Question

find the derivative of f(x)=cosx/1+sinx w.r.t 'x'

Answer 1

Hey

Hope it helps ^_^

Answer 2

Step-by-step explanation:

given

f(x)=cosx/1+sinx

we know that

d/dx (a/b)=b× d/dx(b)-a×d/dx(b)/b^2

so,

{cosX/1+sinX=1+sinX. d/dx(cosX)- cosx. d/dx(1+sinX)}/(1+sinX)^2

d/dX(cosx)= -sinX

d/dx(1+sinX)=cosX

By solving

1+sinX (-sinX)-cosX(cosX)/(1+sinX)^2

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

-sinX-(sin^2X+cos^2X)/(1+sinX)^2

(-sinX-1)/(1+sinx) (1+sinX)

-1(1+sinX)/(1+sinX) (1+sinX)

= -1/(1+sinX)