d/dx(sin⁻¹ x) =

To find the derivative of $y = \sin^{-1} x$, we use implicit differentiation: 1. **Rewrite the function:** Let $y = \sin^{-1} x$. This is equivalent to saying $\sin y = x$. 2. **Differentiate both sides:** Apply the derivative with respect to $x$ to both sides of $\sin y = x$. Using the Chain Rule on the left side, we get: $$\frac{d}{dx}(\sin y) = \frac{d}{dx}(x)$$ $$\cos y \cdot \frac{dy}{dx} = 1$$ 3. **Solve for $\frac{dy}{dx}$:** $$\frac{dy}{dx} = \frac{1}{\cos y}$$ 4. **Convert back to $x$:** We know from the Pythagorean identity that $\sin^2 y + \cos^2 y = 1$. Therefore, $\cos y = \sqrt{1 - \sin^2 y}$. Since we defined $\sin y = x$, we substitute $x$ in: $$\cos y = \sqrt{1 - x^2}$$ 5. **Final Result:** Substitute this back into the expression for $\frac{dy}{dx}$: $$\frac{d}{dx}(\sin^{-1} x) = \frac{1}{\sqrt{1 - x^2}}$$ Note: This is valid for $-1 < x < 1$.