Listen Show that \( \frac{d}{d x}\left(\sin ^{-1} x\right)=\frac{1}{\sqrt{1-x^{2}}} \)

Let \( y = \sin^{-1}(x) \). Then, by the definition of the inverse function, we have \[ \sin y = x. \] Differentiate both sides with respect to \( x \): \[ \frac{d}{dx} (\sin y) = \frac{d}{dx} (x). \] Using the chain rule on the left-hand side: \[ \cos y \cdot \frac{dy}{dx} = 1. \] Solve for \(\frac{dy}{dx}\): \[ \frac{dy}{dx} = \frac{1}{\cos y}. \] To express \(\cos y\) in terms of \( x \), recall the Pythagorean identity: \[ \sin^2 y + \cos^2 y = 1. \] Since \(\sin y = x\), substitute to obtain: \[ x^2 + \cos^2 y = 1. \] Thus, \[ \cos^2 y = 1 - x^2. \] Taking the square root of both sides (noting that \( \cos y \geq 0 \) when \( y = \sin^{-1}(x) \) because \( y \in \left[-\frac{\pi}{2}, \frac{\pi}{2}\right] \)): \[ \cos y = \sqrt{1-x^2}. \] Now substitute back into the derivative: \[ \frac{dy}{dx} = \frac{1}{\sqrt{1-x^2}}. \] Thus, we have shown that \[ \frac{d}{dx}\left(\sin^{-1}(x)\right) = \frac{1}{\sqrt{1-x^2}}. \]