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Homework Help: Determine (dy/dx) using implicit differentiation

  1. Oct 16, 2007 #1
    Determine (dy/dx) using implicit differentiation.

    cos(X^2Y^2) = x

    I'm really confused what to do now..i think the next steps are:

    d/dx [cos(X^2*Y^2)] = d/dx [x]
    = -sin(X^2*Y^2)* ((X^2*2Y dy/dx) + (Y^2*2X)) = 1
    = -2YX^2 sin(X^2*Y^2) dy/dx + -2XY^2sin(X^2*Y^2) = 1
    = -2YX^2 sin(X^2*Y^2) dy/dx = 1 + -2XY^2sin(X^2*Y^2)
    = dy/dx = (1 + -2XY^2sin(X^2*Y^2))/ (-2YX^2 sin(X^2*Y^2))

    Can someone tell me if this is correct?
  2. jcsd
  3. Oct 16, 2007 #2
    [tex]cos(x^2 y^2)=x[/tex]

    [tex]\frac{d}{dx}[cos(x^2 y^2)]=\frac{d}{dx}[x][/tex]

    [tex]-sin(x^2 y^2) [\frac{d}{dx}(x^2 y^2)] = 1[/tex]

    [tex]-sin(x^2 y^2) [2y^2x + 2x^2y\frac{dy}{dx}] = 1[/tex]

    [tex]-sin(x^2 y^2)2y^2x -sin(x^2 y^2)2x^2y\frac{dy}{dx} = 1[/tex]

    [tex]-sin(x^2 y^2)2x^2y\frac{dy}{dx} = 1+sin(x^2 y^2)2y^2x [/tex]

    [tex]\frac{dy}{dx} = \frac{1+sin(x^2 y^2)2y^2x}{-sin(x^2 y^2)2x^2y} [/tex]
  4. Oct 16, 2007 #3
    cool thanks
  5. Oct 17, 2007 #4


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    atqamar, please note that one should not provide full solutions in the homework forums, but rather should offer guidance and hints as to how to proceed.
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