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Math Help - Implict differentiation: 2cosysinx=1

  1. #1
    Member Furyan's Avatar
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    Implict differentiation: 2cosysinx=1

    Hello,

    The question is to find the equation of the normal to the curve 2\cos y\sin x = 1 at (\tfrac{\pi}{4},\tfrac{\pi}{4}).

    I'm not confident with implicit differentiation, but using the product rule I got:

    2\sin x\dfrac{dy}{dx}(-\sin y) + 2\cos y\cos x = 0

    \dfrac{dy}{dx}(2\sin x\sin y) - 2\cos y\cos x = 0

    \dfrac{dy}{dx} = \dfrac{2\cos y\cos x}{2\sin x\sin y}

    \dfrac{dy}{dx} = \cot x\cot y

    My question is, is it correct to write the derivative of \cos y with respect to x as \dfrac{dy}{dx}(-\sin y) as I have done.

    Thank you.
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  2. #2
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    Re: Implict differentiation: 2cosysinx=1

    Hey Furyan.

    Your working looks good for the implicit differentiation.
    Thanks from Furyan
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  3. #3
    Member Furyan's Avatar
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    Re: Implict differentiation: 2cosysinx=1

    Hey chiro,

    That's a relief, thank you very much for looking it over
    Last edited by Furyan; January 26th 2013 at 07:13 PM.
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