1. Apr 6, 2010

kenshaw93

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. Apr 6, 2010

lanedance

hi kenshaw93 - have you had an attempt?

3. Apr 6, 2010

kenshaw93

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. Apr 6, 2010

kenshaw93

i meant right not write*-

5. Apr 6, 2010

kenshaw93

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. Apr 6, 2010

lanedance

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
$$\frac{d}{dx}(1+3sec^2(y)) = \frac{d}{dx}(1+3cos^{-2}(y))$$
then differentiate from there using the power & chain rules

also take try differentiating both sides of the equation implicitly & things should simplify...
$$\frac{d}{dx}( (\frac{dy}{dx})^2) = \frac{d}{dx}(1+3sec^2(y))$$

7. Apr 6, 2010

kenshaw93

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