What is the derivative of ?

#y=sinxcos^2 2x# ?

#y=sinxcos^2 2x# ?

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Feb 9, 2018

Answer:

#(dy)/(dx)=cos^3x-2sinxsin4x#

Explanation:

We use here product rule. Here #y=f(x)g(x)# and according to product rule #(dy)/(dx)=(df)/(dx)g(x)+f(x)(dg)/(dx)#

Here we have #f(x)=sinx# and #g(x)=cos^2 2x#

and therefore #(df)/(dx)=cosx# and #(dg)/(dx)=2cos2x xx (-sin2x)xx2=-4sin2xcos2x=-2sin4x#

Hence #(dy)/(dx)=cosx xx cos^2x+sinx xx(-2sin4x)#

= #cos^3x-2sinxsin4x#

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