Partial Derivatives

theleftside

Hey,

Little confused by something:

if we have u=x+y and v=xy what is the partial derivative w.r.t. u of

y^2=uy-v

I am told it is 2y (dy/du) = u (dy/du) + y

And I can see where these terms come from. What I don't understand is why there is no (dv/du) term, as v and u aren't independent.

Any ideas?

ehild

Homework Helper
The variables x and y have been exchanged with u and v:They are the independent variables, and y is a function of them, y(u,v).
Any function f(u,v) have partial derivatives with respect to u and with respect to v, but the derivatives of u or v with respect to each other is zero.

Sashwat Tanay

Definition: A function f is said to be a function of the independent variables (a, b, c) if specifying a, b, c fully determines f (but not overdetermine: i.e. specification of all the three a, b, c are necessary for the determination of f; non-specification of any one of the a, b, c will leave f undetermined). This is our so called 'CONSTRUCTION'. Also, a, b, c are said to be independent variables.

Now, take y^2 be z. Then z=z(u, v, y) - as per the definition above, and all three, i.e. u, v, y are independent (as per above). Don't go by 'u=x+y and v=xy' to conclude that u and v aren't independent.

Also, I don't think it's correct to write y=y(u,v), since specification of u, v doesn't fully determine y (it's a quadratic equation-hence two roots)
and y is a function of them, y(u,v).
But, y^2=y^2(u, v, y) is correct, since y^2 is fully determined by u, v and y.

ehild

Homework Helper
Definition: A function f is said to be a function of the independent variables (a, b, c) if specifying a, b, c fully determines f (but not overdetermine: i.e. specification of all the three a, b, c are necessary for the determination of f; non-specification of any one of the a, b, c will leave f undetermined). This is our so called 'CONSTRUCTION'. Also, a, b, c are said to be independent variables.

Now, take y^2 be z. Then z=z(u, v, y) - as per the definition above, and all three, i.e. u, v, y are independent (as per above). Don't go by 'u=x+y and v=xy' to conclude that u and v aren't independent.

Also, I don't think it's correct to write y=y(u,v), since specification of u, v doesn't fully determine y (it's a quadratic equation-hence two roots)

But, y^2=y^2(u, v, y) is correct, since y^2 is fully determined by u, v and y.
is it not overdetermined????

Sashwat Tanay

I see. True.
By the way, can we even legitimately talk of partial derivative of y^2=uy-v wrt u? Before doing that, we I think we should decide our independent variables on which y^2 depends.
Also, in
I am told it is 2y (dy/du) = u (dy/du) + y
'dy/du' indicates a total derivative, so y should be fully determined by u. But we see, it's not.