Partial Derivatives and using the chain rule

  • #1

Homework Statement



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

Answers and Replies

  • #2
623
0
Use the chain rule.
 
  • #3
ok,
let u = f(y/x)
then du/dx = -f(y/x2)

am i doing this right?
 
  • #4
623
0
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
yes its definately f(y/x) thats whats throwing me...
 
  • #6
16
0
can you write out the question exactly as it's worded?
 
  • #7
its worded exactly like i worded it at the start...do you think the question is wrong?
 
  • #8
Dick
Science Advisor
Homework Helper
26,260
619
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.
 
  • #9
HallsofIvy
Science Advisor
Homework Helper
41,833
956
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
[tex]\frac{\partial f(y/x)}{\partial x}= f'(y/x)\frac{\partial (y/x)}{\partial x}[/tex]
[tex]= f'(y/x)\frac{\partial yx^{-1}}{\partial x}= - f'(y/x)yx^{-2}= -f'(y/x)\frac{y}{x^2}[/tex]
where "f'(y/x)" is the ordinary derivative of f(x) evaluated at y/x.
 
  • #10
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
 

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