calculus

find the equation of the tangent line to y=4^((x^2)-6x+3) at x=4

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  1. when x=4
    y = 4^(16-24+3) = 4^-5 = 1/1024
    so the point is (4,1/1024)

    dy/dx = (2x - 6)(ln4)(4^(x^2 - 6x + 3)
    when x=4, dy/dx = 2ln4(1/1024) = ln4/512

    so y - 1/1024 = (ln4/512)(x-4)

    512y - 1/2 = (ln4)x - 4ln4

    (ln4)x - 512y = 4ln4 - 1/2

    I'll leave it up to you to convert to y = mx+b form if it is needed.

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