Question

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

Answer 1

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`