As we know, for implicit differentiation we have (del y)/(del x)=−(F_x)/(F_y) but if I use a trick that (del y)/(del x) = ((del F)(del x))/((del F)/(del y))= (del F)/(del x)(del y)/(del F)= (del y)/(del x) as del F cancels each other out, so it is like del y del x=(F_x)/(F_y) without the "minus", where did I go wrong?

ljudskija7s

ljudskija7s

Answered question

2022-08-10

As we know, for implicit differentiation we have y x = F x F y
but if I use a trick that y x = F x F y = F x y F = y x as F cancels each other out, so it is like y x = F x F y without the "minus", where did I go wrong?

Answer & Explanation

Mariam Hickman

Mariam Hickman

Beginner2022-08-11Added 13 answers

This is a classic case of "differentials are not fractions" and is a marked downside of Leibniz notation. While the notation d y d x or F y looks like a fraction, we can't simply "cancel the F" out. In effect, you have just proved a difference between differentials and fractions.

For another, consider the chain rule in several variables. Let F = F ( x , y ) and let x = x ( s , t ) and y = y ( s , t ). Then
F t = F x x t + F y y t
If we were allowed to cancel terms like you suggest, then we would have that F t = 2 F t , which is clearly not true.

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