# Partial differentiaton question, how to distinguish between u and f (u )

1. Aug 23, 2011

### mohsin031211

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

V=xf(u) and u=y/x, show that x^2. d2v/dx2 + 2xy. d2v/dxdy + y^2. d2V/dy2 = 0
Is u the same as f(u)

2. Relevant equations

V=xf(u) and u=y/x

3. The attempt at a solution
do you just differentiate the product of x.y/x

2. Aug 23, 2011

### vela

Staff Emeritus
No, u is not the same as f(u).

You need to use the product and chain rules. For example, to calculate ∂V/∂x, you'd start by applying the product rule to get
\begin{equation*}
\frac{\partial V}{\partial x} = \frac{\partial}{\partial x}[xf(u)] = f(u) + x \frac{\partial f(u)}{\partial x}
\end{equation*}
To calculate the second term, you need to use the chain rule, which says
\begin{equation*}
\frac{\partial f}{\partial x} = \frac{\partial f}{\partial u}\frac{\partial u}{\partial x} = -\frac{y}{x^2}\frac{\partial f}{\partial u}
\end{equation*}
So you have
\begin{equation*}
\frac{\partial V}{\partial x} = f(u) - \frac{y}{x} \frac{\partial f}{\partial u} = f(u) - u\frac{\partial f}{\partial u}
\end{equation*}

Last edited: Aug 23, 2011
3. Aug 23, 2011

### mohsin031211

Thankyou very much but i still cannot understand how that may lead to the form which the question wants us to show, thanks.

4. Aug 23, 2011

### Staff: Mentor

Well, you have a way to go. Starting from vela's advice, you need to calculate ∂2V/∂x2, ∂V/∂y, ∂2V/∂y2, and ∂2V/(∂x∂y), and then substitute them into the left side of your equation, which should simplify to zero.