Calculus

posted by Luke

find d^2y/dx^2 (second derivative) for y=cos^2 4x

  1. bobpursley

    y=cos^2 4x
    y'=2cos 4x * -sin 4x*4= -8cos4x*sin4x
    = -4sin(8x)
    y"= -4cos(8x)(8)=-32 cos(8x)
    y"'=+32*8 sin(8x)
    y""=32*64*cos(8x)

    check my work

  2. Luke

    how did you get -4sin8x from -8cos4x*sin4x?

    The answer is -32(cos^2 4x-sin^2 4x) but not sure how to get there

  3. What Zit Tooya

    y'=-8sin(4x)cos(4x)
    y"=-8[sin(4x)(-sin(4x))*4+cos(4x)(cos(4x))*4]
    y"=-8[-4sin^2(4x)+4cos^2(4x)]
    y"=-32[cos^2(4x)-sin^2(4x)]

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