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Implicit Differentiation issue

  1. Oct 2, 2005 #1
    Doing fine until I reached a trig function where I know i've done the work correctly but the answer does not match up exactly with the one in the back of the book.


    I do the work using product and chain rule

    [tex]2xy^2+2x^2yy' = \frac {1} {\cos(x^2y^2)}[/tex]
    [tex]y'=\frac {2xy^2} {2x^2y\cos(x^2y^2)}[/tex]

    But the answer in the back of the book says

    [tex]\frac {1-2xy^2\cos(x^2y^2)} {2x^2y\cos(x^2y^2)}[/tex]

    Is there a theorem i'm missing?
  2. jcsd
  3. Oct 2, 2005 #2


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    Staff Emeritus
    Science Advisor

    No theorem- it's algebra! You are fine up to
    [tex]2xy^2+2x^2yy' = \frac {1} {\cos(x^2y^2)}[/tex]
    but then you have:
    [tex]y'=\frac {2xy^2} {2x^2y\cos(x^2y^2)}[/tex]
    You appear to have multiplied by 2xy2 on the right.
    Since 2xy2 is added on the left you need to subtract it on both sides:
    [tex]2x^2yy'= \frac{1}{\cos(x^2y^2)}- 2xy^2[/tex]
    which is also
    [tex]2x^2yy'= \frac{1- 2xy^2\cos(x^2y^2)}{\cos(x^2y^2)}[/tex].
    Now, divide both sides by 2x2y.
  4. Oct 2, 2005 #3

    Nevermind! I got the algebraic rule, the trig just threw me off.
    Last edited: Oct 2, 2005
  5. Oct 3, 2005 #4
    yuck. implicit differentiation is a mess. check out the Devil's Rule by Cramer and The Folium of Descartes. <<< good to work with for finding out slopes at various points on the graph and understanding differentiation.
    Devil's Curve y^4 -4y^2 = x^4 - 9x^2.
    Folium of Descartes x^3 + y^3 - 9xy =0
    Then find tangents, normals, and also parametrics!! FUNN
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