- #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)

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- Thread starter kenshaw93
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- #1

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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

lanedance

Homework Helper

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hi kenshaw93 - have you had an attempt?

- #3

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- #4

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i meant right not write*-

- #5

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so:

dy/dx=2cos(X)/cos(y)

(dy/dx)^2=1+3sec^2(y)=(cos^2y + 3)/cos^2y.

- #6

lanedance

Homework Helper

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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]

- #7

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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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