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

1 Answer
Feb 21, 2016

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)