# Partial derivative

1. Mar 19, 2009

### ak123456

1. The problem statement, all variables and given/known data
what s the wrong with the following arguments
suppose that w=f(x,y)and y=x^2
by the chain rule
(for partial derivative )
Dw/Dx=(Dw/Dx)( Dx/Dx)+(Dw/Dy)(Dy/Dx)=Dw/Dx+2x( Dw/Dy)
hence 2x( Dw/Dy)=0 ,and so Dw/Dy=0

2. Relevant equations

3. The attempt at a solution
i think the argument is not right because we can not write Dx/Dx and w has the relationship with x and y ,so we don't need to use Dy/Dx Dx/Dx to get x ,we can do it directly
is there any counterexample ?

Last edited: Mar 19, 2009
2. Mar 19, 2009

### Staff: Mentor

You can't write dw/dx because f is a function of two variables. The derivatives you have to work with are
$$\frac{\partial w}{\partial x}$$
and
$$\frac{\partial w}{\partial y}$$

I don't see that it makes any difference that y happens to be equal to x^2. If both x and y were functions of a third variable, say t, then you could talk about dw/dt, but to get it you would still need both partial derivatives and would need to use the chain rule.

3. Mar 19, 2009

### ak123456

all of them are partial derivative ,because i don't know how to type the symbol for partial derivative

4. Mar 19, 2009

### Staff: Mentor

I can't see any justification for expanding
$$\frac{\partial w}{\partial x}$$
the way you did. Certainly
$$\frac{\partial w}{\partial x} = \frac{\partial w}{\partial x} \frac{\partial x}{\partial x}$$
since the latter partial derivative is 1, but I don't see any way that you can add all the other stuff (i.e., the other partials).

5. Mar 19, 2009

### ak123456

the problem is that i have to prove the argument is false

6. Mar 19, 2009

### Staff: Mentor

Come up with a counterexample for which the statement isn't true. Pick a function of two variables f(x, y), such as f(x, y) = 3x + 2y, where y = x^2.

7. Mar 19, 2009