How do you differentiate #g(x) = (1 + sin(2-x))(1 +cos^2(x)) # using the product rule?

1 Answer
Feb 12, 2017

Answer:

#(dg)/(dx)=-cos(2-x)-cos^2xcos(2-x)-sin2x--sin2xsin(2-x)#

Explanation:

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)#