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Implicit Differentiation help please

  • Thread starter kenshaw93
  • Start date
  • #1
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Homework Statement


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)


Homework Equations





The Attempt at a Solution

 

Answers and Replies

  • #2
lanedance
Homework Helper
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hi kenshaw93 - have you had an attempt?
 
  • #3
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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
 
  • #4
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i meant right not write*-
 
  • #5
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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.
 
  • #6
lanedance
Homework Helper
3,304
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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
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]
 
  • #7
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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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