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Find dy/dx

  1. Jun 27, 2007 #1
    assistance asap is needed !!

    Using implicit differentiation, find dy/dx given that:
    cos(xy^2) - (x^3+y) / (x+1) = sec(x)sin(y)

    i am horrid at these, i came up with a few lines, please check them

    -sin(xy^2)[(y)^2+(2y)(dy/dx)(x)] - [ {(3x^2)(x+1)-(x^3+y)} / (x+1)^2] = sec(x)tan(x)sin(y) + cos(y)sin(y)(dy/dx)

    after this line i have no idea what else i can do with this, it looks jsut looks like a huge mess to me.
     
  2. jcsd
  3. Jun 27, 2007 #2
    You seem to have left out the y (in the numerator) while using the quotient rule, and you need a sec(x) (not sin(y) - second term RHS).

    Try to bring the dy/dx s to one side and see if the rest of the expressions simplify, somehow.
     
  4. Jun 27, 2007 #3
    so it would look this so:

    -sin(xy^2)[(y)^2+(2y)(dy/dx)(x)] - [ {(3x^2)(x+1)(dy/dx)-(x^3+y)} / (x+1)^2] = sec(x)tan(x)sin(y) + cos(y)sec(y)(dy/dx)


    so many dy/dx 's should i still try and bring them all on one side?
     
  5. Jun 27, 2007 #4
    You missed the dy/dx again...

    {(3x^2 + dy/dx)(x+1) - (x^3+y)} / (x+1)^2

    Yes, if the question requires it. I think many terms will cancel if you expand the expression on both sides.
     
  6. Jun 27, 2007 #5
    ok i get that now, but when i try to isolate the dy/dx the big bracket is too hard to get the (dy/dx) out of..

    this is what i got when i expanded it.

    [{(3x^3)+(3x^2)+(x)(dy/dx)+(dy/dx)-(x^3)-(y)} / (x+1)^2 ]

    then only 2 terms in then can be groups the 3x^3 - x^3 and i dont know where to do for the rest of it..
     
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