Let f(x) = frac{x^2 - x}{x^2 + 2x},x ne 0, -2 | vedclass

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(B) Let $y = \frac{x^2 - x}{x^2 + 2x} = \frac{x(x - 1)}{x(x + 2)} = \frac{x - 1}{x + 2}$ for $x 
e 0$.To find $f^{-1}(x)$,we solve $y = \frac{x - 1}{

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$\frac{3}{(1 - x)^2}$

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(B) Let $y = \frac{x^2 - x}{x^2 + 2x} = \frac{x(x - 1)}{x(x + 2)} = \frac{x - 1}{x + 2}$ for $x 
e 0$.To find $f^{-1}(x)$,we solve $y = \frac{x - 1}{x + 2}$ for $x$ in terms of $y$:$y(x + 2) = x - 1$$xy + 2y = x - 1$$xy - x = -2y - 1$$x(y - 1) = -(2y + 1)$$x = \frac{2y + 1}{1 - y}$Thus,$f^{-1}(x) = \frac{2x + 1}{1 - x}$.Now,differentiate $f^{-1}(x)$ with respect to $x$ using the quotient rule:$\frac{d}{dx}[f^{-1}(x)] = \frac{d}{dx}\left( \frac{2x + 1}{1 - x} ight)$$= \frac{(1 - x)(2) - (2x + 1)(-1)}{(1 - x)^2}$$= \frac{2 - 2x + 2x + 1}{(1 - x)^2}$$= \frac{3}{(1 - x)^2}$

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

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