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

Textbook Question

Chapter 3, 3.4 - Problem 48 - Calculus: Early Transcendentals (7th Edition, James Stewart).
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gsarora17's profile pic

gsarora17 | (Level 2) Associate Educator

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

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


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tiffanynelson | In Training Educator

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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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vineetchaurasia | (Level 1) eNoter

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Please refer to image with thia answer for solution.

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