Question

y=1-sinx/1+sinx in differentiate​

Answer 1

Step-by-step explanation:

y = (1-sinx)/(1+sinx)

division rule :-

d(v/u)/dx = (udv/dx - vdu/dx)/u^2

d(1-sinx/1+sinx)/dx = [(1+sinx){d(1-sinx)/dx} - (1-sinx){d(1+sinx)/dx}] / (1+sinx)^2

d(1-sinx/1+sinx)/dx = {-cosx(1+sinx)-cosx(1-sinx)}/(1+sinx)^2

d(1-sinx/1+sinx)/dx = -2cosx / (1+sinx)^2

dy/dx = -2cosx / (1+sinx)^2