How do you differentiate g(x) = [1 + sin(2x)]/[1 - sin(2x)] using the product rule?

1 Answer
Jan 9, 2016

In order to forcedly use the product rule, we'll rewrite it as g(x)=(1+sin(2x))(1-sin(2x))^-1

Explanation:

Also, we'll need chain rule to differentiate the second term.

  • Chain rule: (dy)/(dx)=(dy)/(du)(du)/(dv)(dv)/(dx)

  • For the first term, we'll rename u=2x

  • For the second term, we'll rename v=(1-sin(w)) and w=2x (but the logic follows the same way that of u, v, and so on)

  • Product rule: (ab)'=a'b+ab'

(dg(x))/(dx)=2cos(2x)(1-sin(2x))^-1+(1+sin(2x))-v^-2(-2cos(2x))

Substituting v and rearranging negative exponents:

(dg(x))/(dx)=(2cos(2x))/(1-sin(2x))+(2cos(2x)(1+sin(2x)))/(1-sin(2x))^2

(dg(x))/(dx)=(2cos(2x)(1-sin(2x))+2cos(2x)(1+sin(2x)))/(1-sin(2x))^2

(dg(x))/(dx)=(2cos(2x)(1cancel(-sin(2x))+1+cancel(sin(2x))))/(1-sin(2x))^2

(dg(x))/(dx)=(4cos(2x))/(1-sin(2x))^2