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Implicit Differentiation help please

  1. Apr 6, 2010 #1
    1. The problem statement, all variables and given/known data
    If siny=2sinx and (dy/dx)^2=1+3sec^2(y) show that:
    by differentiating 1+3sec^2(y) with respect to x, d^2y/dx^2=3sec^2(y)tan(y)


    2. Relevant equations



    3. The attempt at a solution
     
  2. jcsd
  3. Apr 6, 2010 #2

    lanedance

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    hi kenshaw93 - have you had an attempt?
     
  4. Apr 6, 2010 #3
    sorry i didn't write it, i thought it would be useless but i tried differentiating 1+3sec^2(y) and all i got was 3tany(dy/dx)... if thats write then i dont know how to continue
     
  5. Apr 6, 2010 #4
    i meant right not write*-
     
  6. Apr 6, 2010 #5
    i also tried finding dy/dx of siny=2sinx and i got 2cos(X)/cos(y), then i squared that to get (dy/dx)^2 and it matched the given one ie. 1+3sec^2(y) which i also found it to be equal to (cos^2y + 3)/cos^2y.
    so:
    dy/dx=2cos(X)/cos(y)
    (dy/dx)^2=1+3sec^2(y)=(cos^2y + 3)/cos^2y.
     
  7. Apr 6, 2010 #6

    lanedance

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    maybe show you working for the derivative here, i don't think that derivative is quite right, i find it helps to do the following
    [tex] \frac{d}{dx}(1+3sec^2(y)) = \frac{d}{dx}(1+3cos^{-2}(y)) [/tex]
    then differentiate from there using the power & chain rules


    also take try differentiating both sides of the equation implicitly & things should simplify...
    [tex]\frac{d}{dx}( (\frac{dy}{dx})^2) = \frac{d}{dx}(1+3sec^2(y))[/tex]
     
  8. Apr 6, 2010 #7
    thank you that really helped. this is what i have managed to do with your help:
    d/dx(1+3sec^2(y))= 6sec^3y.siny.dy/dx
    d/dx(dy/dx)^2=6sec^3y.siny.dy/dx
    i cancelled the dy/dx from the RHS and LHS to get:
    d2y/dx2= 6 sec^2y tany

    thats a great step forward but the answer needed is 3sec^2y tany... I've probably done a silly mistake somewhere there but i cant find it. Thank you so much for your help and tips :D
     
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