\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)