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

1 Answer
Feb 21, 2016

Answer:

#2\cos (2x)(\cos ^2(2x)-2\sin (2x)(\sin (2x)+1)+1)#

Explanation:

#\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)#