Question

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

• A. $-\frac{3}{{(1-x)}^2}$
• B. $\frac{3}{{(1-x)}^2}$
• C. $\frac{1}{{(1-x)}^2}$
• D. None of these

Answer 1

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

We have, $f\left(x\right)=\frac{x^2-x}{x^2+2x}$

Clearly, $f\left(0\right)$ and $f(-2)$ are not defined.

So, domain of $fisR-\{0,-2\}$

Then, in this domain, we have

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

or, $x=\frac{2y+1}{1-y},$

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

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