If f(x) = cosx/(1+sinx) find f'(x).

f(x) = cosx / (1+ sinx)

To
differentiate we will will assume that:

u= cosx   
==>  we know that  du = -sinx

v= 1+ sinx  
==>    v' = cosc

Then, f(x) =
u/v

We know that , f'(x) = (u'v-
uv')/v^2

Let us subsitute with u and v
:

==> f'(x) = [-sinx (1+sinx) -
(cosx*cosx)]/(1+sinx)^2

Now let us expand the
brackets:

==> f'(x) = (-sinx - sin^2 x - cos^2
x)/(1+sinx)^2

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

But we know that : sin^2 x + cos^2 x =
1

==> f'(x) = (-sinx
-1)/(1+sinx)^2

==> f'(x) =
-(sinx+1)/(1+sinx)^2

==> f'(x) =
-1/(1+sinx)