Results 1 to 8 of 8

Math Help - Implicit Differentiation ???

  1. #1
    Newbie
    Joined
    Mar 2008
    Posts
    22

    Question Implicit Differentiation ???

    Ok... I have gotten this far with this problem:

    xy+y^3=1

    [x'*y+x*y'] +[3y^2]=0

    My reasoning:
    To get [x'*y+x+y'] I used the product rule
    To get [3y^2] I used the product rule
    to get "0" I used the constant rule

    so... what's next ?

    I know I am solving for y'

    but can't figure out the next step

    How to get y' alone one side or should it be taken away at this point ?

    VERY VERY confused
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Junior Member
    Joined
    Apr 2008
    From
    Pittsburgh, PA
    Posts
    41
    Quote Originally Posted by kbgemini16 View Post
    Ok... I have gotten this far with this problem:

    xy+y^3=1

    [x'*y+x*y'] +[3y^2]=0

    My reasoning:
    To get [x'*y+x+y'] I used the product rule
    To get [3y^2] I used the product rule
    to get "0" I used the constant rule

    so... what's next ?

    I know I am solving for y'

    but can't figure out the next step

    How to get y' alone one side or should it be taken away at this point ?

    VERY VERY confused
    Sorry not that simple..will correct and edit in a minute.

    First, you wanna move the xy' and 3y^2 terms over.

    Just to give you a tip, it's easier if you keep track of your differentials in this situation, because the 3y^2 needs a y' term that is easily lost in this situation, if you don't do so.

    So now that you have y=-xy'-3y^2y'

    You can factor out the y' and solve for it, to get the final answer of y'=\frac{y}{-x-3y^2}
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Newbie
    Joined
    Mar 2008
    Posts
    22

    Question Confused...what happened ?

    Quote Originally Posted by Kalter Tod View Post
    Sorry not that simple..will correct and edit in a minute.

    First, you wanna move the xy' and 3y^2 terms over.

    Just to give you a tip, it's easier if you keep track of your differentials in this situation, because the 3y^2 needs a y' term that is easily lost in this situation, if you don't do so.

    So now that you have y=-xy'-3y^2y'

    You can factor out the y' and solve for it, to get the final answer of y'=\frac{y}{-x-3y^2}

    ok..
    So the step with [x'*y+x*y']+[3y62]=0 is wrong?

    what step did I miss ?
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Junior Member
    Joined
    Apr 2008
    From
    Pittsburgh, PA
    Posts
    41
    Quote Originally Posted by kbgemini16 View Post
    ok..
    So the step with [x'*y+x*y']+[3y62]=0 is wrong?

    what step did I miss ?
    Err...No...you didn't miss anything. I just worked from that step...

    Edit..if it helps clear it up, I did not include the x' term in the factor of x'y because it's more or less irrelevant when solving for y'
    Follow Math Help Forum on Facebook and Google+

  5. #5
    MHF Contributor Mathstud28's Avatar
    Joined
    Mar 2008
    From
    Pennsylvania
    Posts
    3,641

    Ok

    Quote Originally Posted by kbgemini16 View Post
    Ok... I have gotten this far with this problem:

    xy+y^3=1

    [x'*y+x*y'] +[3y^2]=0

    My reasoning:
    To get [x'*y+x+y'] I used the product rule
    To get [3y^2] I used the product rule
    to get "0" I used the constant rule

    so... what's next ?

    I know I am solving for y'

    but can't figure out the next step

    How to get y' alone one side or should it be taken away at this point ?

    VERY VERY confused
    xy+y^3=1 differentiating implicity we get xy'+y+3y^2\cdot{y'}=0 now solve for y'
    Follow Math Help Forum on Facebook and Google+

  6. #6
    Junior Member
    Joined
    Apr 2008
    From
    Pittsburgh, PA
    Posts
    41
    Quote Originally Posted by Mathstud28 View Post
    xy+y^3=1 differentiating implicity we get xy'+y+3y^2\cdot{y'}=0 now solve for y'
    Yeah, Stud. The OP had already gotten to that point. They wanted help going from there.
    Follow Math Help Forum on Facebook and Google+

  7. #7
    Newbie
    Joined
    Mar 2008
    Posts
    22

    Question Is it ???

    Quote Originally Posted by Mathstud28 View Post
    xy+y^3=1 differentiating implicity we get xy'+y+3y^2\cdot{y'}=0 now solve for y'

    y'= -3y^2
    x
    Follow Math Help Forum on Facebook and Google+

  8. #8
    Junior Member
    Joined
    Apr 2008
    From
    Pittsburgh, PA
    Posts
    41
    Quote Originally Posted by kbgemini16 View Post
    y'= -3y^2
    x
    Check the work I already posted. I ended up with the negative in the denominator because I did the problem slightly different, but it's the same answer.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Replies: 2
    Last Post: July 26th 2010, 05:24 PM
  2. implicit differentiation
    Posted in the Calculus Forum
    Replies: 3
    Last Post: May 1st 2010, 07:42 AM
  3. Implicit Differentiation
    Posted in the Calculus Forum
    Replies: 4
    Last Post: April 28th 2010, 03:19 PM
  4. Implicit differentiation
    Posted in the Calculus Forum
    Replies: 3
    Last Post: April 26th 2010, 05:41 PM
  5. Implicit Differentiation
    Posted in the Calculus Forum
    Replies: 3
    Last Post: March 25th 2008, 07:33 PM

Search Tags


/mathhelpforum @mathhelpforum