Homework Help: Partial derivatives

"Complete partial derivatives"?

Perhaps you mean the total derivative?

 

using the z-function as now given,

get [tex]\frac{\partial z}{\partial x}[/tex] and [tex]\frac{\partial z}{\partial y}[/tex]
Create the lhs using these values of [tex]\frac{\partial z}{\partial x}[/tex] and [tex]\frac{\partial z}{\partial y}[/tex] and you will end up with the rhs. viz [itex]x[/itex]

Edit: it's fairly simple - not complicated.

 

Those partial derivatives are quite wrong, I'm afraid.

e.g. the partial derivative of y², wrt x, is zero!!

z = y + f(x²- y²)

dz/dx = 0 + (f'')*(2x) where f' is the derivative of f wrt its argument.

and dz/dx is the partial derivative.

 

Here's how to do those partial derivatives.

[tex]\mbox{If }z = f(\phi),\mbox{ where } \phi = x^2 + y^2[/tex]

then

[tex]\frac{\partial z}{\partial x} = \frac{\partial f}{\partial \phi}.\frac{\partial \phi}{\partial x}[/tex]

where

[tex]\frac{\partial f}{\partial \phi} \mbox{ is } f'[/tex]

and

[tex]\frac{\partial \phi}{\partial x} = 2x + 0[/tex]

So,

[tex]\frac{\partial z}{\partial x}= f'.(2x)[/tex]

 

Not quite right yet.
Have you seen my earlier post?

Bob, could you post a new message rather than sinmply editing an old post? No one can tell if you have responded or not unless they actually read the old message. If you post a new reply, it will show up as such in the forum index page.
Thanks

 

I'm afraid these partial derivatives are still wrong.

Have you read my earlier post, post #7, showing how to do the partial derivatives.

There should be a [itex]f'[/itex] in the expression for [tex]\partial z/\partial x[/tex]

 

Almost there,

[tex]\mbox{The }\frac{\partial z}{\partial y} \mbox{ should be }\frac{\partial z}{\partial y}= 1- f'.2y[/tex]

[tex]z = y + f(\phi) \mbox{ where } \phi = x^2 - y^2[/tex]

[tex]\frac{\partial z}{\partial y} = \frac{\partial y}{\partial y} + \frac{\partial f}{\partial \phi}.\frac{\partial \phi}{\partial y}[/tex]
[tex]\frac{\partial z}{\partial y} = 1 + f'.(-2y)[/tex]
[tex]\frac{\partial z}{\partial y} = 1 - f'.(2y)[/tex]

 

[tex]\frac{\partial z}{\partial x} = 2x.f'[/tex]

and

[tex]\frac{\partial z}{\partial y} = 1 - 2y.f'[/tex]

So,

[tex]x.\frac{\partial z}{\partial y} + y.\frac{\partial z}{\partial y} = x.(1 - 2y.f') + y.2x.f' = ??[/tex]

 

Bob, can you not see that,

[tex]x.(1 - 2y.f') + y.2x.f' = x - 2xy.f' + 2xy.f' = x\ ?[/tex]