If f: [1, ∞) → [2, ∞) is given by f(x) = x + 1/x then, find d/(dx)f⁻¹(x).

$y = x + \frac{1}{x} = \frac{x^2+1}{x}$ $xy = x^2 + 1$ $x^2 - xy + 1 = 0$ $x = \frac{y \pm \sqrt{y^2 - 4}}{2}$ $f^{-1}(y) = \frac{y \pm \sqrt{y^2-4}}{2}$, $f^{-1}(x) = \frac{x \pm \sqrt{x^2-4}}{2}$ Since range of inverse function is $[1, \infty)$, then $f^{-1}(x) = \frac{x + \sqrt{x^2-4}}{2}$ $\frac{d}{dx}f^{-1}(x) = \frac{1}{2}\left[1 + \frac{2x}{x\sqrt{x^2-4}}\right] = \frac{1}{2}\frac{(x + \sqrt{x^2-4})}{\sqrt{x^2-4}}$