Calculus

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arcsinxy = 2/3 arctan4x find dy/dx at the point (1/4, 2)

  • Calculus -

    using implicit differentiation,

    (xy'+y)/√(1+x^2y^2) = 8/3 * 1/(1+16x^2)

    a little rearranging of terms yields

    y' = 8√(1+x^2y^2) / (3x(1+16x^2)) - y/x

  • Calculus -

    oops that's 1-x^2y^2 throughout

  • Calculus -

    when I plug in the ordered pair (.25, 2)I get a zero in the denominator, was I doing that wrong?

  • Calculus -

    √(1-x^2y^2) = 0, all right, but it's not in the denominator.

    So, all you end up with is -y/x = -16

  • Calculus -

    oops: -2/.25 = -8

  • Calculus - final correction -

    Rats! I keep being inconsistent in the squaring! √(1-x^2y^2) is not zero.

    at (1/4,2)

    y' = 8√(1-1/4)/(3*1/4*2) - 2/.25
    = (8√6)/3 - 8

    judging from my other posts, you'd better double-check this one too!

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