Determine whether a function with these partial derivatives exist

[autodidude]

Homework Statement

Determine whether a function with partial derivatives f_x(x,y)=x+4y and f_y(x+y)=3x-y exist.

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)

 

[Dick]

Homework Statement

Determine whether a function with partial derivatives f_x(x,y)=x+4y and f_y(x+y)=3x-y exist.

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)

Homework Statement

Homework Equations

The Attempt at a Solution

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.

 

[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..

 

[Dick]

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..

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.