We can use product formula according to which, if #f(x)=g(x)h(x)#
then #(df)/(dx)=(dg)/(dx)xxh(x)+(dh)/(dx)xxg(x)#
We also use Chain Rule according to whicg a function of a function, say #y, =f(g(x))#, where we have to find #(dy)/(dx)#, we need to do (a) substitute #u=g(x)#, which gives us #y=f(u)#. Then we need to use a formula called Chain Rule, which states that #(dy)/(dx)=(dy)/(du)xx(du)/(dx)#
Now, as such as #g(x)=(1+sin(2-x))(1+cos^2(x))=u(x)v(x)#,
where #u(x)=1+sin(2-x)# and #v(x)=1+cos^2x# and
#(dg)/(dx)=(du)/(dx)xxv(x)+(dv)/(dx)xxu(x)#
Now= #(du)/(dx)=cos(2-x)xx(-1)=-cos(2-x)#, using chain rule as #d.(dx)(2-x)=-1# and
#(dv)/(dx)=0+2cosx xx (-sinx)=-2sinxcosx#
Hence #(dg)/(dx)=-cos(2-x)xx(1+cos^2x)+(-2sinxcosx)xx(1+sin(2-x))#
= #-cos(2-x)-cos^2xcos(2-x)-2sinxcosx--2sinxcosxsin(2-x)#
= #-cos(2-x)-cos^2xcos(2-x)-sin2x--sin2xsin(2-x)#
