Category: Derivatives

We are now in the real meat of calculus….rates and derivatives. Essentially, rates are simply slopes. Depending on the field you are working in or the representation of the data or function, it might have a different terminology. Keeping all of these straight is the biggest challenge in the class. Let’s try to put what we know into a table.

Common Description	Graphical Interpretation	Mathematical Terminology
Average Rate of Change of f(x) from x=a to x=b	Slope between (a,f(a)) and (b,f(b))	Slope of the Secant Line from x=a to x=b
Instantaneous Rate of Change of f(x) at x=a	Slope between (a,f(a)) and (a+h,f(a+h)) where h is very small	Slope of the Tangent Line on f(x) at x=a

 

If it feels like we use slope to calculate everything…you are right. The only difference is where the two points are located. If they are located fairly far apart, then we are calculating an average rate of change. If they are located an infinitesimal amount apart, then we are calculating the instantaneous rate of change.

But here is where it gets hazy…if the representation of the function does not allow us to pick the points infinitely close together, we may approximate the instantaneous rate of change with an average rate of change between two points that are as close as the representation allows. I think this point is bothering many of you. It occurs most often when we try to find the instantaneous rate from a data table. In that case, the points are where they are at and we can pick pairs that are closer and closer together. We pick them as close as we can near the place we want the instantaneous rate of change and say we are estimating the instantaneous rate of change.

If we are given a graph, we can draw a secant line between the points to calculate the average rate of change. The only estimating done in this case is estimating where the points are on the graph. Our eyes are only so good in reading off the values. For the instantaneous rate of change, we draw the tangent line where we want the rate and eyeball its slope. Again, since we are making educated guesses about the slopes, the numbers are estimates based on our ability to read the slope.

If we are given the function’s formula, we can calculate the average rate or the instantaneous rate exactly. For the average rate of change of f(x) over x = a to x = b, we calculate [latex]{{f(b) – f(a)} \over {b – a}}[/latex].

For the instantaneous rate of change, we use the function’s formula to calculate the limit [latex]\mathop {lim }\limits_{h \to 0} {{f(a + h) – f(a)} \over h}[/latex].

The estimating comes from the fact that we may not be able to find two points infinitesimally close or the fact that we cannot calculate the slope perfectly from how the function is given to us.

Another cautionary note…different disciplines call the derivative by different names. Mathematicians call it the derivative of course, but in physics it might be called the instantaneous rate. In finance and economics, it will be called the marginal function. And in other fields you’ll here other terms.

A few of you had trouble with the problems referring to slopes of tangent lines. Remember, the slope of a tangent line to a function is given by the derivative. So by setting the derivative equal to the desired slope value, we can solve for the x value.

Problem 1 For the function $latex f(x)=3{{x}^{2}}-4x+1$.

1. 1. Find where the tangent line is horizontal.
   2. Find where the tangent line’s slope is equal to -1.

Problem 2 For the function $latex f(x)=2{{x}^{2}}-5x+7$.

1. 1. Find where the tangent line is horizontal.
   2. Find where the tangent line’s slope is equal to 1.

Problem 3 For the function $latex f(x)=2{{x}^{3}}-11{{x}^{2}}-8x+1$.

1. 1. Find where the tangent line is horizontal.

In this last example, you need to factor the quadratic to find the two places where the tangent line is horizontal (has a slope of zero). In some instances, you might need to use the quadratic formula to solve the equation.

In many business problems, we are interested in knowing the rate at which some quantity is changing. If this quantity is modeled by a function, the rate is modeled by the derivative of the function.

In the problem below, we are given a model of revenue for Verizon, a national wireless carrier. To find the rate, we’ll need to take the derivative of the revenue function.

Questions that ask about rate or refer to “marginal” are questions about the derivative. In this case, it is referring to the value of the derivative at c = 80. The derivative of the revenue function is

The value of the derivative at 80 million connections is

To find the units on this rate, recall that the derivative is also the slope of the tangent line at c = 80. The slope of this line has vertical units of billions of dollars and horizontal units of millions of connections. This may be simplified to

Since a billion divided by a million is one thousand. So  means that an additional connection at this level results in an increase in revenue of 0.31265 thouusand dollars or $321.65.

A tangent line to a function is a line that looks most like the function at a point. In common terms, it just grazes the function.

To find its equation, we need to locate the point where the two meet as well as the slope of the function at that point. Then we can use the slope-intercept form or point-slope form of a line to get the equation.

Since this problem is asking for the equation of a line, let’s start with the point-slope form

This requires a point (x₁, y₁) and slope m. We’ll use the function to get the point and the derivative to get the slope of the tangent line.

Find the point: We are given a point x = 3. To find the corresponding y value, put the x value into the function

Find the slope of the tangent line: We need h′(3) to get the slope of the tangent line. We’ll use the Power Rule to take the derivative,

The slope of the tangent is

Write the equation of the tangent line: Putting the point (3, 10) and the slope 9 into the line yields

If you are asked to write this in slope-intercept form, you’ll need to solve this for y to give

If you graph h(x) and the tangent line together, it should be obvious that your tangent line is correct (ie. tangent).

The most difficult part of finding a derivative is evaluating the limit involved in the definition of the derivative at a point. Often there is some algebra and simplifying involved as the example below demonstrates.

Solution The definition of the derivative of g(x) at x = 3 is

The function value g(3) is calculated to be

The function value g(3 + h) must be calculated carefully.

Form the difference quotient and simplify:

The derivative is completed by taking the limit as h approaches zero,

The derivative of g(x) evaluated at x = 3, g´(3), is 11.