Differentiate the function

Question

y=cosx/1-sinx

Answer 1

Given that y= cosx/ (1 + sinx)

Defferentiating on both sides we get,

dy/dx =d/dx (cosx/ (1 + sinx))

This is in the form d/dx(u/v) = {(v * du/dx) - (u * dv/dx)}/v^2

=> dy/dx = [(1+ sinx) d/dx(cosx) - cosx d/dx(1 + sinx)]/ (1+ sinx)^2

=> dy/dx = [(1+ sinx) (-sinx) - cosx (0 + cosx)]/ (1+ sinx)^2

=> dy/dx= [-sinx -sin^2x - cos^2x]/ (1+ sinx)^2

=> dy/dx = [ -sinx -(sin^2x + cos^2x)]/(1+ sinx)^2

=> dy/dx = [-sinx -1]/(1+ sinx)^2                                      [ Since sin^2x + cos^2x = 1]

=> dy/dx= -(sinx+1)/(1+ sinx)^2  

=> dy/dx= -1/(1+ sinx)

Answer 2

solution to the problem is