Let f(x)=x^2-xx^2+2x, x0,-2.

Question

Let $f(x)=\dfrac{x^2-x}{x^2+2x}$, $x
e0,-2$. Then $\dfrac{d}{dx}[f^{-1}(x)]$ (wherever it is defined) is equal to:

Accepted Answer

Solving $y=\dfrac{x^2-x}{x^2+2x}=\dfrac{x-1}{x+2}$ for $x$: $x(y-1)=-(2y+1)\Rightarrow x=\dfrac{2y+1}{1-y}\Rightarrow f^{-1}(x)=\dfrac{2x+1}{1-x}$.

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

Solutions · 1