Lesson Explainer: Rate of Change and Derivatives Mathematics • Higher Education

In this explainer, we will learn how to find the instantaneous rate of change for a function using derivatives and apply this in real-world problems.

We begin by recalling the definition of a derivative.

Definition: Derivative of a Function

Given a function 𝑓(π‘₯), the derivative of 𝑓(π‘₯) at π‘₯=π‘Ž is defined by 𝑓′(π‘Ž)=𝑓(π‘Ž+β„Ž)βˆ’π‘“(π‘Ž)β„Ž.limο‚β†’οŠ¦

The expression inside the limit in the definition of a derivative is called the difference quotient. Let us examine the structure of the difference quotient more closely.

For instance, say that the function value 𝑓(π‘₯) represents the temperature of a steak on a grill, where the input value π‘₯ represents time since the steak began cooking. First, we consider the meaning of the difference quotient when 𝑑>0. In this case, the numerator of the difference quotient 𝑓(π‘Ž+β„Ž)βˆ’π‘“(π‘Ž) represents the change in the temperature of the steak at time π‘Ž+β„Ž as compared to the temperature at time π‘Ž. We note that the length of this time interval is given by (π‘Ž+β„Ž)βˆ’π‘Ž=β„Ž. Hence, the difference quotient 𝑓(π‘Ž+β„Ž)βˆ’π‘“(π‘Ž)β„Ž represents the average rate of change of the temperature of the steak on the grill over the time interval [π‘Ž,π‘Ž+β„Ž].

If β„Ž<0, then π‘Ž+β„Ž<π‘Ž. In this case, the average rate of change in the temperature of the steak over the time interval [π‘Ž+β„Ž,π‘Ž] is given by 𝑓(π‘Ž)βˆ’π‘“(π‘Ž+β„Ž)π‘Žβˆ’(π‘Ž+β„Ž)=𝑓(π‘Ž)βˆ’π‘“(π‘Ž+β„Ž)βˆ’β„Ž=𝑓(π‘Ž+β„Ž)βˆ’π‘“(π‘Ž)β„Ž.

We note that this is the same expression as when β„Ž>0. Thus, for any β„Žβ‰ 0, the difference quotient of a function gives the average rate of change of the temperature over the time interval between π‘Ž and π‘Ž+β„Ž.

When we take the limit as β„Ž approaches zero, the difference quotient measures the average rate of change of a shorter and shorter interval containing π‘₯=π‘Ž. If this limit exists, then the limit would symbolically represent the average rate of change of the temperature over an interval of length zero containing π‘Ž, that is, the single element set {π‘Ž}. We refer to this quantity as an instantaneous rate of change at the point π‘₯=π‘Ž. We note that this definition coincides with that of the derivative of the function.

Theorem: Instantaneous Rate of Change of a Function

Given a function 𝑓, the instantaneous rate of change of the function 𝑓 with respect to the input variable π‘₯ at π‘₯=π‘Ž is given by its derivative 𝑓′(π‘Ž).

Note: The instantaneous rate of change of a function is also known as the rate of change of a function at a point.

Let us consider several examples where we use rules of differentiation to compute the derivative of a function and use the derivative to evaluate the instantaneous rate of change of the function at a given point.

Example 1: Evaluating the Rate of Change of a Polynomial Function at a Point

Evaluate the instantaneous rate of change of 𝑓(π‘₯)=7π‘₯+9 at π‘₯=π‘₯.

Answer

We recall that the instantaneous rate of change of a function at a point is the same as the derivative of the function evaluated at the given point. Thus, the instantaneous rate of change will be given by evaluating 𝑓′ at π‘₯=π‘₯.

Using the power rule, (π‘₯)β€²=𝑝π‘₯ for any real number 𝑝, and the constant rule (𝐢)β€²=0, we can compute 𝑓′: 𝑓′(π‘₯)=2Γ—7π‘₯+0=14π‘₯.

Substituting π‘₯=π‘₯, we obtain the instantaneous rate of change of 𝑓 at π‘₯=π‘₯ to be 14π‘₯.

Let us consider another example of the instantaneous rate of change where we will use the chain rule to compute the derivative.

Example 2: Evaluating the Rate of Change of a Radical Function at a Point

Evaluate the instantaneous rate of change of 𝑓(π‘₯)=√6π‘₯+7 at π‘₯=3.

Answer

We recall that the instantaneous rate of change of a function at a point is the same as the derivative of the function evaluated at the given point. Thus, the instantaneous rate of change will be given by 𝑓′(3). So, we need to compute the derivative 𝑓′(π‘₯) and evaluate it at π‘₯=3 to find the answer.

Recall the chain rule for two differentiable functions 𝑔 and β„Ž: (𝑔(β„Ž(π‘₯)))β€²=𝑔′(β„Ž(π‘₯))Γ—β„Žβ€²(π‘₯).

For our example, we note that 𝑓=π‘”βˆ˜β„Ž where the outside function is 𝑔(π‘₯)=√π‘₯ and the inside function is β„Ž(π‘₯)=6π‘₯+7. We can use the power rule (π‘₯)=𝑝π‘₯ to compute the derivatives of 𝑔. Since 𝑔(π‘₯)=π‘₯, we have 𝑔′(π‘₯)=12π‘₯=12√π‘₯.οŽͺ

For β„Ž(π‘₯), we have β„Žβ€²(π‘₯)=6Γ—1Γ—π‘₯+0=6.

Applying the chain rule, 𝑓′(π‘₯)=𝑔′(β„Ž(π‘₯))Γ—β„Žβ€²(π‘₯)=12√6π‘₯+7Γ—6=3√6π‘₯+7.

Evaluating at π‘₯=3, we get 𝑓′(3)=3√6Γ—3+7=3√25=35.

The instantaneous rate of change of function 𝑓 at π‘₯=3 is 35.

Let us consider another example for the instantaneous rate of change where we use the quotient rule for obtaining the derivative function.

Example 3: Differentiating Rational Functions at a Point Using the Quotient Rule

If the function 𝑓(π‘₯)=5π‘₯+74π‘₯+2, determine its rate of change when π‘₯=2.

Answer

We recall that the rate of change of a function at a point is the same as the derivative of the function evaluated at the given point. Thus, the instantaneous rate of change in this example will be given by 𝑓′(2) once we have computed the derivative function 𝑓′(π‘₯).

To compute the derivative of 𝑓, we need to apply the quotient rule: 𝑔(π‘₯)β„Ž(π‘₯)=𝑔′(π‘₯)β„Ž(π‘₯)βˆ’π‘”(π‘₯)β„Žβ€²(π‘₯)(β„Ž(π‘₯)).

Applying the quotient rule to the given function, we get ο€Ό5π‘₯+74π‘₯+2=(5π‘₯+7)β€²(4π‘₯+2)βˆ’(5π‘₯+7)(4π‘₯+2)β€²(4π‘₯+2)=5(4π‘₯+2)βˆ’4(5π‘₯+7)(4π‘₯+2)=20π‘₯+10βˆ’20π‘₯βˆ’28(4π‘₯+2)=βˆ’18(4π‘₯+2).

Evaluating the derivative function at π‘₯=2, 𝑓′(2)=βˆ’18(4Γ—2+2)=βˆ’18100=βˆ’950.

Hence, the rate of change of 𝑓 when π‘₯=2 is βˆ’950.

In the previous examples, we considered the instantaneous rate of change of an algebraic function. However, the interpretation of the derivative as the instantaneous rate of change is more significant when applied to a function with a real-world meaning. In such contexts, we should be careful to use the correct unit for the instantaneous rate of change.

For instance, let us recall the example from earlier where we let 𝑓(π‘₯) represent the temperature of a steak on a grill at time π‘₯. We can assign 𝑓(π‘₯) the temperature unit Celsius, and 𝑑 the time unit second. Then, the numerator of the difference quotient 𝑓(π‘₯+β„Ž)βˆ’π‘“(π‘₯) carries the temperature unit Celsius. On the other hand, the denominator of the quotient β„Ž carries the time unit second. Together, we can see that the difference quotient has the unit Celsius/second. In other words, this average rate of change measures how many degrees Celsius the temperature of the steak is changing per second. We note that taking the limit as β„Ž approaches zero does not change the unit of the expression. Hence, the unit of the instantaneous rate of change is Celsius/second.

In general, the unit of the instantaneous rate of change is given by unitofthefunctionvalueunitoftheinputvalue𝑓(π‘₯)π‘₯.

In our next example, we will consider the instantaneous rate of change of a biological function.

Example 4: Finding the Rate of Change of a Polynomial Function Representing the Biomass of a Bacterial Culture at a Certain Time

The biomass of a bacterial culture in milligrams as a function of time in minutes is given by 𝑓(𝑑)=71𝑑+63. What is the instantaneous rate of growth of the culture when 𝑑=2minutes?

Answer

We recall that the instantaneous rate of change of a function at a point is the same as the derivative of the function evaluated at the given point. Thus, the instantaneous rate of change will be given by 𝑓′(2). So, we need to compute the derivative 𝑓′(𝑑) and evaluate it at 𝑑=2 to find the answer.

We can use the power rule, (π‘₯)β€²=𝑝π‘₯, to compute the derivative of 𝑓: 𝑓′(𝑑)=213𝑑.

Evaluating the derivative function at 𝑑=2, we obtain 𝑓′(2)=213Γ—2=852.

We recall that the unit of the instantaneous rate of change of a function is unitofthefunctionvalueunitoftheinputvalue𝑓(𝑑)𝑑.

In this example, the function value represents the biomass of a bacterial culture in milligrams, while the input variable 𝑑 represents time in minutes. Hence, the unit of the instantaneous rate of change is milligrams per minute (mg/min).

The instantaneous rate of growth of the culture when 𝑑=2 is 852 mg/min.

In our final example, we will find the instantaneous rate of change of a function that is described in words.

Example 5: Finding the Rate of Change in the Area of a Shrinking Circular Disc Using Related Rates

A circular disk preserves its shape as it shrinks. What is the rate of change of its area with respect to radius when the radius is 59 cm?

Answer

We recall that the rate of change of a function at a point is the same as the derivative of the function evaluated at the given point. In this example, we are looking for the rate of change of the area of a circle with respect to its radius. Hence, we need to begin by defining the function representing the area of a circle with the input variable as its radius. Using the variable π‘Ÿ for the radius in centimetres, we denote the area of the circle with radius π‘Ÿ by 𝑓(π‘Ÿ). Then, 𝑓(π‘Ÿ)=πœ‹π‘Ÿ.

To find the instantaneous rate of change, we need to find the derivative function. Since πœ‹ is a constant, we can obtain the derivative of 𝑓 by using the power rule, (π‘₯)β€²=𝑝π‘₯: 𝑓′(π‘Ÿ)=πœ‹ο€Ήπ‘Ÿο…β€²=πœ‹(2π‘Ÿ)=2πœ‹π‘Ÿ.

Since we are looking for the rate of change when the radius is 59 cm, we evaluate 𝑓′ at π‘Ÿ=59: 𝑓′(59)=2πœ‹Γ—59=118πœ‹.

We recall that the unit of the instantaneous rate of change is unitofthefunctionvalueunitoftheinputvalue𝑓(𝑑)𝑑.

In this example, the function value 𝑓(π‘Ÿ) is the area of the circle when the radius is measured in centimetre. Thus, 𝑓(π‘Ÿ) carries the unit square centimetre (cm2). The input variable is the radius, which has the unit centimetre (cm). So, the unit of 𝑓′ is square centimetre per centimetre (cm2/cm).

The rate of change of the area of a circle with respect to its radius when the radius is 59 cm is 118πœ‹ cm2/cm.

Let us finish by recapping a few important concepts from this explainer.

Key Points

  • Given a function 𝑓, the instantaneous rate of change of the function 𝑓 with respect to the input variable π‘₯ at π‘₯=π‘Ž is given by its derivative 𝑓′(π‘Ž).
  • The instantaneous rate of change of a function is approximated by its average rate of change over an interval that is shrinking toward a single point. For this reason, this term is also referred to as the rate of change of a function at a point.
  • In application problems, the unit of the instantaneous rate of change is given by unitofthefunctionvalueunitoftheinputvalue𝑓(π‘₯)π‘₯.

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