Partial derivatives of a function of two variables with respect to different variable

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Homework Statement



Let f(x,y)=[itex]e^{xy}[/itex]

Variables u and v are defined by u=[itex]x^{3}[/itex]-[itex]y^{3}[/itex] , v=[itex]x^{2}+xy[/itex]

Find the values of [itex]\delta[/itex]f/[itex]\delta[/itex]u and [itex]\delta[/itex]f/[itex]\delta[/itex]v at the point where x=-1 and y=2

Homework Equations



N/A

The Attempt at a Solution



At first I thought that I'd have to write x and y in terms of u and v. So, I started by factorising using the difference of two cubes for u and then just taking out a factor of x for v. However, I pretty much hit a brick wall there.

And, if I can't write x and y in terms of u and v, I can't see how I could find either [itex]\delta[/itex]f/[itex]\delta[/itex]u or [itex]\delta[/itex]f/[itex]\delta[/itex]v?
 

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  • #2
I like Serena
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Homework Statement



Let f(x,y)=[itex]e^{xy}[/itex]

Variables u and v are defined by u=[itex]x^{3}[/itex]-[itex]y^{3}[/itex] , v=[itex]x^{2}+xy[/itex]

Find the values of [itex]\delta[/itex]f/[itex]\delta[/itex]u and [itex]\delta[/itex]f/[itex]\delta[/itex]v at the point where x=-1 and y=2

Homework Equations



N/A

The Attempt at a Solution



At first I thought that I'd have to write x and y in terms of u and v. So, I started by factorising using the difference of two cubes for u and then just taking out a factor of x for v. However, I pretty much hit a brick wall there.

And, if I can't write x and y in terms of u and v, I can't see how I could find either [itex]\delta[/itex]f/[itex]\delta[/itex]u or [itex]\delta[/itex]f/[itex]\delta[/itex]v?

Hi Unredeemed! :smile:

Do you have the chain rule for multi variable functions in your notes?
That is:
$${\partial f \over \partial u} = {\partial f \over \partial x}{\partial x \over \partial u} + {\partial f \over \partial y}{\partial y \over \partial u}$$

Edit: Oopsie. The following does not hold. See next posts.
Oh, and are you also aware of the inverse rule for derivatives?
That is:
$${\partial u \over \partial x} = \frac 1 {\partial x \over \partial u}$$
 
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  • #3
Dick
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Hi Unredeemed! :smile:

Do you have the chain rule for multi variable functions in your notes?
That is:
$${\partial f \over \partial u} = {\partial f \over \partial x}{\partial x \over \partial u} + {\partial f \over \partial y}{\partial y \over \partial u}$$

Oh, and are you also aware of the inverse rule for derivatives?
That is:
$${\partial u \over \partial x} = \frac 1 {\partial x \over \partial u}$$

Your 'inverse rule' doesn't hold for partial derivatives. You need to invert the whole jacobian matrix to get the inverse derivatives.
 
  • #4
I like Serena
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Your 'inverse rule' doesn't hold for partial derivatives. You need to invert the whole jacobian matrix to get the inverse derivatives.

Oopsie.
That's what I get for writing something as an afterthought.

@OP: Do you know what the Jacobian matrix is and how to use it?
 
Last edited:
  • #5
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Hi Unredeemed! :smile:

Do you have the chain rule for multi variable functions in your notes?
That is:
$${\partial f \over \partial u} = {\partial f \over \partial x}{\partial x \over \partial u} + {\partial f \over \partial y}{\partial y \over \partial u}$$

Edit: Oopsie. The following does not hold. See next posts.
Oh, and are you also aware of the inverse rule for derivatives?
That is:
$${\partial u \over \partial x} = \frac 1 {\partial x \over \partial u}$$

I am aware of $${\partial f \over \partial u} = {\partial f \over \partial x}{\partial x \over \partial u} + {\partial f \over \partial y}{\partial y \over \partial u}$$

But, the issue with the "inverse rule" for partial derivatives led me to think it was unhelpful?

I have that if f(x,y)=F(u(x,y),v(x,y))

Then $$({\partial f \over \partial x} , {\partial f \over \partial y})= ({\partial F \over \partial u} , {\partial F \over \partial v}) . \begin{pmatrix} {\partial u \over \partial x} {\partial u \over \partial y}\\ {\partial v \over \partial x} , {\partial v \over \partial y}\end{pmatrix}$$

But, inverting the Jacobian matrix on the RHS only gives me $${\partial F \over \partial u} and {\partial F \over \partial v}$$ not $${\partial f \over \partial u} and {\partial f \over \partial v}$$ as required?
 
  • #6
I like Serena
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But, inverting the Jacobian matrix on the RHS only gives me $${\partial F \over \partial u} and {\partial F \over \partial v}$$ not $${\partial f \over \partial u} and {\partial f \over \partial v}$$ as required?

It's the same thing.
What you write is the proper way to write it.

The form ##{\partial f \over \partial u}## is short hand for ##{\partial F \over \partial u}##.
 
  • #7
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It's the same thing.
What you write is the proper way to write it.

The form ##{\partial f \over \partial u}## is short hand for ##{\partial F \over \partial u}##.

Ah, I see. I'd wondered about that but didn't just want to assume it!

So I've said that $$({\partial f \over \partial x} , {\partial f \over \partial y}) . \begin{pmatrix} {\partial v \over \partial y} -{\partial u \over \partial y}\\ -{\partial v \over \partial x} , {\partial u \over \partial x}\end{pmatrix}= ({\partial f \over \partial u} , {\partial f \over \partial v})$$

Is that correct?
 
  • #8
I like Serena
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Almost.
There is a determinant missing.
 
  • #9
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Ah, good point!
 
  • #10
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Think I've got the answer now. Fingers crossed an algebraic error hasn't eluded me!

Thanks very much for everyone's help. Probably would have been head-scratching for hours without it.
 
  • #11
I like Serena
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Cheers.
 

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