# Question about second-order partial derivatives

## Homework Statement

If V=xf(u) and u=y/x, show that

x^2.d2V/dx2 + 2xy.d2V/dxdy + y^2.d2V/dy2= 0

(This a partial differentiation problem so all the d's are curly d's)

## The Attempt at a Solution

I have tried to work out d2V/dx2 and the other derivatives, then multiply them by x^2 or 2xy or whatever is in front of them and then add them all together, but the answer I get is not 0.

I think the issue is I don't know how to work out d2V/dx2

to work out d2V/dx2 I am trying d/dx (dV/du . du/dx). But it does not seem to work

Any help would be appreciated
thanks
cluivee

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Hootenanny
Staff Emeritus
Gold Member

## Homework Statement

If V=xf(u) and u=y/x, show that

x^2.d2V/dx2 + 2xy.d2V/dxdy + y^2.d2V/dy2= 0

(This a partial differentiation problem so all the d's are curly d's)

## The Attempt at a Solution

I have tried to work out d2V/dx2 and the other derivatives, then multiply them by x^2 or 2xy or whatever is in front of them and then add them all together, but the answer I get is not 0.

I think the issue is I don't know how to work out d2V/dx2

to work out d2V/dx2 I am trying d/dx (dV/du . du/dx). But it does not seem to work

Any help would be appreciated
thanks
cluivee
Welcome to Physics Forums.

You are correct in saying that (chain rule)

$$\frac{\partial}{\partial x} f(u) = \frac{\partial}{\partial u} f(u)\frac{\partial u}{\partial x}\;.$$

However, you need to be careful here since $V(u)=xf(u)$. Therefore,

\begin{aligned}\frac{\partial}{\partial x} V(u) & = \frac{\partial}{\partial x} xf(u) \\ &= f(u) + x\frac{\partial}{\partial x} f(u) \\ & = f(u) + x \frac{\partial}{\partial u} f(u)\frac{\partial u}{\partial x}\;.\end{aligned}

Do you follow?

I think so, but I need a bit of clarification, I've seen you've used the product rule to get

=f(u)+x∂∂xf(u)
=f(u)+x∂∂uf(u)∂u∂x.

but f(u)= u = y/x right? I know that may be a dumb question.

Then d/du(f(u)) = 1 always? And du/dx = -y/x^2?

Thanks for your help though Hootenanny

Hootenanny
Staff Emeritus
Gold Member
I think so, but I need a bit of clarification, I've seen you've used the product rule to get

=f(u)+x∂∂xf(u)
=f(u)+x∂∂uf(u)∂u∂x.

but f(u)= u = y/x right? I know that may be a dumb question.

Then d/du(f(u)) = 1 always? And du/dx = -y/x^2?
Why is $f(u)=u$?
Thanks for your help though Hootenanny
No problem