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Homework Help: Differentiation problem

  1. Jan 31, 2009 #1
    1. The problem statement, all variables and given/known data

    differentiate y = (3x2-4)1/2+(x2+4)1/2
    with respect to x

    2. Relevant equations

    y = (3x2-4)1/2+(x2+4)1/2

    3. The attempt at a solution

    dy/dx = (3x2-4)1/2dy/dx+(x2+4)1/2dy/dx

    Is it correct...???

    for the first part
    let u = (3x2-4)1/2
    du/dx = 6x
    y = u1/2
    dy/du = 1/2u-1/2
    dy/dx = du/dx dy/du
    dy/dx = 6x . 1/2u-1/2
    = 3x/u1/2

    for the second part
    and v = (x2+4)1/2
    dv/dx = 2x
    y = v1/2
    dy/dv = 1/2v-1/2
    dy/dx = dv/dx dy/dv
    dy/dx = 2x . 1/2v-1/2
    = x/v1/2

    dy/dx = 3x/u1/2 + x/v1/2

    = x/(3x2-4)1/2 + x/(x2+4)1/2
    Last edited: Jan 31, 2009
  2. jcsd
  3. Jan 31, 2009 #2


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    Staff Emeritus
    Science Advisor
    Gold Member

    Looks good up to here. However, there's a somewhat minor slip between the line above and the line below:
  4. Jan 31, 2009 #3
    Your solution looks good (apart from the typo):

  5. Jan 31, 2009 #4


    Staff: Mentor

    The other folks on this thread didn't seem to notice the line above, which is not correct.

    Perhaps what you meant to say was
    dy/dx = d/dx[(3x2-4)1/2]+d/dx[(x2+4)1/2]

    The difference is between the function dy/dx that you show on the right side of the equation, and the operator d/dx that I show on the right side. The symbol dy/dx represents the derivative itself, while the symbol d/dx represents the intent to take the derivative at some later time.
  6. Oct 28, 2010 #5
    can someone help me solve:
    Find the dx/dy for y=3x[2]
  7. Oct 28, 2010 #6


    Staff: Mentor

    Instead of "highjacking" a thread that's almost two years old, please start a new thread. Since this is your first post, you are probably not aware of the rules of this forum (see https://www.physicsforums.com/showthread.php?t=414380), which say that you must make some effort at solving your problem before we can give any help.
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