How do you differentiate #2cos^2(x)#?

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Answer 1 · 2017-01-19T08:10:51

#(dy)/(dx) =-4cosxsinx#

To differentiate #y=2cos^2x# we need the chain rule

#(dy)/(dx)=(dy)/(du)xx(du)/(dx)#

let #u=cosx=>y=2u^2#

#(du)/(dx)=-sinx#

#(dy)/(du)=4u#

#:.(dy)/(dx)=4uxx(-sinx)#

substitute back for #u#

#(dy)/(dx) =-4cosxsinx#