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Implicit Differentiation

  1. Oct 2, 2008 #1
    1. The problem statement, all variables and given/known data
    Implicit Differentiation:

    I Was given the equation find dy/dx:
    (3x3y2 + 7x)
    (x2y3 + 3xy)-
    3. The attempt at a solution
    Ok, i know i have to use the product rule on top, and on bottom and the quotient rule for the fraction so... if i set

    s = 3x3
    t = y2
    u = (3x3y2 + 7x)
    w = x2
    n = y3
    z = 3x
    q = y
    v = (x2y3 + 3xy)

    ds = 9x2
    dt = 2y(dy/dx)
    du = (s(dt) + t(ds)) + 7)
    dw = 2x
    dn = 3y2(dy/dx)
    dz = 3
    dq = 1(dy/dx)
    dv = (w(dn) + n(dw)) + (z(dq) + q(dz))

    Quotient rule: v(du) - u(dv)/(v2)

    ok, I'm just making sure i didn't make a mistake.

    so now i do:

    (x2y3 + 3xy)((3x3(2y(dy/dx)) + y2(9x2)) + 7)) - (3x3y2 + 7x)((x2(3y2(dy/dx)) + y3(2x)) + ((3x(1(dy/dx)) + 3y)
    (x2y3 + 3xy)2


    I had a tutor help me out in Implicit Diff. but i still don't really remember this stuff unless i look at it, and i almost always get different from the professor, is the above correct ( without factoring?) it looks alot easier on my paper lol. and i'm supposed to be ready for this kind of magnitude questions on my exam :S i also don't know how i am going to be able to get dy/dx out of my equation, any tips there?
     
  2. jcsd
  3. Oct 2, 2008 #2
    you should group all the terms that has dy/dx in it on the other side. then factor out dy/dx.
     
  4. Oct 2, 2008 #3
    ok, but once i have (dy/dx)*(long equation) i can't just subtract dy/dx to the other side?
     
  5. Oct 2, 2008 #4
    you wont subtract it. you just have to divide the other side by the long equation you said.
     
  6. Oct 2, 2008 #5
    well then, i feel stupid lol i even knew that ^.^ i differentiate the other side as well, which was = 7x -> 7
     
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