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Homework Help: D2y / dx2

  1. Feb 27, 2010 #1
    d2y/dx2 = d/dy [1/2 (y'^2)]

    so how do we show that the LHS = RHS? starting from the LHS?

    i don't really understand.

    my lecturer made it become d2y/dx2 = dy/dx * d/dy * y' and he showed it equals RHS

    i understand how he get to dy/dx * d/dy * y'

    but shouldn't that become dy/dx *y' * d/dy which is (y')^2 * d/dy? so where did the 1/2 from the RHS come from?

    any help? thanks!
  2. jcsd
  3. Feb 27, 2010 #2


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    hi quietrain! :smile:

    (try using the X2 tag just above the Reply box :wink:)
    (first, never write " d/dy * ", it makes no sense, either write " d/dy(y') " or " dy'/dy " :wink:)

    now convert dy/dx to to y', and it's y' * d/dy (y'), = d/dy (1/2 y'2) :smile:
  4. Feb 28, 2010 #3
    oh i see.

    so d2y / dx2 = d/dx dy/dx = d/dy dy/dx *dy/dx = y' * d/dy y'

    so i integrate y' * d/dy y' = 1/2 y'2

    so the differiential form of 1/2 y'2 = y' *d/dy y'

    but i don't understand this, what is d/dy y'?

    i know d/dx of y' = d/dx of dy/dx which means differientiate dy/dx one more time with respect to x

    but if i have d/dy of y' then i am differientiating dy/dx with respect to y ?

    so for example y = x3

    dy/dx = 3x2

    so d/dx dy/dx = 6x

    but what is d/dy dy/dx ?
  5. Feb 28, 2010 #4


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    d/dy dy/dx = d/d(x3) 3x2

    = d/dx3 3[x3]2/3

    = 3*(2/3)*[x3]-1/3

    = 2x-1

    so y' d/dy dy/dx = 3x2 2x-1 = 6x :wink:
  6. Mar 1, 2010 #5
    oh isee...

    so thats why we need to put in a y' for partial differientiation...

    so we have to convert the x2 term into a x3 term so that we can differientiate.

    thanks a lot tim!!!
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