Question

How do i find the answer using the quotient rule for this question?

`f(x)=(x)/(1-x)^2`

Answer 1

`d/dx(u/v)=(1+x)/(1-x)^3`

Explanation:

Given:
`f(x)=(x)/((1-x)^2`

Here,
Let `u=x`
and
`v=(1-x)^2`
By quotient rule, we have
`d/dx(u/v)=(v(du)/dx-u(dv)/dx)/v^2`

Then
`(du)/dx=1`
`(dv)/dx=2(1-x)(0-1)=-2(1-x)`
Thus,
`d/dx((x)/((1-x)^2))=((1-x)^2(1)-x(-2(1-x)))/(((1-x)^2)^2`
`=((1-x)^2+2x(1-x))/(1-x)^4`
`=(1-2x+x^2+2x-2x^2)/(1-x)^4`
`=(-x^2+1)/(1-x)^4`
`(1-x^2)/(1-x)^4`
`((1-x)(1+x))/(1-x)^4`
`=(1+x)/(1-x)^3`

Thus,

`d/dx(u/v)=(1+x)/(1-x)^3`

Answer 2

`f'(x)=(1+x)/(1-x)^3`

Explanation:

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

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

`g(x)=xrArrg'(x)=1`

`h(x)=(1-x)^2rArrh'(x)=-2(1-x)larrcolor(blue)"chain rule"`

`rArrf'(x)=((1-x)^2+2x(1-x))/(1-x)^4`

`color(white)(rArrf'(x))=(1-2x+x^2+2x-2x^2)/(1-x)^4`

`color(white)(rArrf'(x))=(1-x^2)/(1-x)^4`

`color(white)(rArrf'(x))=((1-x)(1+x))/(1-x)^4`

`color(white)(rArrf'(x))=(1+x)/(1-x)^3`