If \(\mathrm{f}(x)=\frac{x^2-x}{x^2+2 x}\) then \(\frac{\mathrm{d}}{\mathrm{dx}}\left(\mathrm{f}^{-1}(x)\right)\) at \(x=2\) is

\(\begin{aligned} & \mathrm{f}(x)=y=\frac{x^2-x}{x^2+2 x}=\frac{x(x-1)}{x(x+2)} \\ & \Rightarrow y=\frac{x-1}{x+2} \\ & \Rightarrow y x+2 y=x-1 \quad \Rightarrow \frac{2 y+1}{1-y}=x\end{aligned}\)
\(\begin{aligned} & \Rightarrow \mathrm{f}^{-1}(x)=\frac{2 x+1}{1-x} \\ & \therefore \quad \frac{\mathrm{~d}}{\mathrm{~d} x} \mathrm{f}^{-1}(x)=\frac{\mathrm{d}}{\mathrm{d} x}\left(\frac{2 x+1}{1-x}\right) \\ &=\frac{(1-x)(2)-(2 x+1)(-1)}{(1-x)^2} \\ &=\frac{2-2 x+2 x+1}{(1-x)^2}=\frac{3}{(1+x)^2} \\ & \therefore \quad \frac{\mathrm{~d}}{\mathrm{dx}} \mathrm{f}^{-1}(x) \text { at } x=2=\frac{3}{(1-2)^2}=3\end{aligned}\)