Question

what is the differentiation of ( 1-cos/1+cosx)

Answer 1

Answer:

y′=−2sinx(1−cosx)2

Explanation:

You can differentiate this function by using the quotient rule and the derivative of cosx, which is

ddx(cosx)=−sinx

For a function that can be written as

y=f(x)g(x), where g(x)≠0

the quotient rule allows you to find its derivative by using the formula

ddx(y)=[ddx(f(x))]⋅g(x)−f(x)⋅ddx(g(x))(g(x))2

In your case,

f(x)=1+cosx  and  g(x)=1−cosx

This means that you can write

ddx(y)=[ddx(1+cosx)]⋅(1−cosx)−(1+cosx)⋅ddx(1−cosx)(1−cosx)2

y′=−sinx⋅(1−cosx)−(1+cosx)⋅(−(−sinx))(1−cosx)2

y′=−sinx+sinx⋅cosx−sinx−sinx⋅cosx(1−cosx)2

y′=−2sinx(1−cosx)2


I hope it's help U :-))

Answer 2

di difference is not same