# Determine whether a function with these partial derivatives exist

1. Dec 14, 2012

### autodidude

1. The problem statement, all variables and given/known data
Determine whether a function with partial derivatives $$f_x(x,y)=x+4y$$ and $$f_y(x+y)=3x-y$$ exist.

3. The attempt at a solution

The method I've seen is to integrate $$f_x$$ with respect to x, differentiate with respect to y, set it equal to the given $$f_y$$ and show that it can't be possible.

So after integrating $$f_x$$, we get $$f(x, y) = \frac{1}{2}x^2+4xy+g(y)$$

Then differentating that w.r.t y gives

$$f_y(x,y)=4x+g'(y)$$

So

$$3x-y = 4x+g'(y)$$
$$g'(y)=-x-y$$

Why would this be a contradiction? Is it because g'(y) can only be in terms of y and not x? Couldn't x be treated as a constant?

Would integrating both derivatives and showing that they're not equal be a valid method?

So with respect to x would be $$f(x,y)=\frac{1}{2}x^2+4xy+g(y)$$ and with respect to y would be $$f(x,y)=\frac{3}{2}x^2-\frac{1}{2}y^2+g(x)$$
1. The problem statement, all variables and given/known data

2. Relevant equations

3. The attempt at a solution

2. Dec 14, 2012

### Dick

If your function is sufficiently well behaved, then you ought to have $f_{xy}=f_{yx}$. See http://en.wikipedia.org/wiki/Symmetry_of_second_derivatives Does that hold? And you don't really mean $f_y(x+y)$, right? And sure, integrating them both and showing the partials are not equal would work, but it might be more than you need to show.

Last edited: Dec 15, 2012
3. Dec 16, 2012

### autodidude

Thanks.

I'm not sure what $$f_y(x+y$$ is but it's probably not what I mean!

I'm still not 100% sure why line 9 (tex) is a contradiction..

4. Dec 16, 2012

### Dick

I'm guessing line 9 is g'(y)=(-x-y). Sure, that's a contradiction. If x=0 then that's g'(y)=(-y) and if x=1 that's g'(y)=(-1-y). They can't both be true. The equation can't be true for all x and y if you have an x one side and not the other.