# Partial Derivatives

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.

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.

Stephen Tashi
what is the partial derivative w.r.t. u of

y^2=uy-v

We could equally well ask: What is the partial derivative of y^2 with repsect to v ?

This general type of confusion often occurs in physics problems. Authors establish a complicated relation between several variables and then write a total differential that seems to be missing some partial derivatives. I conjecture the problem lies in ambiguous notation. The notation y^2 = uy - v does not specify the function y^2 as being a function of particular variables. A correct mathematical definition of a function states a particular domain for a function. A relation like y^2 = uy - v does not.

The "given" information to determine a value of a mathematical function is specified by a vector of values for specific variables. The value of the left hand side of an equation might be determined from a variety of information about the right hand side.

• Sashwat Tanay