While I do know that dydx isn’t a fraction and shouldn’t be treated as such, in many situations, doing things like multiplying both sides by dx and integrating, cancelling terms, doing things like dydx=1dxdy works out just fine.
So I wanted to know: Are there any particular cases (in single-variable calculus) we have to look out for, where treating dydx as a fraction gives incorrect answers, in particular, at an introductory level?
Note: Please provide specific instances and examples where treating dydx as a fraction fails
It is because of the extraordinary power of Leibniz’s differential notation, which allows you to treat them as fractions while solving problems. The justification for this mechanical process is apparent from the following general result:
Let y=h(x) be any solution of the separated differential equation
A(y)dydx=B(x)… (i) such that h′(x) is continuous on an open interval I, where B(x) and A(h(x)) are assumed to be continuous on I. If g is any primitive of A (i.e. g′=A) on I, then h satisfies the equation g(y)=∫B(x)dx+c…(ii) for some constant c. Conversely, if y satisfies (ii) then y is a solution of (i).
Also, it would be advisable to say dydx=1dxdy only when the function y(x) is invertible.
Say you are asked to find the equation of normal to a curve y(x) at a particular point (x1,y1). In general you should write the slope of the equation as −1dydx|(x1,y1) instead of simply writing it as −dxdy|(x1,y1) without checking for the invertibility of the function (which would be redundant here). However, the numerical calculations will remain the same in any case.
The Leibniz notation ensures that no problem will arise if one treats the differentials as fractions because it beautifully works out in single-variable calculus. But explicitly stating them as ‘fractions’ in any exam/test could cost one the all important marks. One could be criticised in this case to be not formal enough in his/her approach.
Also have a look at this answer which explains the likely pitfalls of the fraction treatment.