Question

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

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

Answer 1

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

Given that:

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

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

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

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

On differentiating with respect to $x$, we get

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

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

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

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

Hence, this is the required solution.