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

Question

7207 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

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

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

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

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

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

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

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

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