Question

How do you differentiate `f(x)= (1 + sin^2x)/(1 - sin2x)` using the quotient rule?

Answer 1

`(dy)/(dx)=(2cosxsinx)/(1-sin2x)+(2cos2x(1+sin^2x))/(1-sin2x)^2`

Explanation:

We must recall two things to make this simpler. First, `f (x)=g (x)/(h (x)) -> f'(x) = (hg'-h'g)/h^2`.

Here, `g(x)=1+sin^2x=1+(sinx)^2`

`g'(x)=d/(dx)[1]+d/(dx)[(sinx)^2]`
`g'(x)=2cosxsinx`

`h(x)=1-sin(2x)`
`h'(x)=d/(dx)[1]-d/(dx)[sin2x]`
`h'(x)=-2cos2x`

`(dy)/(dx)=(2cosxsinx(1-sin2x)+2cos2x(1+sin^2x))/(1-sin2x)^2`

`(dy)/(dx)=(2cosxsinx)/(1-sin2x)+(2cos2x(1+sin^2x))/(1-sin2x)^2`