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Math Help - Calculus - Implicit differentiation

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    Calculus - Implicit differentiation

    consider: e^(xy)-ycosh(4x)=-sinx

    use implicit differentiation to find dy/dx and then determine the equation of the tangent to the curve at the point (0,1)

    i did the implicit differentiation but it turned out to b really messy plz help! thx
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    Quote Originally Posted by ssdimensionss View Post
    consider: e^(xy)-ycosh(4x)=-sinx

    use implicit differentiation to find dy/dx and then determine the equation of the tangent to the curve at the point (0,1)

    i did the implicit differentiation but it turned out to b really messy plz help! thx
    e^{xy} - y~cosh(4x) = -sinx

    ye^{xy} + e^{xy} \cdot xy^{\prime} - y^{\prime}~cosh(4x) - 4y~sinh(4x) = -cos(x)

    (xe^{xy} - cosh(4x)) y^{\prime} = 4y~sinh(4x) - cos(x) - ye^{xy}

    y^{\prime} = \frac{4y~sinh(4x) - cos(x) - ye^{xy}}{xe^{xy} - cosh(4x)}

    At the point (0, 1) we get
    y^{\prime} = \frac{4sinh(0) - cos(0) - e^{0}}{-cosh(0)} = 2

    So what is the line with a slope of 2 that passes through the point (0, 1)?

    -Dan
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