Question

If $f\left(x\right)=\frac{x-1}{x+2}$, then $\frac{d{f}^{-1}\left(x\right)}{dx}$ is equal to

• A. $\frac{1}{(1-x{)}^2}$
• B. $\frac{-3}{(1-x{)}^2}$
• C. $\frac{3}{(1-x{)}^2}$
• D. $\frac{-1}{(1-x{)}^2}$

Answer 1

The correct option is C $\frac{3}{(1-x{)}^2}$

$f\left(x\right)=\frac{x-1}{x+2}$

Let $y=\frac{x-1}{x+2}\Rightarrow yx+2y=x-1$

$\Rightarrow (y-1)x=-1-2y\Rightarrow x=\frac{2y+1}{1-y}$

$\frac{d{f}^{-1}\left(x\right)}{dx}=\frac{(1-x)2+(2x+1)(-1)}{{(1-x)}^2}=\frac{3}{{(1-x)}^2}$