A common question on mathematics stack exchange is "why can't derivatives be treated as fractions?" I like the explanation that derivatives are not fractions, they are the limits of fractions, which are not the same type of mathematical object, so it is wrong to assume they behave in the same way. The answers I've seen give examples where treating derivatives as fractions yields correct solutions and other cases where it does not, but they do not proceed to show how to tackle the problems of the latter case in the correct way.

My question is: what is the correct way to solve calculus problems without treating derivatives as fractions?

I expect the answer to be quite complex, as otherwise people wouldn't use treat derivatives as fractions in the first place, they would do the problems properly. If possible, I would appreciate a simplified explanation (at the level of an A-level student) to preface the full explanation which could be useful to more advanced users.

  • $\begingroup$ Can you elaborate where derivatives are used as fractions in a formal setting? $\endgroup$
    – user370967
    Jan 3 '18 at 13:52
  • $\begingroup$ This is rather notational abuse than really acting like the objects are fractions. But indeed in differential equations with separation of the variables we actually treat the derivative as a fraction. So this answers my own question. $\endgroup$
    – user370967
    Jan 3 '18 at 13:54
  • $\begingroup$ @Math_QED In integration problems, I've been taught to separate variables by manipulating derivatives as fractions e.g. multiplying by their denominators. As plumbus says, I've also been taught to treat derivatives as fractions for the chain rule and reciprocal rule. When performing a change of variable in integration, the method I use requires me to treat differentials as fractions as well. $\endgroup$ Jan 3 '18 at 14:01
  • $\begingroup$ @Math_QED, I didn't see your second comment before posting mine, as I hadn't refreshed the page, oops! $\endgroup$ Jan 3 '18 at 14:04
  • $\begingroup$ No I reject your chain rule answer. This really isn't treating derivatives as fractions. I also dislike the notation, but this is just my opinion. $\endgroup$
    – user370967
    Jan 3 '18 at 14:14

This is really an issue of how derivatives and differentials interact. The relationship between the $3$ quantities $dx$, $dy$, and $dy/dx$ is

$$dy = \frac{dy}{dx}\; dx.$$

So if you have a differential equation

$$\frac{dy}{dx} = g(x,y)$$

you can multiply both sides by $dx$

$$\frac{dy}{dx}\; dx = g(x,y)\; dx$$

and then replace the left side by the relationship given above

$$dy = g(x,y)\; dx.$$

This looks like we've "treated $dy/dx$ as a fraction" because it looks like the $dx$'s were canceled.


There is subtle distinction here. Consider $${\rm d}y = y' {\rm d}x$$ The differentials ${\rm d}x$ and ${\rm d}y$ are quantities that can be arranged in fractions if needed, but the partial derivative $y' = \frac{\partial }{\partial x}y$ is never a fraction, but rather a quanitty defined with an operator.

So when you see a typical problem like $$ \frac{{\rm d}^2y}{{\rm d}x^2} + \alpha\, y = A\, {\rm e}^{-\beta x} $$

it is not stated canonically. I would say that within the domain of $x$, and for each infinitesimal ${\rm d}x$ the following is true

$$ \left( \frac{\partial }{\partial x}\frac{\partial }{\partial x} y + \alpha y - A\, {\rm e}^{-\beta x} \right) {\rm d} x =0 $$

In applied mathematics the above notation is common. For example, when deriving the motion of a vibrating string. You must consider a physical small segment ${\rm d}x$ and do the balance of forces for this piece. What comes out is the part in the parenthesis (the partial differential equation) that must be zero if the expression above is zero for every differential ${\rm d}x$.


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