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Math Help - Differentiation Stuff

  1. #1
    Newbie Kiwi's Avatar
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    Red face Differentiation Stuff

    x=y^2-3ln(2y)

    a) show that dy/dx = y/(2y^2-3)

    b) find equation for tangent to curve where y=1/2
    give answer in form ax+by+c=0 where a, b, c are all integers

    Have done a) cannot do b) can someone help me please?

    Should have said, can do b), but my answer is never an integer, can someone see wot i am doing wrong. i get x=1/4 and dy/dx = -1/5 are these wrong?
    Last edited by Kiwi; March 1st 2008 at 01:19 PM. Reason: Rephrasing
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  2. #2
    Flow Master
    mr fantastic's Avatar
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    Quote Originally Posted by Kiwi View Post
    x=y^2-3ln(2y)

    a) show that dy/dx = y/(2y^2-3)

    b) find equation for tangent to curve where y=1/2
    give answer in form ax+by+c=0 where a, b, c are all integers

    Have done a) cannot do b) can someone help me please?
    To get a point on the line:

    Substitute y = 1/2 into \frac{dy}{dx} = \frac{y}{2y^2 - 3} to get the gradient of the tangent.

    Substitute y = 1/2 into x = y^2 - 3 \ln (2y) and solve for x. You now have the coordinates of a point on the tangent.

    So now you have a gradient and a point. Can you use these two things to get the equation of the tangent line?
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