# Is this a correct/good way to think interpret differentials for the beginning calculus student?

I was reading the answers to this question, and I came across the following answer which seems intuitive, but too good to be true:

Typically, the $\frac{dy}{dx}$ notation is used to denote the derivative, which is defined as the limit we all know and love (see Arturo Magidin's answer). However, when working with differentials, one can interpret $\frac{dy}{dx}$ as a genuine ratio of two fixed quantities.

Draw a graph of some smooth function $f$ and its tangent line at $x=a$. Starting from the point $(a, f(a))$, move $dx$ units right along the tangent line (not along the graph of $f$). Let $dy$ be the corresponding change in $y$.

So, we moved $dx$ units right, $dy$ units up, and stayed on the tangent line. Therefore the slope of the tangent line is exactly $\frac{dy}{dx}$. However, the slope of the tangent at $x=a$ is also given by $f'(a)$, hence the equation $$\frac{dy}{dx} = f'(a)$$

holds when $dy$ and $dx$ are interpreted as fixed, finite changes in the two variables $x$ and $y$. In this context, we are not taking a limit on the left hand side of this equation, and $\frac{dy}{dx}$ is a genuine ratio of two fixed quantities. This is why we can then write $dy = f'(a) dx$.

By Brendan Cordy

• – Michael Medvinsky Nov 27 '15 at 20:36
• Ovi, I saw your interesting question on infinitesimals at the physics exchange. I think somebody should worry about historical accuracy there... – Mikhail Katz Dec 10 '15 at 9:55

The conclusion is right, but you should not understand $\frac{dy}{dx}$ that way. When you do what you have done it is written $\frac{\Delta y}{\Delta x}$.

If you do what you have explained and take the fixed values, observe that you can get closer to $a$ on the tangent and do the same again.

A derivative of a sufficiently nice function is saying that no matter how close you get to $a$ using your tangent principle, the result is going to be the same. In that sense you are right, you can take any fixed value on the tangent, but fixing something is less general than saying no matter what fixed value on the tangent you take.

Observe as well that the way you construct a tangent is not something that is logically above the definition of the derivative so you could say: we know how to construct a tangent and then we can argue about the consequences. In general, drawing a tangent and finding first derivative are equivalent.

When a function is sufficiently nice all things are clearer, but you must define differentiability so that it is applicable to a wider range of problems.

You need to notice that turning $\frac{\Delta y}{\Delta x}$ into $\frac{dy}{dx}$ and approaching one and the same value is in the core of the definition of having a derivative.

Believe or not, the way you have defined a derivative is applicable in another theory: the theory of chaos, since for many chaotic curves you cannot draw a tangent. Instead you take two close points find the distance and calculate $\frac{\Delta y}{\Delta x}$. In many cases you get a fixed value as you approach $\frac{dy}{dx}$, although it is not possible to draw a tangent in the classical sense. Even when you cannot get a fixed value you make some averaging and still get something useful.

Basically $\frac{dy}{dx} = \lim\limits_{\Delta x \to 0}\frac{\Delta y}{\Delta x}$ and that is the way you should understand it.

But yes, you can find a derivative the way you have described for many nice behaving functions.

• 'A derivative of a sufficiently nice function is saying that no matter how close you get to a, the result is going to be the same.' That's only true for straight lines! (Hence microstraightness in SIA.) Your approach is based on a misconception. – user117644 Jan 12 '16 at 20:41
• We are talking about straight lines all along: a tangent. If you read carefully it says that no matter how close you get to a on the tangent the result will be the same, thus there is no point in using just one fixed value on the tangent since no matter what segment you take around a you will have the same result. I really cannot believe what conclusion you have drawn for such a simple thing. Approach? What approach are you talking about when all I am using is a very simple explanation. – user195934 Jan 12 '16 at 21:54
• I have changed the text a little bit, so that even if you would like to find something in the text that is not there, it will be, I think, more difficult now. – user195934 Jan 12 '16 at 21:59
• I am still gobsmacked with what precision people try to read the text here and there, not regarding this answer alone, even when there is nothing even remotely complicated to find. – user195934 Jan 12 '16 at 22:02
• It was probably just the wording. – user117644 Jan 13 '16 at 3:32

It's my opinion that beginning calculus students shouldn't be thinking about differentials at all. You should just take $\frac {dy}{dx}$ as a notation for the derivative. Trying to think of $dy$ and $dx$ as independent objects is just going to confuse you at this level -- assuming your course doesn't take Robinson's approach.

I realize that you'll probably have to go through a section on differentials in your calculus class but just reinterpret the questions from that section as questions about the tangent line to $y=y(x)$.

Here's an example problem:

Use differentials to approximate the value of $\sqrt{3.98}$.

Seeing this you should immediately mentally convert the word "differentials" to "tangent line approximations". The line tangent to a differentiable function $f$ at $a$ is given by $$T(a+h) = f(a) + f'(a)h$$ Your professor will probably expect you to use differential notation -- unfortunately -- so when you actually write this on your paper use the notation $$\Delta y = y(x+dx) - y(x) \approx dy(x,dx) = y'(x)dx$$ but just keep in mind that this is only a change of notation -- you're still just doing a tangent line approximation problem.

Then the way to solve the above problem is like so (the blue is the part you should write on your homework, the red is the part you should be using on your scratch paper when actually working out the problem): $$\color{blue}{y(x) + dy(x,dx) =}\ \color{red}{T\big((4)+(-0.02)\big) =} \sqrt{4} + \left.(\sqrt{x})'\right|_{x=4}(-0.02) = 2 + \frac{-0.02}{2\sqrt{4}}=1.995$$ Which does in fact approximate $\sqrt{3.98} = 1.994993...$ pretty well. 😉

• This would normally be seen as approximating the value at $4$, although in this case the result is closer to that for the incremented ($3.98$) argument rather than the actual argument ($4$). – user117644 Sep 29 '16 at 22:52
• What are talking about? This is just the application of the first order Taylor expansion of $y=f(x)$ at $x=4$ to approximate $f(3.98)$. – user137731 Sep 30 '16 at 1:30

It is not clear what is gained by this point of view, because to define the dependent differential you already have to know what the derivative is. If one is looking for a way of interpreting the derivative using differentials, one can use bona fide infinitesimals as Leibniz did. A modern approach to this involves infinitesimals a la Robinson as explained in the calculus textbook by Keisler; see https://www.math.wisc.edu/~keisler/calc.html

In my personal teaching experience, students relate better to Keisler's definition than to any other.