Question

How do you differentiate # g(x) =(1+cosx)/(1-cosx) #?

Answer 1

By simplification

Explanation:

`d/dxg(x)=d/dxcot^2(x/2)`
`2cot(x/2)*cosec^2(x/2)*1/2`
`cot(x/2)*cosec^2(x/2)`

Answer 2

Use the quotient rule :

`(d(f(x)/(h(x))))/dx = (f'(x)h(x)-f(x)h'(x))/(h(x))^2`

Explanation:

Let `f(x) = 1 +cos(x)`, then `f'(x) = -sin(x)`
Let `h(x) = 1-cos(x)`, then `h'(x) = sin(x)`

Substituting into the quotient rule:

`(d((1 +cos(x))/(1-cos(x))))/dx = ((-sin(x))(1-cos(x))-(1 +cos(x))(sin(x)))/(1-cos(x))^2`

Answer 3

`g'(x)=-(2sinx)/(1-cosx)^2`

Explanation:

differentiate using the `color(blue)"quotient rule"`

`"Given " f(x)=(g(x))/(h(x))" then"`

`f'(x)=(h(x)g'(x)-g(x)h'(x))/(h(x))^2larr" quotient rule"`

`"here " g(x)=1+cosxrArrg'(x)=-sinx`

`h(x)=1-cosxrArrh'(x)=sinx`

`rArrf'(x)=((1-cosx)(-sinx)-(1+cosx)(sinx))/(1-cosx)^2`

`color(white)(rArrf'(x))=(-sinx+sinxcosx-sinx-sinxcosx)/(1-cosx)^2`

`color(white)(rArrf'(x))=-(2sinx)/(1-cosx)^2`