How do you differentiate #cos(1-2x)^2#?

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Answer 1 · 2018-05-08T11:03:58

#dy/dx=4cos(1-2x)sin(1-2x)#

First, let #cos(1-2x)=u#

So, #y=u^2#

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

#(dy)/(du)=2u#

#(du)/(dx)=d/dx[cos(1-2x)]=d/dx[cos(v)]#

#(du)/(dx)=(du)/(dv)*(dv)/(dx)#

#dy/dx=(dy)/(du)* (du)/(dv) *(dv)/(dx)#

#(du)/(dv)=-sin(v)#
#(dv)/(dx)=-2#

#dy/dx=2u*-sin(v)*-2#

#dy/dx=4usin(v)#

#dy/dx=4cos(1-2x)sin(1-2x)#