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    Given that x = 4 sin(2y + 6), find dy/dx in terms of x.



    Using the chain rule, I got dx/dy = 8cos (2y + 6) so dy/dx = 1/8cos (2y + 6).

    From here:

     \displaystyle



8cos^2 (2y+6) + 8sin^2(2y+6) = 1



\therefore dy/dx =  \frac {1}\sqrt{8sin(2y+6)-1}



dy/dx = \frac {1}\sqrt{2x-1}

    Which is wrong, I'm pretty sure I used the identity wrong but i can't think of one that links the 8cos(2y+6) to x. Any ideas?
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    (Original post by fait)
    [B]

    



8cos^2 (2y+6) + 8sin^2(2y+6) = 8(cos^2 (2y+6) + sin^2(2y+6))=\mathbf{8}
    Fixed.
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    Cheers.

     \displaystyle \frac {1}{\sqrt(8-sin(2y+6))}





\implies \displaystyle   \frac {1}{\sqrt8-(1/4x))}

    Which is still wrong. I don't really see what I'm doing wrong though. =z
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     x = 4\sin(2y+6) \implies x^2 + 16\cos^2(2y+6)=16

    So, we can rearrange this to get:

     8\cos(2y+6) = 2\sqrt{16-x^2} (taking positive square root)

    Hence,  \frac{dy}{dx} = \frac{1}{2\sqrt{16-x^2}} .

    We can check this is right:

     x = 4\sin(2y+6) \implies y = \frac{1}{2}\left(\sin^{-1}(\frac{x}{4})-6\right)

    Differentiating, we get the same result:

     \displaystyle \frac{dy}{dx} = \frac{1}{2} .\frac{1}{\sqrt{1-(\frac{x}{4})^2}} . \frac{1}{4} =  \frac{1}{2\sqrt{16-x^2}}
 
 
 
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