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Partial derivatives

  • Thread starter Bob19
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  • #1
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On question

If I have the plane z = y + f(x,y) where f(x,y) = sin(x) * y

Is it possible to find the complete partial derivatives for z ?

/Bob
 

Answers and Replies

  • #2
TD
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"Complete partial derivatives"?

Perhaps you mean the total derivative?
 
  • #3
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Help

TD said:
"Complete partial derivatives"?

Perhaps you mean the total derivative?
Yes,

I made a typoo in my last post:

What I meant to write was

I'm presented with [tex] z = y + f(x^2 - y^2) [/tex]

I'm told this can be written as [tex] y \frac{\partial z}{\partial x} + x \frac{\partial z}{\partial y} = x[/tex]

Any idears on how I do that ?

/Bob

p.s. I get [tex]y \frac{\partial x}{\partial x} + \frac{\partial z}{\partial y} + 2x - y^2 \frac{\partial z}{\partial x} + x^2 -2y \frac{\partial z}{\partial y} = 0[/tex]

But how do I go from this result to the expected result ?
 
Last edited:
  • #4
Fermat
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Bob19 said:
...

I'm presented with [tex] z = y + f(x^2 - y^2) [/tex]

I'm told this can be written as [tex] y \frac{\partial z}{\partial x} + x \frac{\partial z}{\partial y} = x[/tex]

Any idears on how I do that ?

...?
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.
 
  • #5
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Fermat said:
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.
Okay

I then get

[tex]\frac{\partial z}{\partial x} = 2x [/tex]
[tex]\frac{\partial z}{\partial y} = - 2y[/tex]

What do I then do next ?

/Bob
 
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  • #6
Fermat
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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.
 
  • #7
Fermat
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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]
 
  • #8
Fermat
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Bob19 said:
Okay

I then get

[tex]\frac{\partial z}{\partial x} = 2x [/tex]
[tex]\frac{\partial z}{\partial y} = - 2y[/tex]

What do I then do next ?

/Bob
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
 
  • #9
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Yes sorry,

Please excuse me if mix up standard differentation and partial differentation

If [tex]z = y + f(x^2 - y^2)[/tex]

Can be written as:

[tex]y \frac{\partial z}{\partial x} + x \frac{\partial z}{\partial y} = x[/tex]

Then to use the notation required in the task [tex]\frac{\partial z}{\partial x}= 2x \frac{\partial f}{\partial x}[/tex]

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

If this is correct then how do I go from these two deriatives to the required result ?

Sincerely

/Bob
 
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  • #10
Fermat
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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]
 
  • #11
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Fermat said:
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]
Is it then ??

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

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

/Bob
 
  • #12
Fermat
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Bob19 said:
Is it then ??

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

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

/Bob
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]
 
  • #13
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Okay

Thank You very much :-)

My final question is than which approach do I have use to obtain that

[tex]z = y + f(x^2 - y^2)[/tex]

can be written as

[tex] y \frac{\partial z}{\partial x} + x \frac{\partial z}{\partial y} = x [/tex]

Do I add the two partial deriatives together ?

Best Regards

Bob

Fermat said:
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]
 
  • #14
Fermat
Homework Helper
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[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]
 
  • #15
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0
Thank You again,

I can see that now that, those two equal each other, but how does it equal x ?

/Bob

Fermat said:
[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]
 
  • #16
Fermat
Homework Helper
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Bob, can you not see that,

[tex]x.(1 - 2y.f') + y.2x.f' = x - 2xy.f' + 2xy.f' = x\ ?[/tex]
 
  • #17
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Now I can ;-)

Thank You ;-)


/BOb

Thank You

Fermat said:
Bob, can you not see that,

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

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