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Help: Implicit differentiation with initial values

  1. Sep 13, 2009 #1
    1. The problem statement, all variables and given/known data
    2*y + sin(y) = x^4 + 4(x)^3 + (2(Pi) - 5), show that dy/dx = 16, when x = 1.


    2. Relevant equations



    3. The attempt at a solution
    So I implicitly differentiated it to be dy/dx(2 + cos(y)) = 4(x)^3 + 12(x)^2, and I end up with
    dy/dx = 16 / (2 + cos (y)) which means that y must be equal to Pi for this to be true, but I do not think this is where I was supposed to go with this problem. Is there a way to factor or substitute out Cos(y) to show that dy/dx = 16, when x = 1?
     
  2. jcsd
  3. Sep 13, 2009 #2
    [tex] \frac{dy}{dx} = \frac{4x^3+12x^2}{2+\cos(y)} [/tex]

    [tex] 2y+\sin(y) = 2\pi [/tex]

    [tex] y=\pi \iff 2\pi +\sin(\pi) = 2\pi [/tex]

    [tex] \frac{dy}{dx} = \frac{16}{2+\cos(\pi)}=16 [/tex]
     
  4. Sep 13, 2009 #3
    This is exactly what I had before. Is there any way of making this proof without first implying that y must = Pi?

    Edit: I see where the proof comes from. I got it correct without realizing it. Thanks for the help
     
    Last edited: Sep 13, 2009
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