A.5 Differentials; Begin to Determine 𝑑𝑓𝑑𝑥 at 𝒙 =𝒂

On this screen we're going to introduce differentials, a key Calculus concept, by building from the ideas you used in your simple calculations on the preceding screens. We'll also use those ideas to lay the groundwork for how to determine the rate at which a function changes at a given point.

Summarizing What We've Done So Far

In the preceding Topic, we developed the method of linear approximations to compute a variety of values for a few different functions. For each calculation, the problem statement provided

  • the function itself. For instance in Example 1, 𝑓(𝑥) =𝑥2.
  • a particular value of 𝑥 for which we can easily compute the function. For instance, at 𝑥 =3, 𝑥2 =9.
  • the rate at which the function changes at this particular value of 𝑥. For instance, 𝑑𝑓𝑑𝑥at 𝑥=3 =6.
Given those pieces of information, we could then use our linear approximation method to compute the value of the function at a nearby value of x. For instance, in Example 1 we computed an approximate value for (3.01)2, a small shift away from the value of 32 =9 that we already know.

The table below shows each of the calculations we considered, starting with Example 1 in row 1, where 𝑓(𝑥) =𝑥2.

Notice that in the second column we're using the letter a to represent the input (x or 𝜃 value) that we used as our "base point" from which we started our approximation. For instance, in row 1 we have 𝑎 =3 and 𝑓(𝑎) =32 =9.

Function Value of 𝑎𝑎 Function's value at a 𝑑𝑓𝑑𝑥at 𝑥=𝑎𝑑𝑓𝑑𝑥at 𝑥=𝑎 We calculated
the approximate value of:
Link to Problem
in Preceding Topic
𝑓(𝑥) =𝑥2 3 32 =9 𝑑𝑓𝑑𝑥at 𝑥=3 =6 (3.01)2 =(3 +0.01)2 Example 1
𝑔(𝑥) =𝑥3 1 13 =1 𝑑𝑓𝑑𝑥at 𝑥=1 =3 (0.99)3 =(1 0.01)3 Example 2
𝑓(𝑥) =𝑥 16 16 =4 𝑑𝑓𝑑𝑥at 𝑥=16 =0.125 16.2 =16+0.2 Practice Problem 1
𝑔(𝜃) =sin(𝜃) 0 sin(0) =0 𝑑𝑔𝑑𝜃at 𝜃=0 =1 sin( 0.13)
=sin(0 0.13)
Practice Problem 2
𝑔(𝜃) =sin(𝜃) 𝜋3 sin(𝜋3) =32 𝑑𝑔𝑑𝜃at 𝜃=𝜋/3 =12 sin(𝜋3+0.18) Practice Problem 3
𝑔(𝜃) =sin(𝜃) 𝜋 sin(𝜋) =0 𝑑𝑔𝑑𝜃at 𝜃=𝜋 = 1 sin(𝜋 +0.07) Practice Problem 4
𝑔(𝜃) =sin(𝜃) 𝜋2 sin(𝜋2) =1 𝑑𝑔𝑑𝜃at 𝜃=𝜋/2 =0 sin(𝜋2+0.04) Practice Problem 5

With that overview of calculations in mind, let's summarize some key points about what we've done so far. While we're using the particular computations we've completed to illustrate the following primary concepts, keep in mind that these ideas will apply to most functions we will encounter.

  1. In each problem, we focus on a point at 𝑥 =𝑎 (column 2) for which we know both (I) the function's value at that point (column 3), and (II) the rate at which the function changes at that point, 𝑑𝑓𝑑𝑥at 𝑥=𝑎 (column 4).
    So far we've had to simply provide that rate for you. That's about to change.
  2. Using our linear approximation method we can calculate approximate values of the function for values of x that are a small distance 𝑑𝑥 away from 𝑥 =𝑎. For instance, in Example 1 we had 𝑑𝑥 =0.01.
  3. We can envision our linear approximation method as starting at the point we know about (𝑎,𝑓(𝑎)), and then walking along the line with slope equal to 𝑑𝑓𝑑𝑥at 𝑥=𝑎 instead of following the function's actual curve.
  4. Walking along this line, the small change in the function's output value 𝑑𝑓 is directly proportional to the small change in x-value 𝑑𝑥. The constant of proportionality is the function's rate of change 𝑑𝑓𝑑𝑥at 𝑥=𝑎: 𝑑𝑓=rate of change at 𝑥=𝑎¯¯¯¯(𝑑𝑓𝑑𝑥at 𝑥=𝑎)𝑑𝑥
  5. The function's rate of change 𝑑𝑓𝑑𝑥at 𝑥=𝑎 is thus a measure of the function's sensitivity at 𝑥 =𝑎 : the rate of change at that point determines how strongly the function reacts when you change its input by the small amount dx. For example, in the the interactive figure below, at (a) 𝑥 =𝑎1 the function's rate of change 𝑑𝑓𝑑𝑥at 𝑥=𝑎1 is larger than at (b) 𝑥 =𝑎2. Using the slider beneath the graph you can vary the size of dx, which causes the identical changes in the horizontal direction in both (a) and (b). Observe that because 𝑑𝑓𝑑𝑥at 𝑥=𝑎 is larger at 𝑎1 than at 𝑎2, the function "reacts more strongly" at 𝑎1, and so the change dy is much larger at 𝑎1 than at 𝑎2 for the same change in dx.

Check Question

The interactive graph below shows the function g(x) versus x. You can use the slider beneath the graph to vary the size of dx near each point.

Choose the correct statement that orders the rates of change at the four points from most negative to most positive.

(a) 𝑑𝑔𝑑𝑥at 𝑥=𝑎1 <𝑑𝑔𝑑𝑥at 𝑥=𝑎2 <𝑑𝑔𝑑𝑥at 𝑥=𝑎3 <𝑑𝑔𝑑𝑥at 𝑥=𝑎4

(b) 𝑑𝑔𝑑𝑥at 𝑥=𝑎2 <𝑑𝑔𝑑𝑥at 𝑥=𝑎4 <𝑑𝑔𝑑𝑥at 𝑥=𝑎3 <𝑑𝑔𝑑𝑥at 𝑥=𝑎1

(c) 𝑑𝑔𝑑𝑥at 𝑥=𝑎2 <𝑑𝑔𝑑𝑥at 𝑥=𝑎1 <𝑑𝑔𝑑𝑥at 𝑥=𝑎4 <𝑑𝑔𝑑𝑥at 𝑥=𝑎3

(d) 𝑑𝑔𝑑𝑥at 𝑥=𝑎1 <𝑑𝑔𝑑𝑥at 𝑥=𝑎3 <𝑑𝑔𝑑𝑥at 𝑥=𝑎4 <𝑑𝑔𝑑𝑥at 𝑥=𝑎2

(e) 𝑑𝑔𝑑𝑥at 𝑥=𝑎4 <𝑑𝑔𝑑𝑥at 𝑥=𝑎3 <𝑑𝑔𝑑𝑥at 𝑥=𝑎2 <𝑑𝑔𝑑𝑥at 𝑥=𝑎1

View/Hide Solution

To answer this question, first choose a positive value for dx, as shown in the figure. (You know dx is positive if its horizontal line extends to the right from the red dot that marks the point 𝑥 =𝑎.) You may have chosen a different value for dx than we did; it doesn't matter, as long as you can easily see the different changes in dg that result.

Then look first at the direction (up or down) of each change dg: in (1) and (3), dg is negative. Furthermore, the most negative change happens at (1): the negative change there is larger than the one in (3). Hence the first item in our ordered list will be 𝑑𝑔𝑑𝑥at 𝑥=𝑎1. We already know, then, that the correct answer choice will be (a) or (d).

Next, notice that the most positive change happens at (2). Hence the last item in our list must be 𝑑𝑔𝑑𝑥at 𝑥=𝑎2. We thus know that the correct answer choice is (d).

But let's continue our reasoning anyway: At (3), dg is negative, but less so than at (1) since the size of its dg is smaller than at (1). And at (4) dg is positive, but less so than at (2). Hence our complete ordering is

most negative <smaller negative <smaller positive <most positive𝑑𝑔𝑑𝑥at 𝑥=𝑎1<𝑑𝑔𝑑𝑥at 𝑥=𝑎3<𝑑𝑔𝑑𝑥at 𝑥=𝑎4<𝑑𝑔𝑑𝑥at 𝑥=𝑎2(d)

Linear approximation means direct proportionality between small change in input and small change in output

You've probably noticed that in each of the graphs above, and in the graph for each linear approximation calculation we've done, you see a triangle. This is an important realization! The triangle comes about by definition of "linear approximation": in this approximation method, each function's small change in output-value df (or dg or d-whatever-output) near the "base point" is always directly proportional to the small change in input dx (or 𝑑𝜃 or dt or d-whatever-input):

𝑑𝑓=(𝑑𝑓𝑑𝑥at 𝑥=𝑎)𝑑𝑥

Defining differentials

With that fundamental idea in mind, let's introduce some new terminology: The small-change quantities df, dg, ds, dx, 𝑑𝜃, and dt are all called differentials. As we've seen,

  • differentials are represented by placing a d in front of the variable;
  • differentials represent small changes in the variable's value;
  • a function's output-differential df (or dg or ...) changes at a constant rate with respect to the function's input-variable's differential dx (or 𝑑𝜃 or ...).

Leibniz notation and Leibniz triangles

Differentials were created by Gottfried Leibniz (1646-1716), which is why you might hear quantities like 𝑑𝑓𝑑𝑥 referred to as Leibniz notation. And the triangles we've been forming, with the small change in input (dx, say) as the base and the small change in output (df, say) as the height, are known as Leibniz triangles. [Link to Wikipedia article about Leibniz.]

Graph of a generic function f(x), with a zoom-in dashed rectangle around the point x=a.  Another rectangle shows the zoomed-in portion, and the Leibniz triangle with differential dx as a short horizontal base of the triangle, and the differential df as its height. The hypotenuse mimics what the curve does near x=a. Text states that from this screen forward, we will focus on how to determine the rate df/dx at x=a.

Flipping the script: Using differentials to determine 𝑑𝑓𝑑𝑥at 𝑥=𝑎

We wrote above (Point #1) that so far we've had to simply provide you with the rate at which a function changes at 𝑥 =𝑎, and that that's about to change. Indeed, a (the?) primary focus of this first part of learning Calculus is figuring out how to determine the value of 𝑑𝑓𝑑𝑥at 𝑥=𝑎. This is the same problem that the founders of Calculus faced.

So let's shift our focus to this question, starting with a reversal of the problem-type we've been considering: in each problem we've examined so far, we've provided the function 𝑓(𝑥), a point of interest, 𝑥 =𝑎, and the value of 𝑓(𝑎) and the rate at which the function changes at that value of x. We've then essentially asked you to compute the value of df using 𝑑𝑓=(𝑑𝑓𝑑𝑥at 𝑥=𝑎)𝑑𝑥 For instance, the first approximation we did was for 𝑓(𝑥) =𝑥2 near 𝑥 =3, so 𝑓(3) =9. We then provided the rate 𝑑𝑓𝑑𝑥at 𝑥=3 =6 and asked you to compute the value of df for 𝑑𝑥 =0.01 so we could estimate 𝑓(3.01), which as you might recall, is (3.01)2 9.06.

Let's switch things up, and provide you with values for df and dx and then ask you to determine the rate 𝑑𝑓𝑑𝑥at 𝑥=𝑎. This is our first step toward finding multiple ways to determine the rate at which a function changes. Said differently, we looking to determine the function's exact "sensitivity" at the the point of interest.

To illustrate the idea, let's return to that very problem of what happens when we vary the function 𝑓(𝑥) =𝑥2 a bit, around the point 𝑥 =3. For the purposes of this Activity, pretend that you do not know already know that 𝑑𝑓𝑑𝑥at 𝑥=3 =6, and instead must determine that value.

Activity 1: For 𝒇(𝒙) =𝒙𝟐, determine 𝑑𝑓𝑑𝑥at 𝑥=3

The interactive graph below shows a zoomed-in version of the (by now familiar) graph of 𝑓(𝑥) =𝑥2 near the point 𝑥 =3, along with the Leibniz triangle for this point and slider to vary the size of dx and hence dy.

If you'd like, you can zoom out to see that this really is part of the curve for 𝑦 =𝑥2. Then hit the "Home" button on the graph to return to the initial view.

Step 1. Zoom in even more and hide/show the red 𝑦 =𝑥2 curve to (again) convince yourself that the hypotenuse of the Leibniz triangle closely mimics the function 𝑓(𝑥) =𝑥2 curve's behavior near 𝑥 =3. As usual, the smaller dx is, the better the green triangle's line tracks the function's red curve; if we extend dx to the ends of its range, we see how the green line deviates more and more from the red curve.

The key point here is to look and see for yourself that, "Yes, the triangle's green line segment tracks the curve in this small region." (Later you'll have to put this green line in place yourself. Here we've done that step for you.)

Once you've decided on that "yes" for yourself, please continue below.

(Remember that for the purposes of this Example, you do not yet know the value of 𝑑𝑓𝑑𝑥at 𝑥=3.)

Step 2. We know that 𝑑𝑓=(𝑑𝑓𝑑𝑥at 𝑥=3)𝑑𝑥 So using the interactive calculator, we can simply read off a set of values for df and dx in order to determine the constant 𝑑𝑓𝑑𝑥at 𝑥=3. For instance, if you set 𝑑𝑥 =0.01 you see 𝑑𝑦 =0.06, and so we have 0.06=(𝑑𝑓𝑑𝑥at 𝑥=3)(0.01) and so we must have 𝑑𝑓𝑑𝑥at 𝑥=3 = ...?

Yes: 6!

You probably did that math in your head, but since the numbers won't always work out so easily let's solve for 𝑑𝑓𝑑𝑥at 𝑥=3: 𝑑𝑓𝑑𝑥at 𝑥=3=0.060.01=6

Hence we see that the Leibniz triangle itself – if we can get it aligned so that it looks to you like it mimics the function-curve's behavior (for now, the only criterion we'll use) – tells us the value of 𝑑𝑓𝑑𝑥at the point of interest 𝑥=𝑎.

Just to double-check, you might choose a different value of dx. For instance, if you set 𝑑𝑥 =0.005, you see 𝑑𝑦 =0.03, so 0.03=(𝑑𝑓𝑑𝑥at 𝑥=3)(0.005) and so again 𝑑𝑓𝑑𝑥at 𝑥=3=0.030.005=6

Time for you to try a problem, this time with a result you don't already know.

Practice Problem 1: Approximate 𝑑𝑓𝑑𝑥at 𝑥=4

The interactive calculator below shows a graph of some function 𝑓(𝑥) versus x.

We're focused on the function's behavior around the highlighted point, (4,106). As before, once you have zoomed in sufficiently around this point, you will see the Leibniz triangle appear.

Use the slider beneath the graph to vary the value of dx, and convince yourself that the triangle's line mimics the function's behavior near the point of interest.

Then use the differential values df and dx to determine: 𝑑𝑓𝑑𝑥at 𝑥=4 = (A) 2.4(B) 2.4(C) 24(D) 24(E) none of these

View/Hide Solution

You can use any value of dx and the corresponding value for df. For instance, if 𝑑𝑥 =0.1, then we see the corresponding value of 𝑑𝑓 = 2.4. Hence 𝑑𝑓2.4=(𝑑𝑓𝑑𝑥at 𝑥=4)𝑑𝑥(0.1)

And so 𝑑𝑓𝑑𝑥at 𝑥=4=2.40.1=24(C)

As a check, does it make sense that we obtained a negative value?

The answer is yes: since the function's values decrease as we move to the right from 𝑥 =4, 𝑑𝑓𝑑𝑥at 𝑥=4 must be negative so that 𝑑𝑓 =(𝑑𝑓𝑑𝑥at 𝑥=4) 𝑑𝑥 is negative for positive values of dx (which is the case as we slide to the right from 𝑥 =4). By contrast, if we got a positive value for 𝑑𝑓𝑑𝑥at 𝑥=4, then the function's values would increase as we move to the right, which is not what the graph shows.


The Upshot

  1. Differentials are small changes in a variable's value. For instance, dx is a small change in input value, and df is the resulting small change in the output value.
  2. By definition, df and dx are related according to 𝑑𝑓 =(𝑑𝑓𝑑𝑥at 𝑥=𝑎) 𝑑𝑥, where 𝑑𝑓𝑑𝑥at 𝑥=𝑎 is the rate at which the particular function changes at 𝑥 =𝑎, and is thus a measure of the function's "sensitivity" to changes in input-value at that point.
  3. When we write this rate as 𝑑𝑓𝑑𝑥, we are using Leibniz notation, named after the person who invented it. The triangle that is formed with dx as its base and df as its height is known as a Leibniz triangle.
  4. While on the preceding screens we provided the value of 𝑑𝑓𝑑𝑥at 𝑥=𝑎, we are laying the groundwork for how to determine it, which will lead to one of the Big Ideas in Calculus.
In the next Topic, we'll take another big step toward your being able to determine this rate for yourself.

For now, if you have a question or thought about what's on this screen, please pop over to the Forum and post it there!