Why is du/dx treated as a fraction in integration by substitution?

[Cheman]

Integration by substitution...

Accroding to my notes, when performing integration by substitution, du/dx= f'(x), and therefore du = f'(x)*dx. But how is this possible? We are treatnig dy/dx as if it were a fraction - but in essence it is not! So why is this statement still true?

Thanks.

 

[Hurkyl]

For better or worse, the notation used for various concepts has been chosen to appear as if you're doing arithmetic. The du and dx mean different things in the different equations.

 

[HallsofIvy]

Yes, you are quite correct that dy/dx is NOT a fraction. But it IS a limit of a fraction. It can be "treated" like a fraction since we can go back "before" the limit, use the fraction properties and the take the limit again. In particular, most calculus books define the "differential" dx, basically as a notational device, and then define
dy= f'(x) dx. Strictly speaking, these "dy" and "dx" are NOT the "dy" and "dx" in "dy/dx" since that is not really a fraction but since dy= f '(x)dx (in terms of differentials), dy/dx = f '(x) and THIS dy/dx really is a fraction (although a symbolic one rather than a fraction of numbers or algebraic terms).