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