Question

differentiation of 1/1+sin x​

Answer 1

Answer:

By the Sum Rule, the derivative of 1−sin(x) 1 - sin ( x ) with respect to x x is ddx[1]+ddx[−sin(x)] d d x [ 1 ] + d d x [ - sin ( x ) ] . Since 1 1 is constant with respect to x x , the derivative of 1 1 with respect to x x is 0 0 .

Explanation:

By the Sum Rule, the derivative of 1−sin(x) 1 - sin ( x ) with respect to x x is ddx[1]+ddx[−sin(x)] d d x [ 1 ] + d d x [ - sin ( x ) ] . Since 1 1 is constant with respect to x x , the derivative of 1 1 with respect to x x is 0 0 .

Answer 2

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

dy/dx = - 1/(1 + sinx)² . d/dx (1 + sinx)

dy/dx = - 1/ (1 + 2sinx + sin²x) . cosx

dy/dx = - cosx / (1 + 2sinx + sin²x) or write it as

f'(x) = - cos/(1 + sinx)²