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Partial derivative

  • Thread starter ak123456
  • Start date
1. Homework Statement
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. Homework 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:
32,573
4,303
You can't write dw/dx because f is a function of two variables. The derivatives you have to work with are
[tex]\frac{\partial w}{\partial x}[/tex]
and
[tex]\frac{\partial w}{\partial y}[/tex]

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

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.
all of them are partial derivative ,because i don't know how to type the symbol for partial derivative
 
32,573
4,303
1. Homework Statement
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
I can't see any justification for expanding
[tex]\frac{\partial w}{\partial x}[/tex]
the way you did. Certainly
[tex]\frac{\partial w}{\partial x} = \frac{\partial w}{\partial x} \frac{\partial x}{\partial x}[/tex]
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).

2. Homework 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 ?
 
I can't see any justification for expanding
[tex]\frac{\partial w}{\partial x}[/tex]
the way you did. Certainly
[tex]\frac{\partial w}{\partial x} = \frac{\partial w}{\partial x} \frac{\partial x}{\partial x}[/tex]
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).
the problem is that i have to prove the argument is false
 
32,573
4,303
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.
 
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.
thx for your help
 

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