# Partial Derivatives

1. Aug 18, 2008

### gtfitzpatrick

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

If V=x$$^{3}$$f(y/x) show that x$$^{2}$$Vxx + 2xyVxy + y$$^{2}$$Vyy = 6V

3. The attempt at a solution
i would normally just use the chain rule to differenciate this with respect to x and then so on but the f(y/x) is throwing me. Do i just treat the f like a constant or is it a whole new function which i have to differenciate first?
thanks
gf

2. Aug 18, 2008

### NoMoreExams

Use the chain rule.

3. Aug 18, 2008

### gtfitzpatrick

ok,
let u = f(y/x)
then du/dx = -f(y/x2)

am i doing this right?

4. Aug 18, 2008

### NoMoreExams

I'm a bit confused now, your f appears to be a function of x and y i.e. f(x,y) but yet you have f(x/y), are you sure that's the problem word for word?

5. Aug 18, 2008

### gtfitzpatrick

yes its definately f(y/x) thats whats throwing me...

6. Aug 18, 2008

### wbrigg

can you write out the question exactly as it's worded?

7. Aug 18, 2008

### gtfitzpatrick

its worded exactly like i worded it at the start...do you think the question is wrong?

8. Aug 18, 2008

### Dick

No, use the chain rule, du/dx=f'(y/x)*d(y/x)/dx=-y*f'(y/x)/x^2.

9. Aug 18, 2008

### HallsofIvy

Staff Emeritus
No. You were told to use the chain rule
$$\frac{\partial f(y/x)}{\partial x}= f'(y/x)\frac{\partial (y/x)}{\partial x}$$
$$= f'(y/x)\frac{\partial yx^{-1}}{\partial x}= - f'(y/x)yx^{-2}= -f'(y/x)\frac{y}{x^2}$$
where "f'(y/x)" is the ordinary derivative of f(x) evaluated at y/x.

10. Aug 19, 2008