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A.5 Differentials; Begin to Determine df/dx 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.

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, $f(x) = x^2.$
  • a particular value of $x$ for which we can easily compute the function. For instance, at $x=3,$ $x^2=9.$
  • the rate at which the function changes at this particular value of $x.$ For instance, $\left.\dfrac{df}{dx}\right|_\text{at $x = 3$}=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 $3^2 = 9$ that we already know.

The table below shows each of the calculations we considered, starting with Example 1 in row 1, where $f(x)=x^2.$

Notice that in the second column we’re using the letter a to represent the x-value that we used as our “base point” from which we started our approximation. For instance, in row 1 we have $a=3$ and $f(a) = 3^2 = 9.$

FunctionValue of $\pmb{a}$$\pmb{f(a)}$$\pmb{\left.\dfrac{df}{dx}\right|_\text{at $x = a$}}$We calculated
the approximate value of:
Link to Problem
in Preceding Topic
$f(x)=x^2$3$3^2=9$$\left.\dfrac{df}{dx}\right|_\text{at $x = 3$}=6$$(3.01)^2 = (3+0.01)^2$Example 1
$g(x)=x^3$1$1^3=1$$\left.\dfrac{df}{dx}\right|_\text{at $x = 1$}=3$$(0.99)^3 = (1-0.01)^3$Example 2
$f(x) = \sqrt{x}$16$\sqrt{16}=4$$\left.\dfrac{df}{dx}\right|_\text{at $x = 16$}=0.125$$\sqrt{16.2} = \sqrt{16 + 0.2}$Practice Problem 1
$g(\theta) = \sin (\theta)$$0$$\sin (0) = 0$$\left.\dfrac{dg}{d \theta}\right|_\text{at $\theta = 0$}=1$$\sin(-0.13) = \sin(0-0.13)$Practice Problem 2
$g(\theta) = \sin (\theta)$$\dfrac{\pi}{3}$$\sin \left(\dfrac{\pi}{3} \right) = \dfrac{\sqrt{3}}{2}$$\left.\dfrac{dg}{d \theta}\right|_\text{at $\theta = \pi/3$}=\dfrac{1}{2}$$\sin(\pi/3 + 0.18)$Practice Problem 3
$g(\theta) = \sin (\theta)$$\pi$$\sin \left(\pi \right) = 0$$\left.\dfrac{dg}{d \theta}\right|_\text{at $\theta = \pi$}=-1$$\sin(\pi + 0.07)$Practice Problem 4
$g(\theta) = \sin (\theta)$$\dfrac{\pi}{2}$$\sin\left(\dfrac{\pi}{2}\right) = 1$$\left.\dfrac{dg}{d \theta}\right|_\text{at $\theta = \pi/2$}=0$$\sin\left(\dfrac{\pi}{2}+0.04\right)$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 $f(a)$ at that point (column 3), and (II) the rate at which the function changes at that point, $\left.\dfrac{df}{dx}\right|_\text{at $x = a$}$ (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 $\big(a, f(a)\big),$ and then walking along the line with slope equal to $\left.\dfrac{df}{dx}\right|_\text{at $x = a$}$ 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 $\left.\dfrac{df}{dx}\right|_\text{at $x = a$}$:
    \[df = \overbrace{\left( \left.\dfrac{df}{dx}\right|_\text{at $x = a$}\right)}^\text{rate of change at $x=a$} \cdot dx\]
  5. The function’s rate of change $\left.\dfrac{df}{dx}\right|_\text{at $x = a$}$ 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) $x=a_1$ the function’s rate of change $\left.\dfrac{df}{dx}\right|_\text{at $x = a_1$}$ is larger than at (b) $x=a_2.$ 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 $\left.\dfrac{df}{dx}\right|_\text{at $x = a$}$ is larger at $a_1$ than at $a_2,$ the function “reacts more strongly” at $a_1,$ and so the change dy is much larger at $a_1$ than at $a_2$ for the same change in dx.
    Graph of f(x) versus x
    to illustrate effect of different $\left.\dfrac{df}{dx}\right|_\text{at $x = a$}$ values:

    Remember: $\color{blue}{df = \left(\left.\dfrac{df}{dx}\right|_\text{at $x = a$}\right) \cdot dx} $

    Use the slider to vary the size of dx:

    When you’re ready, check the box above to show points (c) $x=a_3$ and (d) $x=a_4,$ where the function has negative rates of change. Notice that a positive value of dx produces a negative value for df, and the function’s value decreases.

Check Question 1
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.
Graph of g(x) versus x near different x-values
Use the slider to vary the size of dx:
Choose the correct statement that orders the rates of change at the four points from most negative to most positive.
(a) $\left.\dfrac{dg}{dx}\right|_\text{at $x = a_1$} < \left.\dfrac{dg}{dx}\right|_\text{at $x = a_2$} &lt \left.\dfrac{dg}{dx}\right|_\text{at $x = a_3$} < \left.\dfrac{dg}{dx}\right|_\text{at $x = a_4$}$
(b) $\left.\dfrac{dg}{dx}\right|_\text{at $x = a_2$} < \left.\dfrac{dg}{dx}\right|_\text{at $x = a_4$} &lt \left.\dfrac{dg}{dx}\right|_\text{at $x = a_3$} < \left.\dfrac{dg}{dx}\right|_\text{at $x = a_1$}$
(c) $\left.\dfrac{dg}{dx}\right|_\text{at $x = a_2$} < \left.\dfrac{dg}{dx}\right|_\text{at $x = a_1$} &lt \left.\dfrac{dg}{dx}\right|_\text{at $x = a_4$} < \left.\dfrac{dg}{dx}\right|_\text{at $x = a_3$}$
(d) $\left.\dfrac{dg}{dx}\right|_\text{at $x = a_1$} < \left.\dfrac{dg}{dx}\right|_\text{at $x = a_3$} &lt \left.\dfrac{dg}{dx}\right|_\text{at $x = a_4$} < \left.\dfrac{dg}{dx}\right|_\text{at $x = a_2$}$
(e) $\left.\dfrac{dg}{dx}\right|_\text{at $x = a_4$} < \left.\dfrac{dg}{dx}\right|_\text{at $x = a_3$} &lt \left.\dfrac{dg}{dx}\right|_\text{at $x = a_2$} < \left.\dfrac{dg}{dx}\right|_\text{at $x = a_1$}$
Show/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 $\left.\dfrac{dg}{dx}\right|_\text{at $x = a_1$}.$ 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 $\left.\dfrac{dg}{dx}\right|_\text{at $x = a_2$}.$ 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 \begin{align*} \text{most negative } &
[hide solution]

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 those accompanying 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 $d\theta$ or dt or d-whatever-input):
\[df = \left(\left.\dfrac{df}{dx}\right|_\text{at $x = a$}\right) \cdot dx\]
Defining differentials
With that fundamental idea in mind, let’s introduce some new terminology: The small-change quantities df, dg, ds, dx, $d\theta,$ 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 $d\theta$ or …).

Leibniz notation and Leibniz triangles
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.Differentials were created by Gottfried Leibniz (1646-1716), which is why you might hear quantities like $\dfrac{df}{dx}$ 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.]  

Flipping the script: Using differentials to determine $\left.\dfrac{df}{dx}\right|_\text{at $x = a$}$

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 $\left.\dfrac{df}{dx}\right|_\text{at $x = a$}.$ 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 $f(x)$ (like $f(x) = x^2$), a point of interest, $x=a,$ and the value of $f(a)$ (like “at $x=3,$ $f(3) = 3^2 = 9),$” and the rate at which the function changes at that value of x (like $\left.\dfrac{df}{dx}\right|_\text{at $x = 3$}=6$). We’ve then essentially asked you to compute the value of df using
\[df = \left(\left.\dfrac{df}{dx}\right|_\text{at $x = a$}\right) \cdot dx\] Let’s switch things up, and provide you with values for df and dx and then ask you to determine the rate $\left.\dfrac{df}{dx}\right|_\text{at $x = a$}.$ This is our first step toward finding multiple ways to determine the rate at which a function changes – or, said differently, toward determining 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 $f(x)=x^2$ a bit, around the point $x=3.$ For the purposes of this Example, pretend that you do not know already know that $\left.\dfrac{df}{dx}\right|_\text{at $x = 3$}=6,$ and instead must determine that value.

Example 1: Determine $\left.\dfrac{df}{dx}\right|_\text{at $x = 3$}$

The interactive graph below shows a zoomed-in version of the (by now familiar) graph of $f(x)=x^2$ 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 $y=x^2.$ 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 $y=x^2$ curve to (again) convince yourself that the hypotenuse of the Leibniz triangle closely mimics the function $f(x)=x^2$ 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.

Graph of $f(x) = x^2$ versus x


Use the slider to change the value of dx:

Currently dx = 0.010

(Remember that for the purposes of this Example, you do not yet know the value of $\left.\dfrac{df}{dx}\right|_\text{at $x = 3$}.$)

Step 2. We know that
\[df = \left(\left.\dfrac{df}{dx}\right|_\text{at $x = 3$}\right) \cdot dx\] So using the interactive calculator, we can simply read off a set of values for df and dx in order to determine the constant $\left.\dfrac{df}{dx}\right|_\text{at $x = 3$}.$ For instance, if you set $dx=0.01$ you see $dy = 0.06$, and so we have
\[0.06 = \left(\left.\dfrac{df}{dx}\right|_\text{at $x = 3$}\right) \cdot (0.01)\] and so we must have $\left.\dfrac{df}{dx}\right|_\text{at $x = 3$} = \; … ?$

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 $\left.\dfrac{df}{dx}\right|_\text{at $x = 3$}$:
\[\left.\dfrac{df}{dx}\right|_\text{at $x = 3$} = \dfrac{0.06}{0.01}=6 \quad \quad \cmark\]

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 $\left.\dfrac{df}{dx}\right|_\text{at 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
\[0.03 = \left(\left.\dfrac{df}{dx}\right|_\text{at $x = 3$}\right) \cdot (0.005)\] and so again
\[\left.\dfrac{df}{dx}\right|_\text{at $x = 3$} = \dfrac{0.03}{0.005} = 6 \quad \checkmark \]

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

Practice Problem 1: Approximate $\left.\dfrac{df}{dx}\right|_\text{at $x = 4$}$
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: $\left.\dfrac{df}{dx}\right|_\text{at $x = 4$} =$ \[ \begin{array}{lllll} \text{(A) }-2.4 && \text{(B) }2.4 && \text{(C) }-24 && \text{(D) }24 && \text{(E) none of these} \end{array} \]
Graph of f(x) versus x
Use the slider to change the value of dx: Currently dx = 0.1
Show/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 \[\overbrace{-2.4}^{df} = \left(\left.\dfrac{df}{dx}\right|_\text{at $x = 4$}\right) \cdot \overbrace{(0.1)}^{dx}\] And so \begin{align*} \left.\dfrac{df}{dx}\right|_\text{at $x = 4$} &= \dfrac{-2.4}{0.1} \\[8px] &= -24 \quad \implies \text{(C)} \quad \cmark \end{align*} 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 $x=4,$ $\left.\frac{df}{dx}\right|_\text{at $x = 4$}$ must be negative so that $df = \left( \left.\frac{df}{dx}\right|_\text{at $x = 4$}\right) \cdot 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 $\left.\frac{df}{dx}\right|_\text{at $x = 4$},$ then the function’s values would increase as we move to the right, which is not what the graph shows.
[hide solution]

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 = \left(\left.\dfrac{df}{dx}\right|_\text{at $x = a$}\right) \cdot dx,$ where $\left.\dfrac{df}{dx}\right|_\text{at $x = a$}$ 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 $\dfrac{df}{dx},$ 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 $\left.\dfrac{df}{dx}\right|_\text{at $x = a$},$ 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!

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