# `y = cos^2(x)` Find y' and y''

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### 3 Answers

`y=cos^2x`

`y'=(2cosx)(-sinx)`

`y'=-2sinxcosx`

`y" = -2((sinx)(-sinx) +cosxcosx)`

` `y"=`-2(-sin^2x +cos^2x)`

y"=`2sin^2x-2cos^2x`

There is more than one way to do this problem. You can use the chain rule, or I have chosen to use the product rule.

The first derivative can be found using the product rule by noting that cos²(x)=cos(x)cos(x).

`d/dx` cos(x)cos(x)=( `d/dx` cos(x))cos(x) + cos(x)(`d/dx` cos(x)) = 2cos(x)(`d/dx` cos(x))

`d/dx` cos(x) = -sin(x)

`d/dx` cos²(x)=-2cos(x)sin(x) which can also be denoted by y'=-2cos(x)sin(x).

To find y'', we need to take the derivative of y'.

`dy/dx` -2cos(x)sin(x)=-2cos(x)(`dy/dx` sin(x)) + (`dy/dx` (-2cos(x)))sin(x)

`dy/dx` sin(x) = cos(x), so -2cos(x)(`dy/dx` (sin(x)) = -2cos(x)cos(x) = -2cos²(x)

`dy/dx` (-2cos(x)) = -2(`dy/dx` cos(x)) = -2(-sin(x)) = 2sin(x), so (`dy/dx` (-2cos(x)))sin(x) = 2sin(x)sin(x) = 2sin²(x)

When we add the two together we get y''= -2cos²(x) + 2sin²(x) = 2(sin²(x)-cos²(x))

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Please refer to image with thia answer for solution.