Finding a general form of a function, given a partial derivative

[ksdhart]

Hi all. I'm having some difficulty with a problem from my Calculus IV course, about partial derivatives. The text of the exercise says:

For the partial derivatives given in Exercises 51-54, find the most general form for a function of two variables, f(x,y), with the given partial derivative.

54)$\displaystyle \frac{\partial ^2f}{\partial y\:\partial x}=0$

I began by rewriting the given and combining that with the knowledge that mixed partial derivatives are equal.

$\displaystyle \frac{\partial }{\partial y}\left(\frac{\partial f}{\partial x}\right)=\frac{\partial }{\partial x}\left(\frac{\partial f}{\partial y}\right)=0$

I also know that for a partial derivative with respect to x to be 0, the function must be a constant or a function of y, and the opposite is true of a partial derivative with respect to y. So, now I have:

$\displaystyle \frac{\partial f}{\partial y}=g\left(y\right)$ and $\displaystyle \frac{\partial f}{\partial x}=h\left(x\right)$

But I'm uncertain where to go from here to backtrack to the original function. I haven't yet learned about integrals with more than one variable, so that's not going to help me any.

 

[Steven G]

Hi all. I'm having some difficulty with a problem from my Calculus IV course, about partial derivatives. The text of the exercise says:



I began by rewriting the given and combining that with the knowledge that mixed partial derivatives are equal.

$\displaystyle \frac{\partial }{\partial y}\left(\frac{\partial f}{\partial x}\right)=\frac{\partial }{\partial x}\left(\frac{\partial f}{\partial y}\right)=0$

I also know that for a partial derivative with respect to x to be 0, the function must be a constant or a function of y, and the opposite is true of a partial derivative with respect to y. So, now I have:

$\displaystyle \frac{\partial f}{\partial y}=g\left(y\right)$ and $\displaystyle \frac{\partial f}{\partial x}=h\left(x\right)$

But I'm uncertain where to go from here to backtrack to the original function. I haven't yet learned about integrals with more than one variable, so that's not going to help me any.

Are mixed partial always equal to one another?
I would think that f(x,y) might be ____ in one variable.

 

[Ishuda]

Hi all. I'm having some difficulty with a problem from my Calculus IV course, about partial derivatives. The text of the exercise says:

For the partial derivatives given in Exercises 51-54, find the most general form for a function of two variables, f(x,y), with the given partial derivative.

54)$\displaystyle \frac{\partial^2\, f}{\partial y\, \partial x}\, =\, 0$



{/quote]
I began by rewriting the given and combining that with the knowledge that mixed partial derivatives are equal.

$\displaystyle \frac{\partial }{\partial y}\left(\frac{\partial f}{\partial x}\right)=\frac{\partial }{\partial x}\left(\frac{\partial f}{\partial y}\right)=0$

I also know that for a partial derivative with respect to x to be 0, the function must be a constant or a function of y, and the opposite is true of a partial derivative with respect to y. So, now I have:

$\displaystyle \frac{\partial f}{\partial y}=g\left(y\right)$ and $\displaystyle \frac{\partial f}{\partial x}=h\left(x\right)$

But I'm uncertain where to go from here to backtrack to the original function. I haven't yet learned about integrals with more than one variable, so that's not going to help me any.

Almost. What you have is correct, i.e.
$\displaystyle \frac{\partial f}{\partial x}\, =\, h(x)$
as well as
$\displaystyle \frac{\partial f}{\partial y}\, =\, g(y)$

However, for the final answer, just integrate on of those equations to get
$\displaystyle f(x,y) =\, H(x)\, +\, G(y)$,
where H'(x) = h(x) and, G'(y) = g(y). The integration constants have been absorbed into the H(x) and G(y).

 

[Steven G]

Hi all. I'm having some difficulty with a problem from my Calculus IV course, about partial derivatives. The text of the exercise says:



I began by rewriting the given and combining that with the knowledge that mixed partial derivatives are equal.

$\displaystyle \frac{\partial }{\partial y}\left(\frac{\partial f}{\partial x}\right)=\frac{\partial }{\partial x}\left(\frac{\partial f}{\partial y}\right)=0$

I also know that for a partial derivative with respect to x to be 0, the function must be a constant or a function of y, and the opposite is true of a partial derivative with respect to y. So, now I have:

$\displaystyle \frac{\partial f}{\partial y}=g\left(y\right)$ and $\displaystyle \frac{\partial f}{\partial x}=h\left(x\right)$

But I'm uncertain where to go from here to backtrack to the original function. I haven't yet learned about integrals with more than one variable, so that's not going to help me any.

Functions whose second partials are discontinuous need not have their mixed partials equal. The standard example is

. . . . .$\displaystyle f(x,\, y)\, =\, \begin{cases}\dfrac{xy^3\, -\, x^3y}{x^2\, +\, y^2}&,\, (x,\, y)\, \neq\, (0,\, 0)\\0&,\, (x,\, y)\, =\, (0,\, 0)\end{cases}$

 

[ksdhart]

Functions whose second partials are discontinuous need not have their mixed partials equal. The standard example is
$\displaystyle f\left(x,y\right)=\frac{xy^3-x^3y}{x^2+y^2}\:if\:\left(x,y\right) \ne \left(0,0\right)$
$\displaystyle f\left(x,y\right)=0\:if\:\left(x,y\right)=\left(0,0\right)$

Hm. That's interesting. The professor's lecture in class mentioned that in some rare cases, the mixed partial derivatives won't be equal, but he didn't specify what those cases are. He also said that every homework problem we work from this book will have the mixed partials being equal, so we can assume that it's always true.

Almost. What you have is correct, i.e.
$\displaystyle \frac{\partial f}{\partial x}\, =\, h(x)$
as well as
$\displaystyle \frac{\partial f}{\partial y}\, =\, g(y)$

However, for the final answer, just integrate on of those equations to get
$\displaystyle f(x,y) =\, H(x)\, +\, G(y)$,
where H'(x) = h(x) and, G'(y) = g(y). The integration constants have been absorbed into the H(x) and G(y).

Oh, okay. I see. The partial derivatives are functions of just one variable still, so the standard rules of integration apply. And then when I take the partial derivatives of the full function, the parts with the other variable will fall away, producing the desired behavior. Great, thanks! That was far easier than I thought.