#\frac{d}{dx}((1+\sin (2x))(1+\cos ^2(2x))#
Applying product rule: #(f\cdot g)^'=f^'\cdot g+f\cdot g^'#
#f=1+\sin(2x),g=1+\cos ^2(2x)#
#=\frac{d}{dx}(1+\sin (2x))(1+\cos ^2(2x))+\frac{d}{dx}(1+\cos ^2\(2x))(1+\sin (2x))# ...... equation (i)
Here, #frac{d}{dx}(1+\sin (2x))#= #2\cos(2x)#
{ Applying sum/difference rule; #(f\pm g)^'=f^'\pm g^'#
#=\frac{d}{dx}(1)+\frac{d}{dx}(\sin (2x))#
#\frac{d}{dx}(1)=0# and
#\frac{d}{dx}(\sin (2x))# = #2cos (2x)# by applying chain rule
#\frac{df(u)}{dx}=\frac{df}{du}\cdot \frac{du}{dx}# }
Again,
#\frac{d}{dx}(1+\cos ^2(2x))# = #-4\cos (2x)\sin (2x)#
{ Applying sum/difference rule; #(f\pm g)^'=f^'\pm g^'#
#\frac{d}{dx}(1)+\frac{d}{dx}(\cos ^2(2x))#
#\frac{d}{dx}(1)=0# and #\frac{d}{dx}(\cos ^2(2x))=-4\cos (2x)\sin(2x)#}
Finally,Substituting in equation (i)
#=2\cos (2x)(1+\cos ^2(2x))+(-4\cos (2x)\sin (2x))(1+\sin (2x))#
Simplifying it,we get
#2\cos (2x)(\cos ^2(2x)-2\sin (2x)(\sin (2x)+1)+1)#
