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All right, sorta hard to explain but I will try my best. I am suppose to find the equation for the that lines that are tangent to the curve y=cot^2(x) at the point (pie/4, 1)....(the x is not to the power of 2, a variable right next to the cot^2)

I am not sure how to do this and if you can help, that would be awesome. It's the cot that's really tripping me up :/

  • calculus -

    <rant> AARGH! Who are these people that don't understand that π is PI, not PIE? </rant>

    y = cot^2(x)
    y' = 2cot(x)(-csc^2(x)) = -2cot(x)csc^2(x)

    at x=π/4, the slope is -2(1)(2) = -4

    the line through (π/4,1) with slope -4 is

    (y-1) = -4(x-π/4)

    massage that equation into the form you most like.

  • calculus -

    how did u get -2(1)(2)? did u just plug the PI/4 into the formula?

  • calculus -

    yeah, that's the way one computes the value of f(x) -- plug in the value of x.

  • calculus -

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