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: Need some help with implicit differentiation

  1. Dec 28, 2007 #1
    Hi, I need some help with these question, and would appreciate the help.

    1. The problem statement, all variables and given/known data
    Part 1
    Use implicit differentiation to find y'' if 2xy = y^2 Simplify and Leaev in terms of x and y.

    Part 2
    Use implicit differentiation to find y '' if xy + y^3 = 1
    Simplify your answer and leave it in terms of x and y

    2. Relevant equations

    3. The attempt at a solution
    For part 1;

    I had

    2x dy/dx + 2y = 2y dy/dx
    Took out 2's
    solved for dy/dx

    dy dx = y / [y-x]

    Part 2
    I tried the same, took d/dx to get

    dy/dx + 3y^2 dy/dx = 0 But here I get stuck because of the zero.
  2. jcsd
  3. Dec 28, 2007 #2
    It asked for y'' not y', let me work it but am I correct on that part?
  4. Dec 28, 2007 #3
    Ohh, so go through the process again? Ok, but can you help me along?
  5. Dec 28, 2007 #4
    Sure, so now we're at


    Take the derivative again, quotient rule:



    Last edited: Dec 28, 2007
  6. Dec 28, 2007 #5
    For part 2: You forgot or did the the product rule incorrectly;

    [tex]xy+y^3 =1[/tex]

    [tex]xy'+y+3y^2 y'=0[/tex]

    The zero means nothing. It's the same thing as saying x-2=0, well then x=2.

    Solving for y'


    Now take the 2nd derivative, quotient rule:
    Last edited: Dec 28, 2007
  7. Dec 28, 2007 #6


    User Avatar
    Science Advisor

    You don't need to solve for y' itself. If [itex]2xy= y^2[/itex], then [itex]2y+ 2xy'= 2yy'[/itex]. Now differentiate both sides of that with respect to x: [itex]2y'+ 2y'+ 2xy"= 2y'^2+ yy"[/itex] or [itex]4y'- 2y'^2= (y- 2x)y"[/itex] so [itex] y"= (4y'-2y'^2)/(y- 2x)[/itex]. Personally, I would consider that a perfectly good answer but since your problem specifically says "leave in terms of x and y", NOW use x/(y- x).
    Last edited by a moderator: Dec 28, 2007
  8. Dec 28, 2007 #7
    Wow, thanks for clearing it up for me, I greatly appreciate the help =]
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook