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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

- Thread starter Bob19
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- #1

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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

- #2

TD

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"Complete partial derivatives"?

Perhaps you mean the**total derivative**?

Perhaps you mean the

- #3

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Yes,TD said:"Complete partial derivatives"?

Perhaps you mean thetotal derivative?

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 ?

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- #4

Fermat

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using the z-function as now given,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 ?

...?

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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OkayFermat 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.

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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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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[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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Not quite right yet.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

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

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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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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Is it then ??Fermat said:

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]

[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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Almost there,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

[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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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

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

/Bob

Fermat said:

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

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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]

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

- #17

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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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