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Thread: Another differential equation help...

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

    Assume that the relation $\displaystyle x+20y^{12}+cot(20y^{12})=1$ defines y implicitly as a differentiable function of x. Find a simplified expression for $\displaystyle dy/dx$ 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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    $\displaystyle \begin{gathered}
    \frac{d}
    {{dx}}\left( {x + 20y^{12} + \cot \left( {20y^{12} } \right) = 1} \right) \hfill \\
    \Leftrightarrow 1 + 240y^{11} y' - \frac{{240y^{11} y'}}
    {{\sin ^2 \left( {20y^{12} } \right)}} = 0 \hfill \\
    \end{gathered} $

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

    $\displaystyle
    \Leftrightarrow y' = \frac{{\tan \left( {20y^{12} } \right)}}
    {{240y^{11} }}
    $
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  3. #3
    Junior Member toraj58's Avatar
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    firsr sort it like this:

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

    then differntiate using this approach:

    if $\displaystyle f(x,y) = 0$

    then

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

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

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