What is the derivative of # f(x)=cosx/(1-cosx)#?

Question

6834 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2018-04-01T18:25:04

#f'(x)=-sinx/(1-cosx)^2#

#f'(x)=((1-cosx)d/dxcosx-cosxd/dx(1-cosx))/(1-cosx)^2#

#d/dxcosx=-sinx#

#d/dx(1-cosx)=-(-sinx)=sinx#, so

#f'(x)=((1-cosx)(-sinx)-cosx(sinx))/(1-cosx)^2#

Simplify:

#f'(x)=(cosxsinx-sinx-cosxsinx)/(1-cosx)^2#

#f'(x)=-sinx/(1-cosx)^2#

Answer 2 · 2018-04-01T19:12:59

# -1/2*csc^2(x/2)cot(x/2)#.

#f(x)=cosx/(1-cosx)={1-2sin^2(x/2)}/(2sin^2(x/2))#,

#=1/(2sin^2(x/2))-(2sin^2(x/2))/(2sin^2(x/2))#.

#rArr f(x)=1/2csc^2(x/2)-1#.

#:. f'(x)=1/2*2csc(x/2)*d/dx{csc(x/2)}-0........."[The Chain Rule]"#,

#=csc(x/2)*{-csc(x/2)cot(x/2)}*d/dx(x/2)#.

# rArr f'(x)=-1/2*csc^2(x/2)cot(x/2)#.