How do you find the derivative of #(cos x)^2 - cos x#?

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Answer 1 · 2016-01-24T01:22:11

#sinx-sin2x#

To differentiate the squared cosine function, use the chain rule, in that #d/dxu^2=2u*u'#, where #u=cosx#. This gives us a derivative of

#2cosxd/dxcosx-d/dxcosx#

Which simplifies to be

#2cosx(-sinx)-(-sinx)#

or

#mathbf(-2cosxsinx+sinx#

Note that this can be factored as

#sinx(1-2cosx)#

or recognize that #2cosxsinx=sin2x#, so the derivative equals

#sinx-sin2x#