeros 1 and 3 and vertex at x = 2, where f'(2) = -3. Every feature of the sketch is confirmed.
Day 1: a graph made of segments
A function f on [0, 6] has a graph made of segments joining (0, 1), (2, 5), (4, 5) and (6, -1) (Figure 2, top). Find and graph f'.
SolutionThe tangent to a line is the line itself, so f' is constant along each piece. The slopes are \displaystyle \frac{5 - 1}{2 - 0} = 2, then \displaystyle \frac{5 - 5}{4 - 2} = 0, then \displaystyle \frac{-1 - 5}{6 - 4} = -3:
At x = 2 the difference quotient is 2 for small h < 0 and 0 for small h > 0. The one-sided limits differ, so f'(2) does not exist; likewise at x = 4. At the endpoints only one side exists, so f' is given on the open pieces. Its graph is three steps with open circles at their ends (Figure 2, bottom). The graph of f is unbroken, yet f' jumps; corners are the subject of Differentiability: corners, cusps and piecewise functions.
Day 1: from f’ back to f
The top of Figure 3 is the graph of a derivative, f'(x) = 2x - \dfrac{x^2}{2}. It crosses the x-axis at 0 and 4 and is highest at (2, 2). Sketch f given f(0) = 0. How many functions have this derivative?
SolutionFor x < 0, f' < 0, so f falls; at 0 the tangent is level; on (0, 4), f' > 0, so f rises. Falling, level, rising: f has a low point, at the origin since f(0) = 0. The climb is gentle near 0, steepest at x = 2 (slope 2), and gentle again near 4. At 4 the tangent is level and for x > 4 the graph falls, so f has a high point at x = 4. The solid curve in Figure 3 has these features.
The graph of f' does not show the height of the high point at a glance; the rise from 0 to 4 is built from every slope in between, an idea taken up in Approximating area with Riemann sums. One function that fits is f(x) = x^2 - \dfrac{x^3}{6}: the derivatives of x^2 and x^3 are 2x and 3x^2 (both found from the definition in Finding a derivative from the definition), and the difference quotient of x^2 - \dfrac{x^3}{6} splits into the quotient for x^2 minus \dfrac{1}{6} of the quotient for x^3, so this f has derivative 2x - \dfrac{x^2}{2}, and f(0) = 0. Its high point is at height
Infinitely many functions have this derivative, x^2 - \dfrac{x^3}{6} + C for every constant C (the dashed curve is C = 2); the condition f(0) = 0 picks exactly one.
Day 1: Laima’s snowball
Laima packs a snowball of volume V = \dfrac{4}{3}\pi r^3 cubic centimeters when its radius is r centimeters. (a) Write the derivative of V with respect to r in prime and in Leibniz notation, and find it. (b) Find it at r = 6, in both notations, with units. (c) Explain why \dfrac{d}{dr}\big[V(6)\big] is a different thing.
Solution(a) The derivative is V'(r), or \dfrac{dV}{dr}, or \dfrac{d}{dr}\Big[\dfrac{4}{3}\pi r^3\Big]. The constant \displaystyle \frac{4}{3}\pi multiplies every difference quotient,
and the second factor approaches 3r^2, as for the cube in Finding a derivative from the definition. So
the surface area of the snowball. That fits: a thin new layer of thickness h adds about 4\pi r^2 \cdot h.
(b) Differentiate first, then substitute:
about 452.4 cubic centimeters per centimeter: near r = 6, each extra millimeter of radius adds about 45 cubic centimeters.
(c) V(6) = 288\pi is a single number, and a constant has derivative 0, so \dfrac{d}{dr}\big[V(6)\big] = 0. V'(6) means differentiate, then put in r = 6.
Day 2: Gintare’s snow stake
Gintare reads the snowpack depth on a stake every two weeks through a season. (a) Estimate D'(4) with the forward, backward and centered quotients. (b) Estimate D'(t) at every t. (c) Describe the graph of D' and what it says about the snowpack.
| t (weeks) | D(t) (cm) |
|---|---|
| 0 | 40 |
| 2 | 64 |
| 4 | 92 |
| 6 | 112 |
| 8 | 120 |
| 10 | 116 |
| 12 | 98 |
| 14 | 66 |
| 16 | 30 |
(a) In centimeters per week:
The spacing is even, so the centered value is the average of the other two, and it is the one to use: D'(4) \approx 12 centimeters per week.
(b) Centered at every inner row, one-sided at the ends: for instance \displaystyle \frac{64 - 40}{2} = 12 at t = 0, \displaystyle \frac{98 - 120}{4} = -5.5 at t = 10, and \displaystyle \frac{30 - 66}{2} = -18 at t = 16.
| t | D'(t) \approx |
|---|---|
| 0 | 12 |
| 2 | 13 |
| 4 | 12 |
| 6 | 7 |
| 8 | 1 |
| 10 | -5.5 |
| 12 | -12.5 |
| 14 | -17 |
| 16 | -18 |
(c) This table samples the function D', and plotting it (Figure 4) shows its shape. D' is positive through t = 8 and negative from t = 10, so the snowpack grew until somewhere between weeks 8 and 10 and shrank after; since D'(8) \approx 1 is small, the peak came a little after week 8. D' is largest, about 13, near week 2, when snow piled up fastest. After that D' falls steadily: new snow arrives more slowly, melting overtakes it, and by the end the pack loses about 18 centimeters a week.
Day 2: Laima’s readings, unevenly spaced
In spring Laima reads a second stake when she can reach it. Estimate D'(5) and D'(9).
| t (days) | D(t) (cm) |
|---|---|
| 0 | 120 |
| 4 | 112 |
| 5 | 109 |
| 9 | 95 |
| 12 | 82 |
At t = 5 the gaps are h_1 = 1 and h_2 = 4, and the one-sided quotients are
The quotient across both neighbors is
It leans toward -3.5, the slope over [5, 9], and its interval is centered at 6.5, not 5. The slopes over the four gaps are -2, -3, -3.5 and \displaystyle -\frac{13}{3}, steadily more negative, so the melt is speeding up; if D' keeps falling steadily, D'(5) lies between -3.5 and -3. Weighting the nearer interval more,
centimeters per day.
At t = 9, h_1 = 4, h_2 = 3, m_1 = -3.5 and \displaystyle m_2 = \frac{82 - 95}{3} = -\frac{13}{3}, so D'(9) lies between about -4.33 and -3.5, and
about -4 centimeters per day. The quotient across both neighbors, \displaystyle \frac{82 - 109}{7} \approx -3.86, falls in the same range. With uneven data, report the range as well as the estimate.
Boundaries
- f' exists only where the limit does. Its graph has a gap at every corner, vertical tangent or break in the graph of f.
- A graph gives signs and rough sizes. Exact values need a tangent line through known points or a formula.
- Going back gives f only up to a constant, and only on an interval. On a domain with a gap the constant can differ from piece to piece: the function equal to 0 for x < 0 and 1 for x > 0 has derivative 0 wherever it is defined and is not constant.
- Table estimates are secant slopes. They are approximations, and they say nothing about what f does between rows. A quotient across two neighbors is centered only when the spacing is equal.
- \dfrac{dy}{dx} is one symbol for now, not a quotient of two numbers.
Where this goes next
Because f' is a function, it has a derivative of its own, the Second and higher derivatives, which leads to concavity. The sign reading of Day 1 becomes exact in Increasing, decreasing, constant, positive and negative intervals and the The first and second derivative tests, and it all comes together in Curve sketching and connecting f, f' and f''. Going back from f' to f grows into antiderivatives, starting with Antiderivatives and indefinite integrals. Leibniz notation pays off in The chain rule and Linear approximation and differentials, where dy and dx get meanings of their own.
Exercises
Day 1: from the graph of f to the graph of f’
- Using Figure 5, find where g'(x) = 0, where g'(x) > 0 and where g'(x) < 0.
- In Figure 5 the tangent line at the origin passes through (1, 2). Find g'(0), then sketch g', explaining why g'(x) approaches 0 far to the left and right.
- A function k has a graph made of segments joining (0, 0), (1, 3), (4, 3) and (5, 1). Give k'(x) on each open piece and say where k' does not exist.
- The graph of f(x) = 8x - x^2 is a downward parabola with vertex (4, 16). Describe the graph of f' from this alone, then confirm it by finding f'(x) from the definition.
- In Example 1, put f'(0.5), f'(2) and f'(3.5) in order from the graph, then find them from f'(x) = 3x^2 - 12x + 9.
Day 1: from f’ back to f
- The graph of f' is the line f'(x) = x - 2, and f(2) = 1. Describe the graph of f, then check that \displaystyle f(x) = \frac{1}{2}(x - 2)^2 + 1 fits.
- f'(x) = 3 for every x, and f(0) = -1. Find f.
- In Example 1, suppose the graph of f' is the same but f(1) = 2. Give the new formula for f and its new level points.
- The graph of f' is the parabola f'(x) = 3(x - 1)^2, which touches the x-axis at x = 1. Describe the graph of f, and check that f(x) = (x - 1)^3 fits.
Day 1: notation
- For y = x^3, write the derivative in Leibniz notation in two ways and find \left.\dfrac{dy}{dx}\right\vert_{x = -2}.
- A circle of radius r centimeters has area A = \pi r^2. Find \dfrac{dA}{dr} and its value at r = 5 with units, and say what familiar quantity it is.
- For f(x) = x^2, find f'(3) and \dfrac{d}{dx}\big[f(3)\big].
Day 2: estimating from a table
Aurimas reads the snowpack depth S(t) at a higher stake, t days after the new year.
| t (days) | S(t) (cm) |
|---|---|
| 0 | 150 |
| 10 | 168 |
| 20 | 180 |
| 30 | 186 |
| 40 | 184 |
| 50 | 175 |
| 60 | 160 |
- Estimate S'(20) with the forward, backward and centered quotients.
- Estimate S'(t) at every t, centered where possible.
- From your estimates, between which readings did the depth peak, and when was it falling fastest? ### Day 2: uneven spacing
- For f(x) = x^3, use the values at 1.9, 2 and 2.2 to find the backward quotient, the forward quotient, the quotient across both neighbors and the weighted estimate of f'(2). Compare each with the exact f'(2) = 12.
The next two exercises use this table.
| x | f(x) |
|---|---|
| 1 | 2.0 |
| 1.5 | 2.6 |
| 3 | 3.5 |
| 4 | 3.8 |
- Find the backward quotient, the forward quotient and the quotient across both neighbors at x = 1.5. Show that the last is a weighted average of the first two and say which way it leans. If f' is steadily falling, between which values does f'(1.5) lie?
- Estimate f' at all four rows, using the quotient across both neighbors at the inner rows, and say what the estimates suggest about the graph of f.
Reasoning
- Aurimas says that where the graph of f has a peak, the graph of f' has a peak too. Test the claim on Example 1.
- Gintare says a graph of f' can never tell you f(0). Explain why she is right.
- Laima claims the weighted estimate \dfrac{h_2 m_1 + h_1 m_2}{h_1 + h_2} gives the exact derivative of any quadratic f(x) = Ax^2 + Bx + C at a, from the points a - h_1, a, a + h_2, whatever the gaps. Prove it.
- Show that with unequal gaps h_1 \ne h_2 the quotient across both neighbors equals the average of the backward and forward quotients only when m_1 = m_2.
- Every forward quotient in a table of f at t = 0, 1, 2, \ldots, 10 is positive. Can you conclude that f rose the whole time? What can you conclude?
Answers
- g'(x) = 0 at x = -2 and x = 2, where the tangents are level. g' > 0 on (-2, 2), where the graph rises; g' < 0 for x < -2 and x > 2, where it falls.
- g'(0) = 2, the slope of the line through (0, 0) and (1, 2). The graph is steepest at the origin, so g' is highest there, falls to 0 at x = \pm 2, and is negative beyond. Far out the graph of g flattens toward the x-axis, so g' climbs back toward 0 from below. (The figure is \displaystyle g(x) = \frac{8x}{x^2 + 4}, whose derivative has lowest value \displaystyle -\frac{1}{4} at \displaystyle x = \pm 2\sqrt{3}.)
- k'(x) = 3 on (0, 1), 0 on (1, 4) and -2 on (4, 5). k' does not exist at the corners x = 1 and x = 4, and only one-sided slopes exist at 0 and 5.
- f'(x) = 8 - 2x. The parabola rises left of 4, is level at 4 and falls after, more steeply the farther from 4, so f' is positive, then 0 at x = 4, then negative. The quotient is \displaystyle \frac{8h - 2xh - h^2}{h} = 8 - 2x - h, which approaches 8 - 2x: a line of slope -2 through (4, 0).
- f'(2) < f'(0.5) = f'(3.5). At 2 the graph falls, at 0.5 and 3.5 it rises, and the graph looks the same after a half-turn about (2, 4), which keeps slopes, so the two rising slopes are equal. From the formula, f'(0.5) = 0.75 - 6 + 9 = 3.75, f'(3.5) = 36.75 - 42 + 9 = 3.75 and f'(2) = -3.
- f falls for x < 2, has a low point at (2, 1), and rises for x > 2, steeper the farther from 2: a parabola opening up. Check: \displaystyle \frac{1}{2}\big((x + h - 2)^2 - (x - 2)^2\big) equals \displaystyle (x - 2)h + \frac{h^2}{2}, so the quotient \displaystyle (x - 2) + \frac{h}{2} approaches x - 2; and f(2) = 1.
- f(x) = 3x - 1. The line 3x has derivative 3, so every function with derivative 3 is 3x + C; f(0) = -1 gives C = -1.
- f(x) = x^3 - 6x^2 + 9x - 2, the old f moved down 4 since the old f(1) was 6. Level points (1, 2) and (3, -2).
- f rises everywhere, since f' > 0 except at x = 1, where the tangent is level for an instant; there is no high or low point. Check: (x + h - 1)^3 - (x - 1)^3 is 3(x - 1)^2 h + 3(x - 1)h^2 + h^3, so the quotient approaches 3(x - 1)^2.
- \dfrac{dy}{dx} = \dfrac{d}{dx}\big[x^3\big] = 3x^2, and at x = -2 it is 3 \cdot 4 = 12.
- \dfrac{dA}{dr} = 2\pi r, since the derivative of r^2 is 2r. At r = 5 it is 10\pi \approx 31.4 square centimeters per centimeter. It is the circumference: a thin ring of width h has area about 2\pi r \cdot h.
- f'(3) = 6, since f'(x) = 2x. But f(3) = 9 is a constant, so \dfrac{d}{dx}\big[f(3)\big] = 0.
- Forward \displaystyle \frac{186 - 180}{10} = 0.6, backward \displaystyle \frac{180 - 168}{10} = 1.2, centered \displaystyle \frac{186 - 168}{20} = 0.9, in centimeters per day; the best estimate is S'(20) \approx 0.9.
- 1.8, 1.5, 0.9, 0.2, -0.55, -1.2, -1.5 at t = 0, 10, \ldots, 60. The ends are one-sided, \displaystyle \frac{168 - 150}{10} and \displaystyle \frac{160 - 175}{10}; inner rows are centered, for instance \displaystyle \frac{184 - 180}{20} = 0.2 at t = 30.
- The estimates change sign between t = 30 and t = 40, so the depth peaked between those readings, a little after day 30 since S'(30) \approx 0.2 is small (interpolating between 0.2 and -0.55 gives about day 33). The most negative estimate, -1.5 at t = 60, says it was falling fastest at the last reading.
- Backward \displaystyle \frac{8 - 6.859}{0.1} = 11.41, forward \displaystyle \frac{10.648 - 8}{0.2} = 13.24, across \displaystyle \frac{10.648 - 6.859}{0.3} = 12.63, weighted \displaystyle \frac{0.2(11.41) + 0.1(13.24)}{0.3} = 12.02. The errors are 0.59, 1.24, 0.63 and 0.02: with uneven gaps the quotient across both neighbors is no better than the backward one, while the weighted estimate is far closer.
- Backward \displaystyle \frac{2.6 - 2.0}{0.5} = 1.2, forward \displaystyle \frac{3.5 - 2.6}{1.5} = 0.6, across \displaystyle \frac{3.5 - 2.0}{2} = 0.75. Indeed \displaystyle \frac{0.5(1.2) + 1.5(0.6)}{2} = 0.75: the weights are the gap lengths, so it leans toward 0.6, the slope over the longer interval. If f' is steadily falling, 0.6 \le f'(1.5) \le 1.2.
- 1.2, 0.75, 0.48, 0.3 at x = 1, 1.5, 3, 4, from \displaystyle \frac{2.6 - 2.0}{0.5}, \displaystyle \frac{3.5 - 2.0}{2}, \displaystyle \frac{3.8 - 2.6}{2.5} and \displaystyle \frac{3.8 - 3.5}{1}. All positive and shrinking: f rises, more and more slowly.
- The claim is false. In Example 1 the graph of f peaks at (1, 6), where f'(1) = 0, while f'(0.5) = 3.75 is larger. At a smooth peak the tangent is level, so f' is 0 there, not largest.
- Adding a constant C to f changes no difference quotient, since C cancels. So f and f + C have the same graph of f' but values at 0 that differ by C.
- \displaystyle m_1 = \frac{f(a) - f(a - h_1)}{h_1} simplifies to A(2a - h_1) + B, and \displaystyle m_2 = \frac{f(a + h_2) - f(a)}{h_2} to A(2a + h_2) + B. Then h_2 m_1 + h_1 m_2 is (h_1 + h_2)(2Aa + B) - Ah_1h_2 + Ah_1h_2, which is (h_1 + h_2)(2Aa + B). Dividing by h_1 + h_2 gives 2Aa + B for all gaps, and this is f'(a), since f'(x) = 2Ax + B.
- The quotient across both neighbors is \displaystyle \frac{h_1 m_1 + h_2 m_2}{h_1 + h_2}, and subtracting \displaystyle \frac{m_1 + m_2}{2} leaves \displaystyle \frac{(h_1 - h_2)(m_1 - m_2)}{2(h_1 + h_2)}. With h_1 \ne h_2 this is 0 only when m_1 = m_2.
- No. A positive forward quotient says only that f is higher at the end of each gap than at its start, so f(0) < f(1) < \cdots < f(10). Between readings f could dip and recover, and a table cannot rule that out.