Question

What is the derivative of `f(x)=cosx/(1+sin^2x)`?

Answer 1

`(dy/dx) =- (sin{x})/(1+sin^2{x}) - (2cos^2{x}sin{x})/(1+sin^2{x})^2`

Explanation:

Standard form reference: if `y=u/v` then `(dy)/dx =(v((du)/dx)-u((dv)/dx))/v^2`

Considering section at a time

Let `u=cos(x) -> (du)/dx= -sin(x)`
Let `v=1 +sin^2(x) -> (dv)/dx = 2sin(x)cos(x)`

Thus `" " (dy)/dx= (( 1+sin^2{x})(-sin{x}) - (cos{x})( 2sin{x}cos{x} ))/((1+sin^2{x})^2`

from which you derive

`(dy/dx) =- (sin{x})/(1+sin^2{x}) - (2cos^2{x}sin{x})/(1+sin^2{x})^2`