1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
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: Derivative of the square root of xy

  1. Jan 11, 2009 #1
    1. The problem statement, all variables and given/known data

    What is the deriv. of the square root of (xy)?

    2. Relevant equations

    3. The attempt at a solution

    I used the chain rule:

    (1/2)(xy)^(-1/2) times (y + x(dy/dx))

    i am unsure on how to distribute this correctly
  2. jcsd
  3. Jan 11, 2009 #2


    User Avatar
    Science Advisor

    The derivative of [itex](xy)^{1/2}[/itex] with respect to x, and y is a function of x?

    If that is the question, then yes that is correct. I don't know what you mean "how to distribute this correctly". The distributive law is the distributive law: a(b+ c)= ab+ ac.
    Is it the half powers that concern you? [itex](xy)^{1/2}x= (x^{1/2})(x)(y^{1/2}= x^{3/2}y^{1/2}[/itex] and [itex](xy)^{1/2}y= (x^{1/2})(y^{1/2})y= x^{1/2}y^{3/2}[/itex].

    [itex](1/2)(xy)^{1/2}[y+ x dy/dx]= (1/2)x^{1/2}y^{3/2}+ x^{3/2}y^{1/2} dy/dx[/itex]
  4. Jan 11, 2009 #3
    Are you trying to do implicit differentiation? If so treat

    [tex] \sqrt{xy} = \sqrt{x}\sqrt{y} [/tex]

    Then use the product rule, just remember when you differentiate [tex] \sqrt{y} [/tex] to multiply by y'.
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook