The derivative at a point
A rate of change needs two points, but a speedometer shows a speed at a single instant. How can one point have a slope, and how is that number found exactly?
Where you are 3 in Definition of the derivative
Why this exists
A Rate of change compares two moments: change in output over change in input. That is the slope of the line through two points of a graph, and it is an average. A funicular car climbing a hillside track speeds up as it leaves the bottom station, so “meters climbed per second” over a ten-second stretch says little about how fast the car moves at second 8. Yet the speed at second 8 is a real number; a speedometer would show it.
Rates at an instant from shrinking intervals estimated such a number by computing averages over shorter and shorter intervals and watching where they settle. This lesson replaces the watching with a limit and gets the number exactly. The tool is the The difference quotient, \dfrac{f(a + h) - f(a)}{h}, the average rate of change of f from a to a + h. Let h \to 0 and the average over an interval becomes the rate at one point. That limit is the derivative of f at a, the number at the center of calculus.
Day 1 defines it, reads it as the slope of the tangent line, and uses it as an instantaneous velocity. Day 2 gives a second form of the definition, reads derivatives from graphs, shows a point where no derivative exists, and recognizes a limit that is secretly a derivative.
Definition
A secant line of a graph is a line through two points of it. Through P(a, f(a)) and Q(a + h, f(a + h)) the secant slope is the difference quotient
for h \ne 0. It is also the average rate of change of f on the interval from a to a + h (with h positive or negative).
The derivative of f at a, written f'(a), is
provided this limit exists as a finite number. When it does, f is differentiable at a. The derivative is also called the instantaneous rate of change of f at a.
The tangent line to the graph at P(a, f(a)) is the line through P with slope f'(a): the limit of the secant lines through P as Q slides toward P. When f'(a) does not exist, this definition gives no tangent line (a vertical tangent, which can occur, is a separate case handled in Differentiability: corners, cusps and piecewise functions).
Instantaneous velocity. If s(t) is the position of an object along a line at time t, the average velocity from a to a + h is \dfrac{s(a + h) - s(a)}{h}, and the instantaneous velocity at t = a is s'(a).
The alternate form. Writing x for the second point a + h gives the same number:
The function needs to be defined on an open interval around a, since h approaches 0 from both sides.
Intuition
Zoom in on a smooth curve at a point and it looks more and more like a straight line. The derivative is the slope of that line.
Figure 1 shows the motion that the limit describes. The point P(4, 4) is fixed on y = x^2 - 3x, and Q is a second point of the curve. Each secant PQ cuts across the curve, so its slope is an average over the stretch from P to Q. Move Q toward P and the stretch shrinks, the secants pivot about P, and their slopes (7, then 6, then 5.5) close in on 5. The line they approach touches the curve at P and runs along with it there: the tangent line.
Nothing in the limit asks Q to land on P. At h = 0 the quotient is \displaystyle \frac{0}{0}, which is meaningless; the limit asks only what the slopes approach while h stays nonzero. That is why the derivative needed limits of the form \displaystyle \frac{0}{0} before it could be defined.
Read the other way, the derivative turns a graph into slopes you can see (Figure 2). Where the graph rises, the tangent tilts up and f'(a) > 0. Where it falls, f'(a) < 0. At a smooth top or bottom the tangent is level and f'(a) = 0. A steeper graph means a larger |f'(a)|.
Results
Day 1: the definition
Derivative as slope. If f'(a) exists, it is the slope of the tangent line to y = f(x) at (a, f(a)), and the tangent line is
Average versus instantaneous. The average rate of change on an interval is a secant slope; the instantaneous rate at a point is the limit of those slopes as the interval shrinks to the point. For a position function the first is average velocity, the second is velocity.
Lines. If f(x) = mx + b, then f'(a) = m at every a.
Day 2: the alternate form and its uses
The two forms agree. For every function and every a, \displaystyle \lim_{h \to 0} \frac{f(a + h) - f(a)}{h} and \displaystyle \lim_{x \to a} \frac{f(x) - f(a)}{x - a} either both exist and are equal or both fail to exist.
One-sided test. f'(a) exists exactly when the difference quotient has the same finite limit as h \to 0^- and as h \to 0^+.
Recognizing a derivative. Any limit of the shape \displaystyle \lim_{h \to 0} \frac{f(a + h) - f(a)}{h} or \displaystyle \lim_{x \to a} \frac{f(x) - f(a)}{x - a} is f'(a) for the function and point that fit it.
Why it is true
The two forms agree. Substitute x = a + h. Then x - a = h, so the quotient \displaystyle \frac{f(x) - f(a)}{x - a} is exactly \displaystyle \frac{f(a + h) - f(a)}{h}. Also x \to a exactly when h \to 0, and x \ne a exactly when h \ne 0, so the two limits look at the same quotients and agree, including when they fail to exist. The same substitution on one side at a time gives the one-sided test, by the result in One sided limits and when a limit exists.
Lines. For f(x) = mx + b,
for every h \ne 0, so the limit is m. Every secant of a line is the line itself.
The tangent line. The line through (a, f(a)) with slope m is y - f(a) = m(x - a), the point-slope form; taking m = f'(a) gives the boxed equation.
Worked examples
Day 1: secant slopes closing in
Let f(x) = x^2 - 3x and P = (4, f(4)). Find the slopes of the secant lines from P to Q(4 + h, f(4 + h)) for h = \pm 1, \pm 0.1, \pm 0.01, then find f'(4) exactly.
SolutionFirst f(4) = 16 - 12 = 4, so P = (4, 4). Expanding f(4 + h) once handles every h:
So for h \ne 0 the secant slope is
where canceling h is allowed because h \ne 0.
| h | f(4 + h) | secant slope 5 + h |
|---|---|---|
| 1 | 10 | 6 |
| 0.1 | 4.51 | 5.1 |
| 0.01 | 4.0501 | 5.01 |
| -0.01 | 3.9501 | 4.99 |
| -0.1 | 3.51 | 4.9 |
| -1 | 0 | 4 |
The slopes close in on 5 from both sides, and the algebra says why: the slope is 5 + h, so the limit is
The table suggests the answer; the limit proves it. So f'(4) = 5: the tangent line at (4, 4) has slope 5, and near x = 4 the output grows about 5 times as fast as the input.
Day 1: a reciprocal and its tangent line
Find g'(3) for g(x) = \dfrac{6}{x}, and write the tangent line to y = g(x) at x = 3.
SolutionHere g(3) = 2 and \displaystyle g(3 + h) = \frac{6}{3 + h}. Subtract over the common denominator 3 + h:
Dividing by h (nonzero) cancels it:
This is continuous at h = 0, so the limit is found by substitution:
The graph falls at x = 3, which matches the negative sign. The tangent line passes through (3, 2) with slope \displaystyle -\frac{2}{3}:
Check: at x = 3 the line gives -2 + 4 = 2, the point of contact. Building tangent and normal lines in general is the subject of Tangent lines and normal lines.
Day 1: Gurpreet’s funicular car
Gurpreet rides a funicular car up a hillside track. Its distance along the track from the bottom station, t seconds after it starts, is s(t) = 2t + \dfrac{t^2}{4} meters, for 0 \le t \le 20. Find the average velocity on [8, 12] and on [8, 9], and the instantaneous velocity at t = 8.
SolutionFirst s(8) = 16 + 16 = 32, s(12) = 24 + 36 = 60 and s(9) = 18 + 20.25 = 38.25. The average velocities are
and
in meters per second. They differ because the car is speeding up. For the velocity at the instant t = 8, expand s(8 + h):
so the average velocity from 8 to 8 + h is
With h = 4 this gives 7 and with h = 1 it gives 6.25, matching the two averages. As h \to 0 it approaches 6, so s'(8) = 6: at second 8 the car moves at 6 meters per second. The units of a derivative are always output units per input unit; reading them in context is taken further in The derivative as a rate of change.
Day 2: the alternate form
(a) Find f'(4) for f(x) = x^2 - 3x again, this time with \lim_{x \to a} \dfrac{f(x) - f(a)}{x - a}. (b) Find k'(25) for \displaystyle k(x) = \sqrt{x}.
Solution(a) With f(4) = 4, the numerator is x^2 - 3x - 4, which has x = 4 as a root and so has x - 4 as a factor:
for x \ne 4, so f'(4) = \lim_{x \to 4} (x + 1) = 5, agreeing with Day 1. Geometrically x + 1 is the slope of the secant through (4, 4) and (x, f(x)); the first form named the second point by its distance h, this one names it by its position x.
(b) Here k(25) = 5, and the quotient is \displaystyle \dfrac{\sqrt{x} - 5}{x - 25}. The alternate form pays off because x - 25 factors as a difference of squares, \displaystyle (\sqrt{x} - 5)(\sqrt{x} + 5), for x \ge 0:
for x \ne 25. The last expression is continuous at 25, so k'(25) = \dfrac{1}{5 + 5} = \dfrac{1}{10}. The first form would reach the same number by multiplying by the conjugate \displaystyle \sqrt{25 + h} + 5.
Day 2: reading slopes from a graph
Figure 2 shows the graph of a function p. Its tangent line at A(1, 3) passes through (3, 5), and its tangent at the top point B(3, 4) is horizontal. Find p'(1) and p'(3), give the sign of p'(5), and put the three in order.
Solutionp'(1) is the slope of the tangent line at A, and two points of that line are known:
The tangent at B is horizontal, so its slope is 0 and p'(3) = 0. At C(5, 3) the graph is falling as x increases, so a tangent there tilts down and p'(5) < 0. Hence
A graph read this way gives the sign of f'(a) at a glance and its size to the accuracy of the drawing. An exact value needs either a tangent line through known points, as at A, or a formula.
Day 2: a corner at the origin
Show that f(x) = |x| has no derivative at 0.
SolutionThe difference quotient at 0 is
For h > 0 this is \displaystyle \frac{h}{h} = 1, and for h < 0 it is \displaystyle \frac{-h}{h} = -1. So
while
The one-sided limits differ, so the two-sided limit does not exist and f'(0) does not exist. In the picture, secants from the origin to the right all have slope 1 and secants to the left all have slope -1; they settle on two different lines, so no single tangent line exists. The function is continuous at 0; the graph is unbroken there but has a corner. Corners, cusps and pieced-together functions are treated fully in Differentiability: corners, cusps and piecewise functions.
Day 2: a limit in disguise
Recognize each limit as a derivative f'(a), naming f and a, and find its value. (a) \lim_{h \to 0} \dfrac{(2 + h)^5 - 32}{h} (b) \lim_{x \to 1} \dfrac{x^4 - 1}{x - 1}
Solution(a) The pattern \displaystyle \frac{f(a + h) - f(a)}{h} fits with a = 2 and f(x) = x^5, since f(2) = 32. So the limit is f'(2) for f(x) = x^5. To find it, expand by the binomial pattern:
so for h \ne 0
which approaches 80. The limit is 80. Other readings work too: g(x) = (2 + x)^5 at a = 0 gives the same quotient, so the same limit is g'(0).
(b) The alternate form \displaystyle \frac{f(x) - f(a)}{x - a} fits with a = 1 and f(x) = x^4, since f(1) = 1. Factoring,
so the quotient is x^3 + x^2 + x + 1 for x \ne 1, which approaches 4. The limit is f'(1) = 4.
Boundaries
- The limit must exist. f'(a) is defined only when the difference quotient has a finite limit. It fails at a corner (one-sided limits differ, as for |x| at 0), at a vertical tangent (the quotient grows without bound, as for \displaystyle \sqrt[3]{x} at 0), and at any break in the graph.
- f must be defined at a and on both sides of it. The quotient uses f(a), so a hole at a rules out a derivative there, and the two-sided limit needs a + h in the domain for small h of both signs.
- Continuity is necessary but not enough. If f'(a) exists, then \displaystyle f(a + h) - f(a) = h \cdot \frac{f(a + h) - f(a)}{h} approaches 0 \cdot f'(a) = 0 by the product law for limits, so f is continuous at a; a jump or any other break rules out a derivative. But |x| is continuous at 0 and still has no derivative there.
- A secant slope is not the derivative. Any single secant slope, however short the interval, is an average and only approximates f'(a). Only the limit is exact, and for a line the two agree.
- The derivative is local. f'(a) describes f only in the immediate neighborhood of a. It does not say how much f changes over an interval of length 1.
Where this goes next
Today’s derivatives are numbers, one point at a time. Doing the same computation with a letter x in place of a produces a formula for the slope at every point at once: Finding a derivative from the definition does that by hand, and The derivative as a function turns a graph’s slopes into the graph of a new function, f'. The tangent line found in Example 2 grows into Tangent lines and normal lines, with normal lines and linear approximation, and The derivative as a rate of change reads f'(a) with its units as a rate in a real setting. Points where the limit fails are sorted out in Differentiability: corners, cusps and piecewise functions.
Exercises
Day 1: the definition
- For f(x) = 2x^2 - x, find the slope of the secant line from (1, f(1)) to (1 + h, f(1 + h)) as an expression in h, evaluate it at h = 0.1, 0.01 and -0.01, and find f'(1).
- Find f'(-2) for f(x) = 5 - x^2 from the definition.
- Find f'(-1) for f(x) = x^3 + 1 from the definition.
- Find f'(2) for f(x) = \dfrac{3}{x + 1} from the definition.
- Find f'(10) for f(x) = 7 - 2x from the definition, and explain the answer from the graph.
Day 1: Harjit’s car
Harjit rides a second car that brakes as it nears the top station. Its distance along the track is s(t) = 10t - \dfrac{t^2}{4} meters, t seconds after it starts, for 0 \le t \le 20.
- Find the average velocity on [2, 6] and the instantaneous velocity at t = 6.
- Find s'(14) from the definition and compare it with s'(6). What does the comparison say about the car?
Day 2: the alternate form
- Find f'(3) for f(x) = x^2 + 2x using \lim_{x \to a} \dfrac{f(x) - f(a)}{x - a}.
- Find f'(1) for f(x) = \dfrac{1}{x^2} using the alternate form.
- Find f'(1) for \displaystyle f(x) = \sqrt{x + 3} using the alternate form.
- Find f'(2) for f(x) = x^3 - x using the alternate form.
Day 2: tangent lines and graphs
- The tangent line to the graph of f at (3, 1) passes through (5, -3). Find f(3) and f'(3).
- The line y = 4x - 7 is tangent to the graph of g at x = 2. Find g(2) and g'(2).
- In Figure 2, is p'(0) larger or smaller than p'(1)? Explain from the graph.
Day 2: when the derivative fails
- Show that f(x) = |x - 3| has no derivative at 3.
- Find f'(0) for f(x) = x|x|, or show that it does not exist.
Day 2: limits that are derivatives
For each limit, name a function f and a number a for which it equals f'(a), then find its value.
- \displaystyle \lim_{h \to 0} \dfrac{\sqrt{16 + h} - 4}{h}
- \lim_{x \to 3} \dfrac{x^4 - 81}{x - 3}
- \displaystyle \lim_{h \to 0} \dfrac{\frac{1}{5 + h} - \frac{1}{5}}{h}
- \lim_{h \to 0} \dfrac{(1 + h)^{10} - 1}{h}
Reasoning
- Explain why the two forms of the definition always give the same number.
- Manpreet says that wherever f'(a) = 0, the graph must have a top or a bottom point. Test the claim on f(x) = x^3 at a = 0.
- Gurpreet reasons that since f'(4) = 5 for f(x) = x^2 - 3x, the output must rise by exactly 5 from x = 4 to x = 5. Check this and explain what f'(4) does and does not say.
- Harjit says \lim_{h \to 0} \dfrac{(3 + h)^2 - 9}{h} is f'(3) for f(x) = x^2, and Manpreet says it is g'(0) for g(x) = (3 + x)^2. Explain why both are right and find the value.
- Explain why substituting h = 0 into \dfrac{f(a + h) - f(a)}{h} can never give the derivative, and why, when f is continuous at a, the limit always has the form \displaystyle \frac{0}{0}.
Answers
- Slope 3 + 2h; values 3.2, 3.02, 2.98; f'(1) = 3. Here f(1) = 1 and f(1 + h) = 1 + 3h + 2h^2, so the quotient is \displaystyle \frac{3h + 2h^2}{h} = 3 + 2h, which approaches 3.
- f'(-2) = 4. f(-2) = 1 and f(-2 + h) = 5 - (4 - 4h + h^2), which is 1 + 4h - h^2, so the quotient is 4 - h, which approaches 4.
- f'(-1) = 3. f(-1) = 0 and f(-1 + h) = 3h - 3h^2 + h^3, so the quotient is 3 - 3h + h^2, which approaches 3.
- \displaystyle f'(2) = -\frac{1}{3}. f(2) = 1 and \displaystyle f(2 + h) - 1 = \frac{3 - (3 + h)}{3 + h}, which is \displaystyle \frac{-h}{3 + h}, so the quotient is \displaystyle \frac{-1}{3 + h}, which approaches \displaystyle -\frac{1}{3}.
- f'(10) = -2. The quotient is \displaystyle \frac{7 - 2(10 + h) - (7 - 20)}{h}, which is \displaystyle \frac{-2h}{h} = -2 for every h \ne 0. The graph is a line of slope -2, and every secant of a line is the line itself.
- Average velocity 8 meters per second; s'(6) = 7 meters per second. s(6) = 51 and s(2) = 19, so the average is \displaystyle \frac{32}{4} = 8. Also \displaystyle s(6 + h) = 51 + 7h - \frac{h^2}{4}, so the quotient is \displaystyle 7 - \frac{h}{4}, which approaches 7.
- s'(14) = 3 meters per second, less than s'(6) = 7, so the car is slowing down, as a braking car should. s(14) = 91 and \displaystyle s(14 + h) = 91 + 3h - \frac{h^2}{4}, so the quotient is \displaystyle 3 - \frac{h}{4}, which approaches 3.
- f'(3) = 8. f(3) = 15 and x^2 + 2x - 15 = (x - 3)(x + 5), so the quotient is x + 5 for x \ne 3, which approaches 8.
- f'(1) = -2. \displaystyle \frac{1}{x^2} - 1 = \frac{1 - x^2}{x^2}, which is \displaystyle \frac{-(x - 1)(x + 1)}{x^2}, so the quotient is \displaystyle -\frac{x + 1}{x^2} for x \ne 1, which approaches -2.
- \displaystyle f'(1) = \frac{1}{4}. f(1) = 2 and x - 1 = (x + 3) - 4, which factors as \displaystyle (\sqrt{x + 3} - 2) times \displaystyle (\sqrt{x + 3} + 2), so the quotient is \displaystyle \frac{1}{\sqrt{x + 3} + 2}, which approaches \displaystyle \frac{1}{4}.
- f'(2) = 11. f(2) = 6 and x^3 - x - 6 = (x - 2)(x^2 + 2x + 3), so the quotient is x^2 + 2x + 3 for x \ne 2, which approaches 4 + 4 + 3 = 11.
- f(3) = 1 and f'(3) = -2. The point of contact is on the graph, and the tangent slope is \displaystyle \frac{-3 - 1}{5 - 3} = -2.
- g(2) = 1 and g'(2) = 4. The tangent line meets the graph at the point of contact, so g(2) = 4 \cdot 2 - 7 = 1, and the slope of the tangent line is g'(2) = 4.
- Larger: p'(0) > p'(1). The graph is rising at both points but more steeply at x = 0, and the rise flattens out toward the top at x = 3, so the tangent at x = 0 tilts up more than the one at A.
- The quotient at 3 is \displaystyle \frac{|h|}{h}, which is 1 for h > 0 and -1 for h < 0. The one-sided limits 1 and -1 differ, so f'(3) does not exist; the graph has a corner at (3, 0).
- f'(0) = 0. The quotient is \displaystyle \frac{h|h|}{h} = |h| for h \ne 0, and |h| \to 0 from both sides. Unlike |x|, this graph has no corner at the origin.
- \displaystyle f(x) = \sqrt{x} at a = 16; value \displaystyle \frac{1}{8}. Multiplying by the conjugate, the quotient is \displaystyle \frac{1}{\sqrt{16 + h} + 4}, which approaches \displaystyle \frac{1}{8}.
- f(x) = x^4 at a = 3; value 108. x^4 - 81 = (x - 3)(x + 3)(x^2 + 9), so the quotient is (x + 3)(x^2 + 9), which approaches 6 \cdot 18 = 108.
- \displaystyle f(x) = \frac{1}{x} at a = 5; value \displaystyle -\frac{1}{25}. The numerator is \displaystyle \frac{-h}{5(5 + h)}, so the quotient is \displaystyle \frac{-1}{5(5 + h)}, which approaches \displaystyle -\frac{1}{25}.
- f(x) = x^{10} at a = 1; value 10. Multiplying out ten factors of (1 + h) gives 1, plus 10h (one h chosen from one of the ten factors), plus terms that each contain h^2. So the quotient is 10 plus terms that each contain h, which approaches 10.
- Put x = a + h. Then x - a = h, so the quotients \displaystyle \frac{f(x) - f(a)}{x - a} and \displaystyle \frac{f(a + h) - f(a)}{h} are the same number, and x \to a with x \ne a exactly when h \to 0 with h \ne 0. The two limits are taken over the same quotients, so they agree, or both fail.
- The claim is false. For f(x) = x^3 the quotient at 0 is \displaystyle \frac{h^3}{h} = h^2, which approaches 0, so f'(0) = 0 and the tangent at the origin is horizontal. But x^3 < 0 for x < 0 and x^3 > 0 for x > 0, so the origin is neither a top nor a bottom: the graph levels off for an instant and keeps rising.
- The claim is false: f(5) - f(4) = 10 - 4 = 6, not 5. That 6 is the average rate on [4, 5]. The derivative f'(4) = 5 is the rate at the single input 4; because the slope of this graph grows as x grows, the average over [4, 5] comes out larger. f'(4) says the change is close to 5h for small h, not over a whole unit.
- Both readings give the identical quotient: with f(x) = x^2, f(3 + h) - f(3) = (3 + h)^2 - 9; with g(x) = (3 + x)^2, g(0 + h) - g(0) = (3 + h)^2 - 9. A limit depends only on the quotient, so both are right. The value is 6, since (3 + h)^2 - 9 = 6h + h^2 and the quotient 6 + h approaches 6.
- At h = 0 the quotient reads \displaystyle \frac{f(a) - f(a)}{0} = \frac{0}{0}, which has no value, for every f defined at a: the quotient is simply not defined there. Continuity at a means f(a + h) \to f(a) as h \to 0, so the numerator approaches 0 while the denominator h approaches 0 too, and the limit has the form \displaystyle \frac{0}{0}. The limit must be found by first rewriting the quotient for h \ne 0 (canceling, combining fractions, using a conjugate).