Question

How do you determine the derivative of `y = ((cosx)/(1 -sinx))^2`?

Answer 1

`y' = (2cosx)/(1 -sinx)^2`

Explanation:

We have:

`y= ((cosx)/(1 - sinx))^2`

`y = (cos^2x)/(1 - 2sinx + sin^2x)`

`y = (1 - sin^2x)/(1 - sinx)^2`

`y = ((1 + sinx)(1 - sinx))/((1 - sinx)(1 - sinx))`

`y = (1 +sinx)/(1 -sinx)`

By the quotient rule:

`y'= (cosx(1 -sinx) - (-cosx(1 +sinx)))/(1 -sinx)^2`

`y' = (cosx- cosxsinx + cosx + cosxsinx)/(1 - sinx)^2`

`y' = (2cosx)/(1 -sinx)^2`

Hopefully this helps!