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-   -   Implicit differentiation vs explicit (https://www.askmehelpdesk.com/showthread.php?t=204488)

  • Apr 10, 2008, 12:01 PM
    geobeeker
    implicit differentiation vs explicit
    I was asked to implicitly differentiate (x^2)y + y = 2, which I did:

    use product rule: [fg]' = f'g + fg'
    f=x^2
    g=y
    f'=2x
    g'=y'

    [(x^2)y + y]' = 2'
    [(x^2)y]' + [y]' = 2'
    2xy +(x^2)y' + y' = 0
    2xy = -(x^2)y' - y'
    2xy = -y'(x^2 + 1)
    2xy
    y' = - -------
    x^2 +1
    which matches the book's answer.

    But then I thought why can't I solve for y and then differentiate explicitly?

    I tried:
    (x^2)y + y = 2 ==> y( (x^2) + 1) = 2 ===> y= 2/((x^2) + 1)

    But my graphing program says
    4x
    y' = - -------------
    (x^2 +1)^2

    What's wrong?
  • Apr 10, 2008, 02:42 PM
    galactus
    It's the same, you just didn't sub in your y.

    From implicit diff. we get

    From doing it the 'regular' way, we get

    Now, when we solved the equation for y we got

    Sub that into your implicit derivative:



    See?
  • Apr 26, 2008, 09:20 AM
    alamleh
    both answers are the same even they look different

    using explicit diff y' = -4x/(x^2+1)^2

    using implicit diff y' = -2xy/(x^2+1) but the replacing y in the answer by 2/(x^2+1)
    well get you the same above y=-4x/(x^2+1)^2





    Quote:

    Originally Posted by geobeeker
    I was asked to implicitly differentiate (x^2)y + y = 2, which I did:

    use product rule: [fg]' = f'g + fg'
    f=x^2
    g=y
    f'=2x
    g'=y'

    [(x^2)y + y]' = 2'
    [(x^2)y]' + [y]' = 2'
    2xy +(x^2)y' + y' = 0
    2xy = -(x^2)y' - y'
    2xy = -y'(x^2 + 1)
    2xy
    y' = - -------
    x^2 +1
    which matches the book's answer.

    But then I thought why can't I solve for y and then differentiate explicitly?

    I tried:
    (x^2)y + y = 2 ==> y( (x^2) + 1) = 2 ===> y= 2/((x^2) + 1)

    But my graphing program says
    4x
    y' = - -------------
    (x^2 +1)^2

    What's wrong?


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