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# implicit function watch

1. Find (d^2)y/dx^2 as a function of x if Sin(y) + Cos(y) = x

Cos (y). dy/dx + [-Sin(y).dy/dx] = 1
dy/dx = 1 / ( Cos (y) - Sin(y) ) => (Cos (y) - Sin(y)) ^-1

(d^2)y/dx^2 = -(Cos (y) - Sin(y))^-2 X [ -Sin(y).dy/dx - Cos(y).dy.dx]

(d^2)y/dx^2 =-(Cos (y) - Sin(y))^-2 X -[Sin(y).dy/dx + Cos(y).dy.dx]

(d^2)y/dx^2 = dy/dx ( x) / (Cos(y) - Sin(y))^2

(d^2)y/dx^2 =(Cos (y) - Sin(y)) ^-1 ( x) / (Cos(y) - Sin(y))^2

(d^2)y/dx^2 = x / (Cos(y) - Sin(y))^3

2. (Original post by Carlos Nim)

(d^2)y/dx^2 = x / (Cos(y) - Sin(y))^3

Now use to obtain in terms of .
3. (Original post by RDKGames)

Now use to obtain in terms of .
i have a feeling that this is connected to RSin(A+B) but i still can't figure it out.

-2Sin(y)Cos(y) => -[Sin ( y + y )]
4. (Original post by Carlos Nim)
i have a feeling that this is connected to RSin(A+B) but i still can't figure it out.

-2Sin(y)Cos(y) => -[Sin ( y + y )]
It is but that's not required.
5. x = Siny + Cosy
x^2 = Sin ^ 2 (y) + 2Sin(y)Cos(y) + Cos^2(y) => 1 + 2 Sin(yCos(y)
how do i turn it into 1-2Sin(y)Cos(y)?
6. (Original post by Carlos Nim)
x = Siny + Cosy
x^2 = Sin ^ 2 (y) + 2Sin(y)Cos(y) + Cos^2(y) => 1 + 2 Sin(yCos(y)
how do i turn it into 1-2Sin(y)Cos(y)?
You only need to replace by something to do with
7. (x-2Cos(y))^2 =[ Sin(y)+ Cos(y) - 2Cos(y) ]^2 = [ Sin(y) - Cos(y) ] ^2 = 1 - 2Sin(y)Cos(y)
??
8. (Original post by Carlos Nim)
(x-2Cos(y))^2 =[ Sin(y)+ Cos(y) - 2Cos(y) ]^2 = [ Sin(y) - Cos(y) ] ^2 = 1 - 2Sin(y)Cos(y)
??
No... just say

so hence

Now substitute it into
9. ah ok i got it now , i was looking at the whole 1 - 2Sin(y)Cos(y) instead of 2Sin(y)Cos(y) alone
Thank you

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Updated: March 11, 2018
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