Unit 3, lesson 20

Derivatives and Rates of Change

The derivative of a function at a point as the limit of difference quotients, read as the slope of the tangent line, an instantaneous rate of change and a velocity, with its units.

  • Theory part 39
  • Practice part 40
  • Definition of the Derivative

Theory, part 39

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Practice, part 40

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The tangent slope of a curve, the velocity of a car and the rate at which a tank drains are all the same limit in different clothes. That limit gets one name, the derivative. Once you can compute it for a single point, you can answer all three kinds of question at once.

Key terms

Derivative at a point

Definition Let ff be defined on an open interval containing aa. The derivative of ff at aa is

f′(a)=lim⁡h→0f(a+h)−f(a)h,f'(a) = \lim_{h \to 0} \frac{f(a + h) - f(a)}{h},

provided this limit exists as a finite number. In that case ff is called differentiable at aa. If the limit does not exist, or is ∞\infty or −∞-\infty, then f′(a)f'(a) does not exist.

In plain words The derivative at aa is the number the slopes of secant lines through (a,f(a))(a, f(a)) settle on as the second point slides in.

Example For f(x)=1xf(x) = \frac{1}{x} at a=2a = 2,

12+h−12h=2−(2+h)2(2+h)h=−12(2+h),h≠0, h≠−2.\frac{\frac{1}{2 + h} - \frac{1}{2}}{h} = \frac{2 - (2 + h)}{2(2 + h)h} = \frac{-1}{2(2 + h)}, \qquad h \neq 0, \ h \neq -2.

As h→0h \to 0 this tends to −14-\frac{1}{4}, so f′(2)=−14f'(2) = -\frac{1}{4}.

The difference quotient limit

Definition Writing x=a+hx = a + h turns the same limit into

f′(a)=lim⁡x→af(x)−f(a)x−a.f'(a) = \lim_{x \to a} \frac{f(x) - f(a)}{x - a}.

The two forms are equivalent. In both, the quotient is undefined at the limit point itself, so you must simplify it algebraically or estimate it before letting h→0h \to 0 or x→ax \to a.

In plain words You cannot just plug in h=0h = 0, because that gives 00\frac{0}{0}. Cancel the factor that causes the trouble first, then let the small number go to zero.

Example For f(x)=x2+3xf(x) = x^2 + 3x at a=1a = 1 the second form gives

x2+3x−4x−1=(x−1)(x+4)x−1=x+4,x≠1,\frac{x^2 + 3x - 4}{x - 1} = \frac{(x - 1)(x + 4)}{x - 1} = x + 4, \qquad x \neq 1,

which tends to 55. So f′(1)=5f'(1) = 5.

Tangent slope

Definition If ff is differentiable at aa, the tangent line to the graph of ff at (a,f(a))(a, f(a)) is the line through that point with slope f′(a)f'(a),

y=f(a)+f′(a)(x−a).y = f(a) + f'(a)(x - a).

In plain words The derivative is the slope of the line that hugs the curve at that point.

Example Let f(x)=xf(x) = \sqrt{x} and a=4a = 4. Multiply by the conjugate,

4+h−2h=(4+h)−4h(4+h+2)=14+h+2,\frac{\sqrt{4 + h} - 2}{h} = \frac{(4 + h) - 4}{h(\sqrt{4 + h} + 2)} = \frac{1}{\sqrt{4 + h} + 2},

which tends to 14\frac{1}{4}. The tangent line is y=2+14(x−4)y = 2 + \frac{1}{4}(x - 4), or y=14x+1y = \frac{1}{4}x + 1. You can compare secant and tangent slopes in the tangent line explorer.

Average and instantaneous rate of change

Definition If y=f(x)y = f(x), the average rate of change of yy with respect to xx over [a,b][a, b], with a<ba < b, is

f(b)−f(a)b−a.\frac{f(b) - f(a)}{b - a}.

The instantaneous rate of change of yy with respect to xx at x=ax = a is f′(a)f'(a), the limit of the average rates over intervals shrinking to aa.

In plain words The average rate describes a whole stretch. The instantaneous rate describes one moment or one input.

Example The area of a circle of radius rr is A(r)=πr2A(r) = \pi r^2. From r=2r = 2 to r=3r = 3 the average rate is

A(3)−A(2)3−2=9π−4π1=5π\frac{A(3) - A(2)}{3 - 2} = \frac{9\pi - 4\pi}{1} = 5\pi

square units per unit of radius. At r=2r = 2 exactly,

π(2+h)2−4πh=π(4+h)→4π,\frac{\pi(2 + h)^2 - 4\pi}{h} = \pi(4 + h) \to 4\pi,

so the instantaneous rate is 4π4\pi, which is the circumference of that circle.

Velocity and speed

Definition If s(t)s(t) is the position of an object moving along a line at time tt, its velocity at time tt is v(t)=s′(t)v(t) = s'(t), when the derivative exists. Its speed is ∣v(t)∣\lvert v(t) \rvert. Positive velocity means motion in the positive direction, negative velocity means motion in the negative direction.

In plain words Velocity has a direction built into its sign. Speed is how fast, with the direction dropped.

Example Let s(t)=t2−4ts(t) = t^2 - 4t meters after tt seconds. For any tt,

(t+h)2−4(t+h)−(t2−4t)h=2th+h2−4hh=2t+h−4→2t−4.\frac{(t + h)^2 - 4(t + h) - (t^2 - 4t)}{h} = \frac{2th + h^2 - 4h}{h} = 2t + h - 4 \to 2t - 4.

So v(t)=2t−4v(t) = 2t - 4. At t=1t = 1 the velocity is −2-2 meters per second and the speed is 22 meters per second, so the object moves backward. At t=3t = 3 the velocity is 22 and it moves forward.

Units of the derivative

Definition If the output of ff is measured in some unit UU and the input in a unit VV, then the difference quotient, and hence f′(a)f'(a), is measured in units of UU per VV.

In plain words A derivative always reads as output units per input unit. Saying the units out loud is a good check on what the number means.

Example A tank holds W(t)=100−2t2W(t) = 100 - 2t^2 liters of water tt minutes after a valve opens. At t=3t = 3,

100−2(3+h)2−(100−18)h=−12h−2h2h=−12−2h→−12.\frac{100 - 2(3 + h)^2 - (100 - 18)}{h} = \frac{-12h - 2h^2}{h} = -12 - 2h \to -12.

So W′(3)=−12W'(3) = -12 liters per minute. Three minutes in, the tank is losing water at 1212 liters per minute.

Normal line

Definition The normal line to the graph of ff at (a,f(a))(a, f(a)) is the line through that point perpendicular to the tangent line. If f′(a)≠0f'(a) \neq 0 its slope is −1f′(a)-\frac{1}{f'(a)}, and its equation is

y=f(a)−1f′(a)(x−a).y = f(a) - \frac{1}{f'(a)}(x - a).

If f′(a)=0f'(a) = 0 the tangent is horizontal and the normal line is the vertical line x=ax = a.

In plain words The normal line sticks straight out of the curve at a right angle to the tangent.

Example For f(x)=xf(x) = \sqrt{x} at (4,2)(4, 2) the tangent slope is 14\frac{1}{4}, so the normal slope is −4-4 and the normal line is y=2−4(x−4)y = 2 - 4(x - 4), or y=−4x+18y = -4x + 18.

Check yourself

  1. Use the limit definition to find f′(3)f'(3) for f(x)=2x2−xf(x) = 2x^2 - x.
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Since f(3)=15f(3) = 15, the difference quotient is

2(3+h)2−(3+h)−15h=11h+2h2h=11+2h→11,\frac{2(3 + h)^2 - (3 + h) - 15}{h} = \frac{11h + 2h^2}{h} = 11 + 2h \to 11,

so f′(3)=11f'(3) = 11.

  1. A stone’s height is s(t)=20t−5t2s(t) = 20t - 5t^2 meters after tt seconds. Find its velocity and speed at t=3t = 3.
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The difference quotient at t=3t = 3 is

s(3+h)−s(3)h=−10h−5h2h=−10−5h→−10.\frac{s(3 + h) - s(3)}{h} = \frac{-10h - 5h^2}{h} = -10 - 5h \to -10.

The velocity is −10-10 meters per second, so it is falling, and the speed is 1010 meters per second.

  1. Find the tangent and normal lines to y=1xy = \frac{1}{x} at (2,12)(2, \frac{1}{2}).
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The slope is f′(2)=−14f'(2) = -\frac{1}{4}, so the tangent is y=12−14(x−2)y = \frac{1}{2} - \frac{1}{4}(x - 2). The normal slope is 44, so the normal is y=12+4(x−2)y = \frac{1}{2} + 4(x - 2).

How this lesson connects

It also uses difference quotient.

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