calculus

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*find the slope of y= 4/x +6
at the point (8,2)
*evaluate the derivative at the point (pi,4)
y=[tan(5x^8)]^3

please help

  • calculus -

    just tel me how to setup

  • calculus -

    I will assume your equation is as you typed it, and not
    y = 4/(x+6)

    y = 4/x + 6
    = 4x^-1 + 6
    dy/dx = -4x^-2
    = -4/x^2

    at (8,2) , dy/dx = -4/64 = -1/16

    equation:
    y-2 = (-1/16)(x - 8)
    16y - 32 = -x + 8
    x + 16y = 40

    y = [tan(5x^8)]^3
    dy/dx = 3[tan(5x^8)]^2 (sec(5x^8) (40x^7)
    at (π/4)
    use your calculator to find the messy answer for dy/dx

  • calculus -

    -4/64
    why 64 not 8

  • calculus -

    notice dy/dx = -4/(x^2)

    = -4/(8^2) = -4/64 or -1/16

  • calculus -

    just plug pi/4
    right

  • calculus -

    thnks Reiny.

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