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

1. Dec 27, 2012

### Unredeemed

1. The problem statement, all variables and given/known data

Let f(x,y)=$e^{xy}$

Variables u and v are defined by u=$x^{3}$-$y^{3}$ , v=$x^{2}+xy$

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

2. Relevant equations

N/A

3. 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 $\delta$f/$\delta$u or $\delta$f/$\delta$v?

2. Dec 27, 2012

### I like Serena

Re: Partial derivatives of a function of two variables with respect to different vari

Hi Unredeemed!

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:
$${\partial u \over \partial x} = \frac 1 {\partial x \over \partial u}$$

Last edited: Dec 27, 2012
3. Dec 27, 2012

### Dick

Re: Partial derivatives of a function of two variables with respect to different vari

Your 'inverse rule' doesn't hold for partial derivatives. You need to invert the whole jacobian matrix to get the inverse derivatives.

4. Dec 27, 2012

### I like Serena

Re: Partial derivatives of a function of two variables with respect to different vari

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: Dec 27, 2012
5. Dec 27, 2012

### Unredeemed

Re: Partial derivatives of a function of two variables with respect to different vari

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. Dec 27, 2012

### I like Serena

Re: Partial derivatives of a function of two variables with respect to different vari

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. Dec 27, 2012

### Unredeemed

Re: Partial derivatives of a function of two variables with respect to different vari

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. Dec 27, 2012

### I like Serena

Re: Partial derivatives of a function of two variables with respect to different vari

Almost.
There is a determinant missing.

9. Dec 27, 2012

### Unredeemed

Re: Partial derivatives of a function of two variables with respect to different vari

Ah, good point!

10. Dec 27, 2012

### Unredeemed

Re: Partial derivatives of a function of two variables with respect to different vari

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. Dec 27, 2012

### I like Serena

Re: Partial derivatives of a function of two variables with respect to different vari

Cheers.