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    x = 4sin(2y+6)
    Find \dfrac{dy}{dx} in terms of x

    Now I can find it in terms of y it's just:
    \dfrac{1}{8cos(2y+6)}
    But idk how to get it in terms of x, do I substitute it in somewhere along the line?
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    (Original post by Retsek)
    x = 4sin(2y+6)
    Find \dfrac{dy}{dx} in terms of x

    Now I can find it in terms of y it's just:
    8cos(2y+6)
    But idk how to get it in terms of x, do I substitute it in somewhere along the line?
    That's only dx/dy
    \cos y = \sqrt{1-\sin^2y} for suitable y, so you can use it here.
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    (Original post by RDKGames)
    That's only dx/dy
    \cos y = \sqrt{1-\sin^2y} for suitable y, so you can use it here.
    Okay that's handy I think? But I don't know how to get the y out of the sine function? Is there a trig rule I'm not remembering?
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    (Original post by Retsek)
    Okay that's handy I think? But I don't know how to get the y out of the sine function? Is there a trig rule I'm not remembering?
    This comes from \sin^2 x + \cos^2 x \equiv 1

    You get 4\cos (2y+6) = \sqrt{16-16\sin^2(2y+6)} so 4\cos (2y+6) = \sqrt{16-x^2}
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    (Original post by RDKGames)
    This comes from \sin^2 x + \cos^2 x \equiv 1

    You get 4\cos (2y+6) = \sqrt{16-16\sin^2(2y+6)} so 4\cos (2y+6) = \sqrt{16-x^2}
    Okay I think I've got what you're getting at, I think you had sine and cos round the wrong way in your post but you just swap them and you get:
    x=\sqrt{16-16cos^2(2y+6)}
    and I know that:
    \dfrac{dx}{dy} = 8cos(2y+6)
    So I manipulate the first equation until it looks like the second to get:
    \dfrac{dx}{dy} = \dfrac{\sqrt{16-x^2}}{2}
    Is this the right idea and have I messed up along the way at all

    (Also thanks a bunch for pointing me in the right direction instead of just giving me the answer)
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    (Original post by Retsek)
    Okay I think I've got what you're getting at, I think you had sine and cos round the wrong way in your post but you just swap them and you get:
    x=\sqrt{16-16cos^2(2y+6)}
    and I know that:
    \dfrac{dx}{dy} = 8cos(2y+6)
    So I manipulate the first equation until it looks like the second to get:
    \dfrac{dx}{dy} = \dfrac{\sqrt{16-x^2}}{2}
    Is this the right idea and have I messed up along the way at all

    (Also thanks a bunch for pointing me in the right direction instead of just giving me the answer)
    It doesn't matter which way they are since the equalities will still be true. In this case, you want to express cos in terms of x which is what I've had it written as.

    You ended up with \dfrac{dx}{dy} = 8 \cos (2y+6) which is just 2\sqrt{16-x^2}, not half.
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    RDKGames;76216720]It doesn't matter which way they are since the equalities will still be true. In this case, you want to express cos in terms of x which is what I've had it written as.

    You ended up with \dfrac{dx}{dy} = 8 \cos (2y+6) which is just 2\sqrt{16-x^2}, not half.
    Okay so I started from the beginning again and got \dfrac{dy}{dx}=\dfrac{1}{2\sqrt{  16-x^2}}

    Is this right (finally?)
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    (Original post by Retsek)
    Okay so I started from the beginning again and got \dfrac{dy}{dx}=\dfrac{1}{2\sqrt{  16-x^2}}

    Is this right (finally?)
    Yes.
 
 
 
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