Finding Partial Derivatives for z in f(x,y)=sin(x)*y - Bob

[Bob19]

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

 

[TD]

"Complete partial derivatives"?

Perhaps you mean the total derivative?

 

[Bob19]

Help

"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 z = y + f(x^2 - y^2)

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

Any idears on how I do that ?

/Bob

p.s. I get 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

But how do I go from this result to the expected result ?

 

[Fermat]

...

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

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

Any idears on how I do that ?

...?

using the z-function as now given,

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

Edit: it's fairly simple - not complicated.

 

[Bob19]

using the z-function as now given,

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

Edit: it's fairly simple - not complicated.

Okay

I then get

\frac{\partial z}{\partial x} = 2x
\frac{\partial z}{\partial y} = - 2y

What do I then do next ?

/Bob

 

[Fermat]

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.

 

[Fermat]

Here's how to do those partial derivatives.

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

then

\frac{\partial z}{\partial x} = \frac{\partial f}{\partial \phi}.\frac{\partial \phi}{\partial x}

where

\frac{\partial f}{\partial \phi} \mbox{ is } f'

and

\frac{\partial \phi}{\partial x} = 2x + 0

So,

\frac{\partial z}{\partial x}= f'.(2x)

 

[Fermat]

Okay

I then get

\frac{\partial z}{\partial x} = 2x
\frac{\partial z}{\partial y} = - 2y

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

 

[Bob19]

Yes sorry,

Please excuse me if mix up standard differentation and partial differentation

If z = y + f(x^2 - y^2)

Can be written as:

y \frac{\partial z}{\partial x} + x \frac{\partial z}{\partial y} = x

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

\frac{\partial z}{\partial y}= 1- 2y \frac{\partial f}{\partial y}

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

Sincerely

/Bob

 

[Fermat]

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 f' in the expression for \partial z/\partial x

 

[Bob19]

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 f' in the expression for \partial z/\partial x

Is it then ??

\frac{\partial z}{\partial x}= f'. 2x

\frac{\partial z}{\partial y}= f'. 1- 2y

/Bob

 

[Fermat]

Is it then ??

\frac{\partial z}{\partial x}= f'. 2x

\frac{\partial z}{\partial y}= f'. 1- 2y

/Bob

Almost there,

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

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

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

 

[Bob19]

Okay

Thank You very much :-)

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

z = y + f(x^2 - y^2)

can be written as

y \frac{\partial z}{\partial x} + x \frac{\partial z}{\partial y} = x

Do I add the two partial deriatives together ?

Best Regards

Bob

Almost there,

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

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

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

 

[Fermat]

\frac{\partial z}{\partial x} = 2x.f'

and

\frac{\partial z}{\partial y} = 1 - 2y.f'

So,

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

 

[Bob19]

Thank You again,

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

/Bob

\frac{\partial z}{\partial x} = 2x.f'

and

\frac{\partial z}{\partial y} = 1 - 2y.f'

So,

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

 

[Fermat]

Bob, can you not see that,

x.(1 - 2y.f') + y.2x.f' = x - 2xy.f' + 2xy.f' = x\ ?

 

[Bob19]

Now I can ;-)

Thank You ;-)


/BOb

Thank You

Bob, can you not see that,

x.(1 - 2y.f') + y.2x.f' = x - 2xy.f' + 2xy.f' = x\ ?