Let f (x) =

cosx

1+sinx

... (i)
Differentiating (i) with respect to x, we get

d

dx

[f (x)] =

(1+sinx)(cosx)′−cosx(1+sinx)′

(1+sinx)2

Since

d

dx

[f (x)] =

(1+sinx)(−sinx)′−cosx(1+sinx)′

(0+cosx)2

=

−sinx−sin2x−cos2x

(1+sinx)2

=

−sinx−−(sin2+cos2x)

(1+sinx)2

∴

d

dx

[

cosx

1+sinx

] =

−1

(1+sinx)

.