If `f(x) = (x^(2)-x)/(x^(2)+2x)`, then `d/dx [f^(-1)(x)] =`

Answer

A, B, C, D

Explanation

`f(x)=(x^(2)-x)/(x^(2)+2x)" ... (i)"`
`f(x)=(x(x-1))/(x(x+2))`
`f(x)=((x-1))/((x+2)),xne0" ... (ii)"`
`D_(f)={x:x^(2)+2xne0}` [from Eq. (i)]
`={x:x inR-{0, -2}}`
Now, let `y=(x-1)/(x+2)`
`implies yx+2y=x-1impliesx(y-1)=-(1+2y)`
implies `x=(1+2y)/(1-y)`
Now, for y = 1, x is not defined.
Now, `x = 0, f(x)=-(1)/(2)`
`therefore R_(f)=R-{1,-(1)/(2)}`
Now, let `x_(1),x_(2)inD_(f)`
Then, `f(x_(1))=f(x_(2))implies(x_(1)-1)/(x_(1)+2)=(x_(2)-1)/(x_(2)+2)` `impliesx_(1)x_(2)+2x_(1)-x_(2)-2=x_(1)x_(2)-x_(1)+2x_(2)-2`
`implies x_(1)=x_(2)`
`therefore` f is one-one function.
Now, let `y=(x-1)/(x+2)`
Then, `x=(1+2y)/(1-y)`
`implies f^(-1)(y)=(1+2y)/(1-y)" "[because f(x)=yimpliesx=f^(-1)(y)]`
Replace y be x, we get `f^(-1)(x)=(1+2x)/(1-x)`
`implies (d)/(dx){f^(-1)(x)}=((1-x)2-(1+2x)(-1))/((1-x)^(2))`
`=(2-2x+1+2x)/((1-x)^(2))`
`implies (d)/(dx){f^(-1)(x)}=(3)/((1-x)^(2))`
`therefore` Domain of `(d)/(dx){f^(-1)(x)}=R-{1}`