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By searching this question, I found this: Can I ever go wrong if I keep thinking of derivatives as ratios?

However, the answers don't have what I'm looking for! (Edit: Meaning, a counterexample. There is one involving partial derivatives, but then the only difference has to do with signs, which means that $dy/dx$ can still be interpreted as a ratio. Thanks, fuglege)

So long as you treat $dx$ as $dx$ (meaning one "object", that is, not d times x), I still have yet to see an example where using the differentials as a fraction yields an incorrect answer (and I have watched almost all the khan academy videos on calculus and read quite a bit out of the ubiquitous Stewart calculus book!)

Note that I am not asking about non-standard analysis. I have found an online textbook and I am currently reading it. I am only trying to find a counterexample to using the differentials as a fraction.

Thanks very much

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    $\begingroup$ Perhaps you could explain what you think the answers are lacking. This one seems to answer your question completely. $\endgroup$
    – fuglede
    Apr 29, 2014 at 12:30

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For functions of one variable, I have never seen a problem, and wouldn't hesitate to treat them as fractions (multiplicatively). However, suppose you have $F(x,y)$ which implicitly defines a function $y=f(x)$, then

$\dfrac{dy}{dx} = -\dfrac{\frac{\partial F}{\partial x}}{\frac{\partial F}{\partial y}}$

If you just straight cancel as fractions, you'd get the wrong sign.

Edit: I just thought; there is a problem with the notation used which makes this error possible. The $\partial F$'s are different! One is given constant $x$ and the other constant $y$, hence they shouldn't necessarily cancel as they do in fractions. I guess it's rather pretty how they do manage to cancel to give a $-1$, but this particular case as just one instance, it's entirely possible for one symbol to represent different things in the same expression, so one would have to be much more careful about cancelling terms.

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  • $\begingroup$ Indeed, this is one of the examples given in the MO thread linked to in xyz's answer to the question the OP links to. $\endgroup$
    – user1729
    Apr 29, 2014 at 13:00
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    $\begingroup$ The problem with this example isn't differentials, it's partial derivatives. Partial derivatives are a really fickle piece of notation whose problems should not be considered problems with differentials. $\endgroup$
    – DanielV
    Apr 29, 2014 at 13:03
  • $\begingroup$ Thanks for the answer! So, apart from partial derivatives, where the notation is a little different anyway, there are no counterexamples? $\endgroup$
    – Mahkoe
    Apr 29, 2014 at 19:24
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I do not know very much. I am still studying like you. One thing I know hope this would help and I might be wrong.

If we consider $\dfrac{dy}{dx}$ as a fraction and say $\dfrac{dy}{dx} = \dfrac{1}{\dfrac{dx}{dy}}$ because it is a fraction then there is a problem.
Consider $y=f(x)=x^0$. Now $\dfrac{dy}{dx}=0$ but $\dfrac{dx}{dy}$ is undefined so we see that $\dfrac{dy}{dx} \neq \dfrac{1}{\dfrac{dx}{dy}}$.

Currently I am reading this book for calculus. The author has explained chain rule very good so you might like it.

I had somewhere read that for chain rule to work $x$ and $y$ should be expressible in form $\phi(x,y)=0$. - I might be wrong.

I had also read somewhere that $y=f(x)$ should be a bijective function. $-$ Again I might be wrong here.

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    $\begingroup$ Technically, if $dx/dy$ were undefined, then $dy/dx \neq 1/(dx/dy)$ is not true: it's undefined. $\endgroup$
    – user14972
    Jun 4, 2014 at 16:45
  • $\begingroup$ @Hurkyl $dy/dx≠1/(dx/dy)$ is true because the right hand side is not equal to the left hand side. I know that $1/(dx/dy)$ is undefined, but we know that $dy/dx$ is finite and can never be equal to an undefined expression. $\endgroup$
    – user103816
    Jun 4, 2014 at 16:59
  • $\begingroup$ If $a$ and $b$ are real numbers such that $\frac ab \neq 0$, then $\frac ab = \frac{1}{\left(\frac ba\right)}$. But if $\frac ab = 0$ then $\frac ba$ is undefined. If one substitutes $dy$ for $a$ and $dx$ for $b$ then it might appear that the facts in this answer are completely explained by treating $\frac{dy}{dx}$ as a fraction (even though they are not). The important difference between $\frac{dy}{dx}$ and a fraction is in how we know that certain formulas are equal, not so much in whether the formulas are equal. $\endgroup$
    – David K
    Aug 28, 2016 at 16:48
  • $\begingroup$ The truth value of "undefined = a" is debatable: It can be seen as neither true nor false, but undefined. Similarly one would consider "1 + 1 = =" as neither true nor false because there is a syntax error. $\endgroup$
    – Trebor
    Sep 13, 2023 at 21:15
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The problem comes when you hit the second-order differentials. When using second-order differentials in the standard notation, they DO NOT cancel.

Example: Let's say that $y = x^3$ and $x = t^2$. Find the second derivative of the first one:

$$y = x^3 \\ \frac{dy}{dx} = 3x^2 \\ \frac{d^2y}{dx^2} = 6x$$

Now, take the derivative of $x = t^2$. You get:

$$\frac{dx}{dt} = 2t$$.

Now, let's try to cancel by squaring the second one:

$$\frac{d^2y}{dx^2}\cdot\left(\frac{dx}{dt}\right)^2 = 6x\cdot (2t)^2 \\ \frac{d^2y}{dx^2}\cdot\frac{dx^2}{dt^2} = 6(t^2) (2t^2) \\ \frac{d^2y}{dt^2} = 24t^4$$

The problem? Let's do it again, but substitute at the beginning. Since $y = x^3$ and $x = t^2$, that means that $y = (t^2)^3 = t^6$. What's the second derivative?

$$\frac{dy}{dt} = 6t^5 \\ \frac{d^2y}{dt^2} = 30t^4$$

Uh oh! Contradiction!

Now, the problem isn't with the concept of treating differentials as fractions, but our implementation of it. See Extending the Algebraic Manipulability of Differentials. When you take the derivative of the derivative, if you are treating $\frac{dy}{dx}$ as a fraction, then the derivative of that fraction should use the quotient rule. Therefore, the notation of the second derivative should be: $$\frac{d^2y}{dx^2} - \frac{dy}{dx}\frac{d^2x}{dx^2}$$ Using this notation, then the higher-order differentials become algebraically manipulable again. I could demonstrate it for you, but it is pretty messy. See the paper if you want to see it happen.

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I'd like to work through the example of $\sf johnnyb$ using differentials.

First calculate the differential of $x$ $$\eqalign{ \def\D#1#2{\frac{d#1}{d#2}} \def\LR#1{\left(#1\right)} \def\c#1{\color{red}{#1}} \def\CLR#1{\c{\LR{#1}}} \def\qiq{\quad\implies\quad} x &= t^2 \\ dx &= 2t\,dt \\ }$$ Then the differential of $y$ $$\eqalign{ y &= x^3 \\ dy &= 3x^2\,\c{dx} \\&= 3x^2\CLR{2t\,dt} \\&= 6x^2t\,dt \\ }$$ From this we identify the derivative, give it a name $(w),\,$ and proceed to calculate its differential $$\eqalign{ w &= \D yt = 6x^2t \\ dw &= 6\CLR{dx^2}t + 6x^2\CLR{dt} \\ &= 6\LR{2x\,\c{dx}}t + 6x^2\LR{dt} \\ &= 12xt\CLR{2t\,dt} + 6x^2\LR{dt} \\ &= \LR{24xt^2 + 6x^2} dt \\ }$$ Finally we obtain various expressions for the second derivative $$\eqalign{ \D{^2y}{t^2} &= \D wt = \LR{24xt^2 + 6x^2} \;\equiv\; 30t^4 \\ }$$ There's nothing tricky here. Just treat each variable as being independent when calculating differentials.

You can substitute variables whenever it is convenient to do so, i.e.

  • at the start of a calculation
  • at the end of a calculation
  • or in an intermediate step during the calculation

There is no "right way" to do things. Once you obtain an expression of the form $$dy = F(y,x,t)\;dt$$ you can rest assured that the derivative is the equal to the ratio $$\D yt = F(y,x,t)$$

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