Where can I find the proof that dy/dx can be treated as a fraction?

It can't. You have to prove the chain rule by entirely different methods; although the initial premises and the end result may look like you've treated it as a fraction, in fact you haven't. If that makes sense.

David

It does. Thankyou. Still where can I find the proof?

If u=g(x) define du/dx as: the limit as dx® 0 of:

g(x+dx)-g(x))/dx

You should have seen this definition before. The key is the precise meaning of 'limit'. Here it means that, for a given value of x, the above expression can be made as close as we like to x. (This is, roughly, the precise definition of 'limit')

More precisely, given e > 0, we can find dx such that the above expression is within e of du/dx.

From now on I'll write g ' (x) for du/dx etc.

Ok, now for the proof:

Let u=g(x), y=f(u). We want to prove that

dy/dx=(dy/du).(du/dx)

Let dx be a change in x, and du and dy be the corresponding changes in u and y.

By playing around with the limit expression above, we get:

du=(g ' (x)+e)dx

where e depends on x and dx and ® 0 as dx® 0.

Similarly for y:

dy=(f ' (u)+e ' )du

where, as before, e ' ® 0 as du® 0.

Combining this with the expression for du, we have:

dy=(f ' (u)+e ' )(g ' (x)+e)dx

NOTE: This expression is EXACT, once e and e ' have been found, and these are uniquely defined by the value of dx.

So now divide both sides by dx and let dx® 0. We get:

dy/dx=(dy/du).(du/dx)

I hope that helps.

wonderful