Lesson Video: Rate of Change and Derivatives | Nagwa Lesson Video: Rate of Change and Derivatives | Nagwa

# Lesson Video: Rate of Change and Derivatives Mathematics

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

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### Video Transcript

In this video, we’ll learn how to find the instantaneous rate of change for a function using derivatives and apply this to real-world problems. We’ll look at the slope of secant and tangent lines before defining a formula for the rate of change of a function. And we’ll look at the application of this formula to a variety of examples, including those involving symbol polynomials, functions involving roots and quotients.

We’ll begin this lesson by recalling some information about the tangent line to a curve. Let’s imagine this curve has the equation 𝑦 is equal to 𝑓 of 𝑥. And we want to find the tangent line to our curve at some point 𝑃, given by the order pair 𝑎, 𝑓 of 𝑎. We consider a nearby point 𝑄, given by the ordered pair 𝑥, 𝑓 of 𝑥, where 𝑥 of course is not equal to 𝑎. And we find the slope of the secant line joining 𝑃 to 𝑄. We find the slope, of course, using the formula change in 𝑦 divided by change in 𝑥 or 𝑦 two minus 𝑦 one over 𝑥 two minus 𝑥 one. Taking the coordinates for 𝑃 and 𝑄, we find the slope of our secant line to be 𝑓 of 𝑥 minus 𝑓 of 𝑎 over 𝑥 minus 𝑎.

We’re then going to let 𝑄 approach 𝑃 along the curve. And we do so by letting the value of 𝑥 approach 𝑎. As we do, the distance between 𝑃 and 𝑄 becomes smaller and smaller. And we approach the slope of the tangent line at 𝑃. And we come to our first definition. The tangent line to the curve 𝑦 equals 𝑓 of 𝑥 at the point 𝑎, 𝑓 of 𝑎 is the line through 𝑃 with the slope 𝑚, given by the limit as 𝑥 approaches 𝑎 of 𝑓 of 𝑥 minus 𝑓 of 𝑎 over 𝑥 minus 𝑎, provided that this limit exists. And this definition actually leads us directly into a second. This time, we let ℎ be equal to the difference between 𝑥 and 𝑎. That’s 𝑥 minus 𝑎. By adding 𝑎 to both sides, we find 𝑥 to be equal to 𝑎 plus ℎ. And the slope of our secant line 𝑃𝑄 is now 𝑓 of 𝑎 plus ℎ minus 𝑓 of 𝑎 over ℎ. This time as 𝑥 approaches 𝑎, ℎ approaches zero. So our second expression for the slope of the tangent line at 𝑃 is now the limit as ℎ approaches zero of 𝑓 of 𝑎 plus ℎ minus 𝑓 of 𝑎 all over ℎ.

When finding slope in real-life situations, we call it the rate of change. It essentially means the same as slope. So we have the definition for the rate of change. Now, in fact, limits of this form arise regularly when we compute rate of change, specifically in the sciences in engineering. And we, therefore, give it a special name. We call it the derivative of the function 𝑓 at 𝑎 and we denote it 𝑓 prime of 𝑎. And this is great. Because by considering these definitions, we can now say that the tangent line to 𝑦 equals 𝑓 of 𝑥 at the point 𝑎 𝑓 of 𝑎 is the line that goes through this point whose slope is equal to 𝑓 prime of 𝑎, the derivative of 𝑓 at 𝑎.

So now, we have all of these definitions. Let’s look how we might apply them in rate of change and derivative problems.

Evaluate the rate of change of 𝑓 of 𝑥 equals seven 𝑥 squared plus nine at 𝑥 equals 𝑥 one.

Remember, the definition for the rate of change of a function or its derivative at some point 𝑥 equals 𝑎 is the limit as ℎ approaches zero of 𝑓 of 𝑎 plus ℎ minus 𝑓 of 𝑎 all over ℎ. In this question, 𝑓 of 𝑥 is equal to seven 𝑥 squared plus nine. And we’re looking to find the rate of change at 𝑥 equals 𝑥 one. So we’re going to let 𝑎 be equal to 𝑥 one. Then, we’re interested in finding the rate of change at 𝑥 one. So that’s 𝑓 prime of 𝑥 one. That’s therefore equal to the limit as ℎ approaches zero of 𝑓 of 𝑥 one plus ℎ minus 𝑓 of 𝑥 one all over ℎ.

Our next job is to substitute 𝑥 one plus ℎ and 𝑥 one into our original function. So 𝑓 of 𝑥 one plus ℎ is seven times 𝑥 one plus ℎ squared plus nine. We distribute 𝑥 one plus ℎ all squared. And we get 𝑥 one squared plus two 𝑥 one ℎ plus ℎ squared. And then, we distribute again and we get seven 𝑥 one squared plus 14𝑥 one ℎ plus seven ℎ squared plus nine. 𝑓 of 𝑥 one is a little simpler. It’s simply seven 𝑥 one squared plus nine. Let’s substitute these back into our expression for the rate of change of our function at 𝑥 one.

It’s the limit shown. And, of course, we can distribute the parentheses. And our final two terms become negative seven 𝑥 one squared minus nine. And this is great because seven 𝑥 one squared minus seven 𝑥 one squared is zero and nine minus nine is also zero. And 𝑓 prime of 𝑥 one is, therefore, the limit as ℎ approaches zero of 14𝑥 one ℎ plus seven ℎ squared over ℎ. We’re still not quite ready to perform direct substitution. But we can divide both of our terms on the numerator by ℎ. And now, the rate of change is the limit as ℎ approaches zero of 14𝑥 one plus seven ℎ. And now, we can perform direct substitution. It’s 14𝑥 one plus seven times zero, which is, of course, just 14𝑥 one. The rate of change of 𝑓 of 𝑥 equals seven 𝑥 squared plus nine at 𝑥 equals 𝑥 one is 14𝑥 one.

In this example, we ended up finding a general equation for the rate of change of the function. We could use this to find the particular rate of change at any point, given a value for 𝑥 one. For instance, let’s say we wanted to find the rate of change of the function at the point where 𝑥 equals two. We let 𝑥 one be equal to two. And the rate of change becomes 14 times two, which is 28.

We’ll now look at an example, where we are looking to find the rate of change at a specific point.

Evaluate the rate of change of 𝑓 of 𝑥 equals the square root of six 𝑥 plus seven at 𝑥 equals three.

Remember, the definition for the rate of change of a function or its derivative at a point 𝑥 equals 𝑎 is the limit as ℎ approaches zero of 𝑓 of 𝑎 plus ℎ minus 𝑓 of 𝑎 over ℎ. In this question, 𝑓 of 𝑥 is equal to the square root of six 𝑥 plus seven. And we’re looking to find the rate of change at 𝑥 equals three. So we’re going to let 𝑎 be equal to three. The rate of change is the derivative of our function evaluated at 𝑥 equals three. Using our earlier definition, we see that it’s the limit as ℎ approaches zero of 𝑓 of three plus ℎ minus 𝑓 of three all over ℎ.

Our job is going to be to work out 𝑓 of three plus ℎ and 𝑓 of three. To find 𝑓 of three plus ℎ, we replace 𝑥 in our original expression for the function with three plus ℎ. So it’s the square root of six times three plus ℎ plus seven. Distributing the parentheses and we get 18 plus six ℎ plus seven, which simplifies to six ℎ plus 25. So 𝑓 of three plus ℎ is the square root of six ℎ plus 25. We repeat this process for 𝑓 of three. This time, it’s the square root of six times three plus seven, which is root 25 or simply five.

Substituting this back into our original definition for 𝑓 prime of three and we find that it’s now equal to the limit as ℎ approaches zero of the square root of six ℎ plus 25 minus five over ℎ. Well, we’re not quite ready to perform direct substitution. If we did, we’d be dividing by zero, which we know to be undefined. So instead, we’re going to need to do something a little bit clever to manipulate our expression. We begin by writing the square root of six ℎ plus 25 as six ℎ plus 25 to the power of one-half. We’re then going to multiply the numerator and denominator of our limit by the conjugate of the numerator. So that’s six ℎ plus 25 to the power of one-half plus five.

Let’s distribute the numerator. We begin by multiplying the first term in each expression. Now, six ℎ plus 25 to the power of one-half times itself. Well, that’s simply six ℎ plus 25. We multiply six ℎ plus 25 to the power of one-half by five. And then, we multiply negative five by six ℎ plus 25 to the power of one-half. And we end up with five lots of six ℎ plus 25 to the power of one-half minus five lots of six ℎ plus 25 to the power of one-half, which is, of course, zero. Finally, we multiply negative five by five and we get negative 25. For now, we’ll leave the denominator as shown. Let’s clear some space for the next step.

We noticed that 25 minus 25 is zero. So our numerator becomes six ℎ. And then, we spot that we can simplify by dividing through by ℎ. And in fact, we’re now ready to perform direct substitution. We’re going to replace ℎ in our limit with zero. Then, our denominator becomes six times zero plus 25 to the power of one-half or 25 to the power of one-half. Well, 25 to the power of one-half is the square root of 25, which is five. So 𝑓 prime of three is six over five plus five, which is, of course, six tenths. That simplifies to three-fifths. The rate of change of our function at 𝑥 equals three is, therefore, three-fifths.

We’ve now seen how this process can work for linear functions and those involving roots. Next, we’ll have a look at using the rate of change function on an example that involves a quotient.

If the function 𝑓 of 𝑥 equals five 𝑥 plus seven over four 𝑥 plus two, determine its rate of change when 𝑥 is equal to two.

We recall the definition for the rate of change of the function or its derivative. It’s the limit as ℎ approaches zero of 𝑓 of 𝑎 plus ℎ minus 𝑓 of 𝑎 over ℎ assuming that limit exists. In this question, 𝑓 of 𝑥 is equal to five 𝑥 plus seven over four 𝑥 plus two. And we want to find the rate of change when 𝑥 is equal to two. So we’re going to let 𝑎 be equal to two. So we need to evaluate 𝑓 prime of two, the rate of change or the derivative of our function, when 𝑥 equals two.

By our definition, it’s the limit as ℎ approaches zero of 𝑓 of two plus ℎ minus 𝑓 of two over ℎ. Let’s work out 𝑓 of two plus ℎ and 𝑓 of two. To find 𝑓 of two plus ℎ, we replace each instance of 𝑥 in our original function with two plus ℎ. And we get five times two plus ℎ plus seven over four times two plus ℎ plus two. And when we distribute our parentheses and simplify, we get 17 plus five ℎ over 10 plus four ℎ. Similarly, 𝑓 of two is five times two plus seven over four times two plus two which is 17 tenths. And we now see that 𝑓 prime of two is the limit as ℎ approaches zero of the difference between these all over ℎ.

There are two fractions in our numerator. So we’re going to simplify by creating a common denominator there. We’ll multiply the numerator and denominator of the first fraction on the numerator by 10 and the second fraction on our numerator by 10 plus four ℎ. And when we do, we achieve the numerator shown. Well, this doesn’t make a lot of sense. But actually, dividing this entire fraction by ℎ is the same as timesing it by one over ℎ. So we rewrite our denominator as a ℎ times 100 plus 40ℎ. And then, we simplify a numerator to negative 18ℎ. And you might now see that we can simplify further by dividing through by ℎ.

And we’re now ready to perform direct substitution. By replacing ℎ with zero, we find that 𝑓 prime of two is negative 18 over 100, which simplifies to negative nine over 50. The rate of change of our function 𝑓 of 𝑥 when 𝑥 is equal to two is negative nine over 50.

In our final example, we’re going to consider the real-world applications for rate of change and the derivative.

The circular disc preserves its shape as it shrinks. What is the rate of change of its area with respect to radius when the radius is 59 centimetres?

We begin by recalling the formula that allows us to calculate the rate of change of a function at a given point when 𝑥 is equal to 𝑎. It’s 𝑓 prime of 𝑎 equals the limit as ℎ approaches zero of 𝑓 of 𝑎 plus ℎ minus 𝑓 of 𝑎 over ℎ, where 𝑓 prime is the derivative of the function. But we don’t seem to have a function here. So let’s consider what we do know about the area of a circle. It’s given by the formula 𝐴 equals 𝜋𝑟 squared. We could write this as 𝐴 of 𝑟. 𝐴 is a function of 𝑟. That means that the rate of change of 𝐴 with respect to 𝑟 is the derivative of 𝐴 with respect to 𝑟.

Now, we’re trying to find the rate of change when the radius is equal to 59. So we’re going to let 𝐴 be equal to 59. We want to find 𝐴 prime of 59. And by definition, that must be equal to the limit as ℎ approaches zero of 𝐴 of 59 plus ℎ minus 𝐴 of 59 all over ℎ. Let’s work out what 𝐴 of 59 plus ℎ and 𝐴 of 59 actually are. 𝐴 of 𝑟 is 𝜋𝑟 squared. So 𝐴 of 59 plus ℎ is 𝜋 times 59 plus ℎ squared. We distribute our parentheses. And we see that this is equal to 𝜋 times 3481 plus 118ℎ plus ℎ squared. Similarly, 𝐴 of 59 is 𝜋 times 59 squared, which is 3481𝜋. We can replace 𝐴 of 59 plus ℎ and 𝐴 of 59 with these two expressions in our definition for the derivative. And when we factor by 𝜋, we see the numerator is 𝜋 times 3481 plus 118ℎ plus ℎ squared minus 3481. Now, of course, these give us zero.

So we’re looking for the limit as ℎ approaches zero of 𝜋 times 118ℎ plus ℎ squared all over ℎ. And you might now spot we can actually divide through by ℎ. And our derivative is now the limit as ℎ approaches zero of 𝜋 times 118 plus ℎ. We’re now ready to perform direct substitution. We let ℎ be equal to zero. And when we do, we find that 𝐴 prime of 59 equals 118𝜋. The rate of change of the circular disc’s area with respect to its radius is 118𝜋 centimetres squared per centimetre. Now, you might be inclined to think that the answer should be negative. We’re told that the circular disc is shrinking. However, that’s a bit of a trick. The area changes in the same positive or negative direction as the radius. So, in fact, it is indeed a positive rate of change.

In this video, we’ve seen that we can use the formula the limit as ℎ approaches zero of 𝑓 of 𝑎 plus ℎ minus 𝑓 of 𝑎 over ℎ to find the rate of change of a function 𝑓 of 𝑥 at the point where 𝑥 equals 𝑎 provided that limit exists. We saw that we often call this the derivative of the function 𝑓 and that we can use this formula to find the general form and a particular solution given a value of 𝑥. But that we need to carefully consider the nature of the function when thinking about contextual examples.