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Math Help - Higher Derivatives & Implicit Differentiation

  1. #1
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    Higher Derivatives & Implicit Differentiation

    Hi MHF,
    There are two questions that are confusing me. I understand the material, but when I was working them out, I would get stuck (with algebra parts) and I do not know how to continue.

    a) Find y'' by implicit differentiation.
    \sqrt{x}+\sqrt{y} = 1

    I found y'= -\sqrt{y} /\sqrt{x}
    (that's supposed to be a fraction)

    What is y''?

    b) If f (\theta) = cot (\theta), find f'''( \frac{\pi}{6})
    f'(x) = csc^2x

    What is f"? f"'?

    Thank you very much!
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  2. #2
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    Quote Originally Posted by whiteslash View Post
    Hi MHF,
    There are two questions that are confusing me. I understand the material, but when I was working them out, I would get stuck (with algebra parts) and I do not know how to continue.

    a) Find y'' by implicit differentiation.
    \sqrt{x}+\sqrt{y} = 1

    I found y'= -\sqrt{y} /\sqrt{x}
    (that's supposed to be a fraction)

    What is y''?

    b) If f (\theta) = cot (\theta), find f'''( \frac{\pi}{6})
    f'(x) = csc^2x

    What is f"? f"'?

    Thank you very much!
    y' = -\frac{\sqrt{y}}{\sqrt{x}}<br />

    y'' = -\frac{\sqrt{x} \cdot \frac{y'}{2\sqrt{y}} - \sqrt{y} \cdot \frac{1}{2\sqrt{x}}}{x}<br />

    y'' = \frac{-\frac{1}{2} - \frac{\sqrt{y}}{2\sqrt{x}}}{x}<br />

    y'' = -\frac{2\sqrt{x}}{2\sqrt{x}} \cdot \frac{-\frac{1}{2} - \frac{\sqrt{y}}{2\sqrt{x}}}{x}<br />

    y'' = \frac{\sqrt{x} + \sqrt{y}}{2x\sqrt{x}}

    y'' = \frac{1}{2x\sqrt{x}}


    f(t) = \cot{t}<br />

    f'(t) = -\csc^2{t}

    f''(t) = -2\csc{t} \cdot (-\csc{t}\cot{t})

    f''{t} = 2\csc^2{t}\cot{t}

    f''{t} = 2(\cot^2{t} + 1)\cot{t}

    f''(t) = 2(\cot{t})^3+2\cot{t}

    f'''(t) = 6(\cot{t})^2(-\csc^2{t}) - 2\csc^2{t}

    f'''(t) = -2\csc^2{t}(3\cot^2{t} + 1)

    now evaluate for t = \frac{5\pi}{6}
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  3. #3
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    Here's a slightly different way to do the first problem. Since x^{1/2}+ y^{1/2}= 1, (1/2)x^{-1/2}+ (1/2)y^{-1/2}y'= 0 (from which you can get y'= -x^{-1/2}y^{1/2}= -\frac{\sqrt{y}{\sqrt{x}})

    Differentiating x^{-1/2}+ y^{-1/2}y'= 0 with respect to x, -(1/2)x^{-3/2}- (1/2)y^{-3/2}y'+ y^{-1/2}y"= 0 so that y^{1/2}y"= (1/2)(x^{-3/2}- y^{-3/2}y'), y"= (1/2)(x^{-3/2}y^{-1/2}- y^{-5/2}y'= (1/2)(x^{-3/2}y^{-1/2}+x^{-1/2}y^{-2}
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