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Math Help - [SOLVED] How to get this 2nd differential as a function of x?

  1. #1
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    [SOLVED] How to get this 2nd differential as a function of x?

    The question is: If \sin y + \cos y = x find \frac{d^2 y}{dx^2} as a function of x.

    I differentiated once to get \frac{dy}{dx}=\frac{1}{\cos y - \sin y}

    Then again to get \frac{d^2y}{dx^2}=\frac{dy}{dx}\frac{\sin y + \cos y}{(\cos y - \sin y)^2} \Rightarrow \frac{d^2y}{dx^2} =\frac{x}{(\cos y - \sin y)^3}

    The book's answer is \frac{d^2y}{dx^2}=\pm\frac{x}{\sqrt{(2-x^2)^3}}. So I guess that means  \cos y -\sin y =\pm\sqrt{(2-x^2)}. But I duno why.

    Did I make a mistake? Can anyone help? Thanks.
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  2. #2
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    Quote Originally Posted by tleave2000 View Post
    The question is: If \sin y + \cos y = x find \frac{d^2 y}{dx^2} as a function of x.

    I differentiated once to get \frac{dy}{dx}=\frac{1}{\cos y - \sin y}

    Then again to get \frac{d^2y}{dx^2}=\frac{dy}{dx}\frac{\sin y + \cos y}{(\cos y - \sin y)^2} \Rightarrow \frac{d^2y}{dx^2} =\frac{x}{(\cos y - \sin y)^3}

    The book's answer is \frac{d^2y}{dx^2}=\pm\frac{x}{\sqrt{(2-x^2)^3}}. So I guess that means  \cos y -\sin y =\pm\sqrt{(2-x^2)}. But I duno why.

    Did I make a mistake? Can anyone help? Thanks.
    Note that x = \cos y + \sin y \Rightarrow x^2 = 1 + 2 \cos y \sin y \Rightarrow x^2 - 1 = 2 \cos y \sin y.

    Therefore (\cos y - \sin y )^2 = 1 - 2 \cos y \sin y = 1 - (x^2 - 1) = 2 - x^2.
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  3. #3
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    Wow, that's pretty difficult to spot...
    Thanks alot.
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