Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Homework Help: Partial derivative

  1. May 26, 2004 #1
    Hi All,

    Can someone refresh my memory and show me how to find the following partial derivate:

    [tex]t=\frac{x}{\sqrt{x^2+y^2}}[/tex]

    with respect to x.

    Thanks
     
    Last edited: May 26, 2004
  2. jcsd
  3. May 26, 2004 #2

    arildno

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member
    Dearly Missed

    The partial derivative of t(x,y) with respect to x is found by differentiating the expression as in 1-variable calculus; just remember that y is to be considered a constant. For example:
    [tex]t(x,y)=xy\rightarrow\frac{\partial{t}}{\partial{x}}=y[/tex]
     
  4. May 26, 2004 #3
    Yeah I understand the concept. It's just that I've forgotten how to find the derivative of a quotient. The textbook that I have doesn't explain it clearly.
     
  5. May 26, 2004 #4

    arildno

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member
    Dearly Missed

    The simplest way is to solve for the derivative of t w.r. to x by differentiating the equivalent equation with respect to x:
    [tex]\sqrt{x^{2}+y^{2}}t(x,y)=x[/tex]

    Another easy way is (I'll give it for one variable):
    [tex]t(x)=\frac{f(x)}{g(x)}[/tex]
    Use the product rule and the chain rule:
    [tex]\frac{dt}{dx}=\frac{f'(x)}{g(x)}-\frac{f(x)}{g^{2}(x)}*g'(x)[/tex]
     
    Last edited: May 26, 2004
  6. May 26, 2004 #5
    I remember the quotient rule as "Lo d Hi minus Hi d Lo over Lo squared," or:

    [tex]\frac{d}{dx}\,\frac{f(x)}{g(x)}=\frac{gf' - fg'}{g^2}[/tex]
     
    Last edited: May 26, 2004
  7. May 26, 2004 #6

    NateTG

    User Avatar
    Science Advisor
    Homework Helper

    Personally I generally use:
    [tex]\frac{a}{b}=a (b)^{-1}[/tex]
    so in your case
    [tex]\frac{x}{\sqrt{x^2+y^2}}=x (x^2+y^2)^{-\frac{1}{2}}[/tex]
     
  8. May 26, 2004 #7
    When I forget how do calculate a particular derivate I always derive it by using the definition of the derivate, i.e.
    [tex]\frac{d}{dx}f(x) = \lim_{h \to \infty} \frac{f(x+h)-f(x)}{h}[/tex]​
    e(ho0n3
     
  9. May 26, 2004 #8
    How do you proceed using this form?
     
  10. May 26, 2004 #9
    Product rule and power rule.

    f'*g + g'*f, where f = x and g = (x^2 + y^2)^(-1/2), so f' = 1 and g' = -x(x^2 + y^2)^(-3/2)

    And e(ho0n3, don't you mean limit as h goes to 0?

    cookiemonster
     
    Last edited: May 26, 2004
  11. May 26, 2004 #10
    Thanks Cookiemonster.
     
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook