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What is the derivative of #(-sinx^2)/2#?

1 Answer

May 29, 2017

#d/dx( -sinx^2/2) = -x cosx^2#

Explanation:

Using the chain rule with #u= x^2#:

#d/dx( -sinx^2/2) = -1/2 d/(du) (sinu) xx (du)/dx#

#d/dx( -sinx^2/2) = -1/2 cosx^2 xx (dx^2)/dx#

#d/dx( -sinx^2/2) = -1/2 cosx^2 xx 2x#

#d/dx( -sinx^2/2) = -x cosx^2#

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