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

To show that \( \frac{d}{dx}(\sin^{-1} x) = \frac{1}{\sqrt{1-x^2}} \), we can use implicit differentiation. Let \( y = \sin^{-1} x \). This means \( x = \sin y \). Now, differentiate both sides with respect to \( x \): 1. \[\frac{dx}{dx} = \cos y \frac{dy}{dx} \] This simplifies to \( 1 = \cos y \frac{dy}{dx} \). Therefore, we have: 2. \[\frac{dy}{dx} = \frac{1}{\cos y}\] Next, we need to express \( \cos y \) in terms of \( x \). Using the identity, \( \sin^2 y + \cos^2 y = 1 \): 3. From \( x = \sin y \), we have \( \sin^2 y = x^2\), which gives us \( \cos^2 y = 1 - x^2 \). Therefore, \( \cos y = \sqrt{1 - x^2} \). Substituting this back into our derivative, we get: 4. \[\frac{dy}{dx} = \frac{1}{\cos y} = \frac{1}{\sqrt{1 - x^2}}\] Thus, we have shown that: \[ \frac{d}{dx}(\sin^{-1} x) = \frac{1}{\sqrt{1 - x^2}} \]