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Math Help - Another differential equation help...

  1. #1
    Junior Member
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    Post Another differential equation help...

    Assume that the relation x+20y^{12}+cot(20y^{12})=1 defines y implicitly as a differentiable function of x. Find a simplified expression for dy/dx in terms of the tangent function.

    Any help will be much appreciated..thanks
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  2. #2
    Senior Member Peritus's Avatar
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    \begin{gathered}<br />
  \frac{d}<br />
{{dx}}\left( {x + 20y^{12}  + \cot \left( {20y^{12} } \right) = 1} \right) \hfill \\<br />
   \Leftrightarrow 1 + 240y^{11} y' - \frac{{240y^{11} y'}}<br />
{{\sin ^2 \left( {20y^{12} } \right)}} = 0 \hfill \\ <br />
\end{gathered}

    \begin{gathered}<br />
   \Leftrightarrow 1 + 240y^{11} y'\left( {1 - \frac{1}<br />
{{\sin ^2 \left( {20y^{12} } \right)}}} \right) = 0 \hfill \\<br />
   \Leftrightarrow 1 - 240y^{11} y'\cot \left( {20y^{12} } \right) = 0 \hfill \\ <br />
\end{gathered}

    <br />
 \Leftrightarrow y' = \frac{{\tan \left( {20y^{12} } \right)}}<br />
{{240y^{11} }}<br />
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  3. #3
    Junior Member toraj58's Avatar
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    Lightbulb

    firsr sort it like this:

    x+20y^{12}+cot(20y^{12})-1=0

    then differntiate using this approach:

    if f(x,y) = 0

    then

    \frac {dy}{dx} = -\frac {f^\prime_x}{f^\prime_y}

    that means when you differntiate in respect to x then y is assumed as constant and in reverse.

    then replace \cot with \frac {1}{\tan}
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