Question

How do you derive `y = (1 + sin(2x)]/[1 - sin(2x))` using the quotient rule?

Answer 1

`f'(x) = (4cos(2x))/(1-sin(2x))^2 = dy/dx`.

Explanation:

By the quotient rule, the derivative of `(u(x))/(v(x))` is `(u'(x)v(x) - u(x)v'(x))/(v(x)^2)`. Let's say `f(x) = (1+sin(2x))/(1-sin(2x))`.

Here, `u(x) = 1 + sin(2x)` and `v(x) = 1 - sin(2x)`. The derivative of `sin(2x)` is `2cos(2x)` by the chain rule, so `u'(x) = 2cos(2x)` and `v'(x) = -u'(x)`.

So, by applying the quotient rule :
`f'(x) = (2cos(2x)(1-sin(2x)) - (1+sin(2x))(-2cos(2x)))/(1-sin(2x))^2`

`f'(x) = (2cos(2x)(1-sin(2x) + 1 + sin(2x)))/(1-sin(2x))^2 = (4cos(2x))/(1-sin(2x))^2`