3 Rates of Change and Derivatives

A progressive guide to average and instantaneous rates of change, derivative definitions and rules, motion applications, units, interpretation, and linear approximation.

From Change over an Interval to a Rate

Rates of change compare how much an output changes with how much its input changes. If ff is a function and the input changes from aa to bb, the is

f(b)−f(a)b−a.\frac{f(b)-f(a)}{b-a}.

The numerator is the change in the dependent variable, and the denominator is the change in the independent variable. Geometrically, this quotient is the slope of the through (a,f(a))(a,f(a)) and (b,f(b))(b,f(b)).

For example, suppose an object's position is

s(t)=t2+3t,s(t)=t^2+3t,

where position is measured in meters and time in seconds. From t=0t=0 to t=2t=2, the average is

s(2)−s(0)2−0=(22+3⋅2)−02=102=5 m/s.\frac{s(2)-s(0)}{2-0} =\frac{(2^2+3\cdot 2)-0}{2} =\frac{10}{2}=5\text{ m/s}.

This means the object's overall change in position was 5 m5\text{ m} per second during the interval. It does not mean that the object necessarily moved at exactly 5 m/s5\text{ m/s} at every moment.

Takeaway: An average rate describes change across an interval and is represented by a secant slope.

Instantaneous Change and the

The describes how a quantity changes at one particular input value. To estimate it at x=ax=a, use average rates over increasingly short intervals:

f(a+h)−f(a)h,\frac{f(a+h)-f(a)}{h},

where h≠0h\neq 0. As hh approaches zero, the secant lines approach the . If the limiting value exists, the 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}.

The gives this rate at every input where it exists:

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

The notation h→0h\to 0 does not mean substituting h=0h=0 into the quotient. At h=0h=0, the quotient would involve division by zero. Instead, the limit examines values for nonzero hh that become arbitrarily close to zero.

A function is at a point when its exists there. Sharp corners, cusps, discontinuities, and certain vertical tangents can prevent differentiability.

Takeaway: The is the limiting version of an average rate and represents the slope of the .

Finding Derivatives

The limit definition can be used directly to find a . Let

f(x)=x2.f(x)=x^2.

Then

f′(x)=lim⁡h→0(x+h)2−x2h=lim⁡h→0x2+2xh+h2−x2h=lim⁡h→0(2x+h)=2x.\begin{aligned} f'(x) &=\lim_{h\to 0}\frac{(x+h)^2-x^2}{h}\\ &=\lim_{h\to 0}\frac{x^2+2xh+h^2-x^2}{h}\\ &=\lim_{h\to 0}(2x+h)\\ &=2x. \end{aligned}

Thus, the of x2x^2 at x=ax=a is 2a2a. At x=3x=3,

f′(3)=2(3)=6.f'(3)=2(3)=6.

The graph therefore has tangent-line slope 66 at x=3x=3.

For routine calculations, rules are more efficient. The constant rule gives

ddx(C)=0,\frac{d}{dx}(C)=0,

and the gives

ddx(xn)=nxn−1.\frac{d}{dx}\left(x^n\right)=nx^{n-1}.

The sum and difference rules allow differentiation term by term. For example, if

f(x)=3x4−5x2+7,f(x)=3x^4-5x^2+7,

then

f′(x)=12x3−10x.f'(x)=12x^3-10x.

Takeaway: Use the limit definition to understand the and standard rules to calculate it efficiently.

Units and Interpretation

The units of a are the output units divided by the input units:

units of f′=units of funits of x.\text{units of }f'=\frac{\text{units of }f}{\text{units of }x}.

For example, position in meters divided by time in seconds gives meters per second. Cost in dollars divided by number of items gives dollars per item. A of 1212 dollars per item means that near the specified production level, one additional item is predicted to increase cost by approximately 1212 dollars, assuming the model is appropriate.

The sign and magnitude also matter. A positive indicates that the output is increasing locally; a negative indicates that it is decreasing locally. The magnitude indicates how rapidly the change occurs.

When interpreting a over a small finite change, use language such as “approximately.” A describes local behavior, so it predicts what happens near the specified input rather than guaranteeing the exact change over a large interval.

For example, if a population function satisfies

P′(5)=240,P'(5)=240,

where population is measured in people and time in years, then at year 55 the population is increasing at approximately 240240 people per year. If

T′(10)=−1.5T'(10)=-1.5

in degrees Celsius per hour, then at hour 1010 the temperature is decreasing at approximately 1.51.5 degrees Celsius per hour.

Takeaway: A meaningful interpretation states the direction, numerical rate, units, and local context.

Derivatives in Motion

For motion, let s(t)s(t) represent position as a function of time. Then is the first :

v(t)=s′(t),v(t)=s'(t),

and is the second :

a(t)=v′(t)=s′′(t).a(t)=v'(t)=s''(t).

The signs give qualitative information:

  • If v(t)>0v(t)>0, position is increasing and the object moves in the positive direction.

  • If v(t)<0v(t)<0, position is decreasing and the object moves in the negative direction.

  • If a(t)>0a(t)>0, is increasing.

  • If a(t)<0a(t)<0, is decreasing.

An object is speeding up when and have the same sign. It is slowing down when they have opposite signs.

For

s(t)=t2+3t,s(t)=t^2+3t,

we obtain

v(t)=s′(t)=2t+3v(t)=s'(t)=2t+3

and

a(t)=v′(t)=2.a(t)=v'(t)=2.

At t=2t=2 seconds,

v(2)=7 m/s,a(2)=2 m/s2.v(2)=7\text{ m/s}, \qquad a(2)=2\text{ m/s}^2.

The object is moving in the positive direction, and its is increasing. Since and are both positive at this time, the object is speeding up.

Takeaway: Position, , and form a chain of successive derivatives, and their signs must be interpreted together.

Local Approximation and Problem Solving

A can be used to estimate a nearby function value through a . If f(a)f(a) and f′(a)f'(a) are known, then for a small change hh,

f(a+h)≈f(a)+f′(a)h.f(a+h)\approx f(a)+f'(a)h.

This is the equation of the tangent-line model near x=ax=a.

Suppose a machine has produced f(100)=250f(100)=250 units after 100100 minutes and has production rate f′(100)=4f'(100)=4 units per minute. After approximately 33 more minutes,

f(103)≈f(100)+f′(100)(3)=250+4(3)=262.f(103)\approx f(100)+f'(100)(3) =250+4(3)=262.

The estimate is most reliable when the change in input is small and the production rate does not change too dramatically during those three minutes.

For rate-of-change problems, use this sequence:

  1. Define what the input and output represent.

  2. Decide whether the question concerns an interval or a single point.

  3. Choose an average-rate quotient or a accordingly.

  4. Attach the correct units.

  5. Interpret the sign and magnitude in context.

  6. Consider whether the model is reliable over the interval or near the point being studied.

Takeaway: Derivatives provide both exact local rates and practical predictions, but local approximations should not be extended carelessly over large changes.