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Math Help - find the slope of the curve

  1. #1
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    find the slope of the curve

    The equation sin xy = y defines y implicitly as a function of x.
    Find the slope y'(pi/3,1/2) at the point x=pi/3, y=1/2.

    i know that y' = (ycos(xy))/(1-xcos(xy))

    but i just dont know how to solve the problem when i plug in x and and y to find the slope

    please help

    ty
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  2. #2
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    Quote Originally Posted by vtong View Post
    The equation sin xy = y defines y implicitly as a function of x.
    Find the slope y'(pi/3,1/2) at the point x=pi/3, y=1/2.

    i know that y' = (ycos(xy))/(1-xcos(xy))

    but i just dont know how to solve the problem when i plug in x and and y to find the slope

    please help

    ty
    y=sin(xy) so y'=\frac{y\cos{xy}}{1-x\cos{xy}} and plug in to get y'=\frac{\frac{1}{2}\cos{\frac{\pi}{6}}}{1-\frac{\pi}{3}\cos{\frac{\pi}{6}}} which is the slope of your function
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  3. #3
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    Quote Originally Posted by vtong View Post
    The equation sin xy = y defines y implicitly as a function of x.
    Find the slope y'(pi/3,1/2) at the point x=pi/3, y=1/2.

    i know that y' = (ycos(xy))/(1-xcos(xy))

    but i just dont know how to solve the problem when i plug in x and and y to find the slope

    please help

    ty
    The derivative of a function evaluated at a given point is the slope of the line tangent to the graph of the function at that point.

    So, You've got a derivative, and you've got a point. Evaluate the derivative at that point and that will give the slope.

    i.e. Plug in \frac{\pi}{3} wherever you see an x, and \frac{1}{2} whereever you see a y.

    Note* \cos\frac{\pi}{6}=\frac{\sqrt{3}}{2}.
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