# 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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