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Math Help - implicit differentiation

  1. #1
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    implicit differentiation

    x^-1 +y^-1 = pi^-1

    How do I differentiate this? dy/dx should be -(y^2)/x^2.
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  2. #2
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    Re: implicit differentiation

    Quote Originally Posted by Stuck Man View Post
    x^-1 +y^-1 = pi^-1
    How do I differentiate this? should be \frac{dy}{dx}=\frac{-y^2}{x^2}.
    YES
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  3. #3
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    Re: implicit differentiation

    That is the answer in the book. I have not obtained it.
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  4. #4
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    Re: implicit differentiation

    Quote Originally Posted by Stuck Man View Post
    That is the answer in the book. I have not obtained it.
    Start with x^{-1}+y^{-1}=\pi^{-1}
    STEP I: -x^{-2}-y^{-2}y^/=0

    STEP II: Solve for y^/
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  5. #5
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    Re: implicit differentiation

    Thanks. Is the second derivative (2y^3 +2xy^2)/x^4?
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  6. #6
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    Re: implicit differentiation

    Hello, Stuck Man!

    \text{Is the second derivative: }\:y'' \:=\:\frac{2y^3 +2xy^2}{x^4}\:?

    Yes . . . but it can be further simplified.

    We have: . y'' \;=\;\frac{2y^2(y + x)}{x^4} .[1]


    From the original equation: . \frac{1}{x} + \frac{1}{y} \:=\:\frac{1}{\pi}

    . . we have: . \frac{y+x}{xy} \:=\:\frac{1}{\pi} \quad\Rightarrow\quad y+x \:=\:\frac{xy}{\pi}

    Substitute into [1]: . y'' \;=\;\frac{2y^2\left(\frac{xy}{\pi}\right)}{x^4} \;=\;\frac{2y^3}{\pi x^3}


    This is something to watch for . . .
    Quite often, the second derivative contains part (or all) of the original equation.
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