Partial Derivatives and using the chain rule

[gtfitzpatrick]

Homework Statement

If V=x$$^{3}$$f(y/x) show that x$$^{2}$$V_xx + 2xyV_xy + y$$^{2}$$V_yy = 6V

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

 

[NoMoreExams]

Use the chain rule.

 

[gtfitzpatrick]

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

am i doing this right?

 

[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?

 

[gtfitzpatrick]

yes its definitely f(y/x) that's what's throwing me...

 

[wbrigg]

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

 

[gtfitzpatrick]

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

 

[Dick]

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

am i doing this right?

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

 

[HallsofIvy]

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

am i doing this right?

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.

 

[gtfitzpatrick]

thanks lads,
got it worked out, when i looked at the first line in my log tables it all became clear. Thanks for the help