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

Question

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

Aicharya · Accepted Answer

Let $y=\frac{x^2-x}{x^2+2 x}$ $ \begin{aligned} & \Rightarrow \quad\left(x^2+2 xight) y=x^2-x \ & \Rightarrow \quad x(x+2) y=x(x-1) \ & \Rightarrow \quad x[(x+2) y-(x-1)]=0 \ & \because \quad x 
eq 0, \quad 	herefore \quad(x+2) y-(x-1)=0 \ & \Rightarrow \quad x y+2 y-x+1=0 \ & \Rightarrow \quad x(y-1)=-(2 y+1) \ & 	herefore \quad x=\frac{2 y+1}{1-y} \Rightarrow f^{-1}(x)=\frac{2 x+1}{1-x} \ & \frac{d}{d x}\left(f^{-1}(x)ight)=\frac{2(1-x)-(2 x+1)(-1)}{(1-x)^2} \ & \quad=\frac{2-2 x+2 x+1}{(1-x)^2}=\frac{3}{(1-x)^2} \end{aligned} $

Let $f(x)=\frac{x^2-x}{x^2+2 x} x \neq 0,-2$. Then $\frac{d}{d x}\left[f^{-1}(x)\right]$ (wherever it is…

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