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Math Help - implicit differentiation

  1. #1
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    implicit differentiation

    xy^2+yx^2=1

    i am not good at solving these problems.
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  2. #2
    Senior Member Peritus's Avatar
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    <br />
\begin{gathered}<br />
  xy^2  + yx^2  = 1 \hfill \\<br />
  \frac{d}<br />
{{dx}}\left( {xy^2  + yx^2  = 1} \right) \hfill \\<br />
   \Rightarrow y^2  + 2xyy' + 2xy + x^2 y' = 0 \hfill \\ <br />
\end{gathered}

    all is left to do is isolate y', I hope you're good at this.
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  3. #3
    MHF Contributor
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    Quote Originally Posted by Handro View Post
    xy^2+yx^2=1

    i am not good at solving these problems.
    x(y^2) +y(x^2) = 1

    If it is with respect to x, or if you want dy/dx or y', then,
    very basically,
    [x(2y dy/dx) +(y^2)(1 dx/dx)] +[y(2x dx/dx) +(x^2)(1 dy/dx)] = 0

    You see, "divide everything" by dx if it is with rexpect to x.

    To continue,
    [(2xy dy/dx ) +(y^2)] +[(2xy) +(x^2 dy/dx)] = 0

    It does not have to be as slow as that. It's just mny way of explaining the basics or the details to you.

    (2xy)y' +y^2 +2xy +(x^2)y' = 0
    Collect the (y')s,
    (2xy +x^2)y' = -2xy -y^2
    x(2y +x)y' = -y(2x +y)
    y' = -y(2x +y) / x(x +2y) -----------answer.
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