Let y = x sin⁻¹x + √(1 - x²) for -1 < x < 1…

Step 1. The first term. By the product rule, $\dfrac{d}{dx}(x\sin^{-1}x) = \sin^{-1}x + \dfrac{x}{\sqrt{1 - x^2}}$.

Step 2. The second term. $\dfrac{d}{dx}\sqrt{1 - x^2} = \dfrac{-2x}{2\sqrt{1 - x^2}} = -\dfrac{x}{\sqrt{1 - x^2}}$.

Step 3. Add. The two fractions cancel, so $\dfrac{dy}{dx} = \sin^{-1}x$.

Step 4. Put in the values. At $x = 0.5$, $\sin^{-1}0.5 = \dfrac{\pi}{6} = 0.5236$. At $x = 0$, $\sin^{-1}0 = 0$.