Question

Let $f\left(x\right) = \frac{x^{2}-x}{x^{2}+2x}x \ne 0, -2.$ Then $\frac{d}{dx}\left[f^{-1}\left(x\right)\right]$ (wherever it is defined) is equal to :

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

Answer 1

Answer: (B)

Let $y = \frac{x^{2}-x}{x^{2}+2x}$
$\Rightarrow \quad\left(x^{2} + 2x\right)y = x^{2}-x$
$\Rightarrow \quad x\left(x + 2\right)y = x\left(x - 1\right)$
$\Rightarrow \quad x\left[\left(x + 2\right)y - \left(x -1\right)\right] = 0$
$\because\quad x \ne 0,\quad\therefore \quad\left(x + 2\right)y-\left(x-1\right) = 0$
$\Rightarrow \quad xy + 2y-x + 1=0$
$\Rightarrow \quad x\left(y-1\right) = -\left(2y+ 1\right)$
$\therefore \quad x = \frac{2y+1}{1-y} \Rightarrow f^{-1}\left(x\right) = \frac{2x+1}{1-x}$
$\frac{d}{dx}\left(f^{-1}\left(x\right)\right) = \frac{2\left(1 -x\right)-\left(2x-+1\right)\left(-1\right)}{\left(1-x\right)^{2}}$
$= \frac{2 - 2x + 2x +1}{\left(1-x\right)^{2}} = \frac{3}{\left(1-x\right)^{2}}$