1. **State the problem:** Find the first derivative of the function $$y = x \arcsin(x) + \sqrt{1 - x^2}$$. 2. **Recall the formulas and rules:** - Derivative of product: $\frac{d}{dx}[u v] = u' v + u v'$ - Derivative of $\arcsin(x)$: $\frac{d}{dx} \arcsin(x) = \frac{1}{\sqrt{1 - x^2}}$ - Derivative of $\sqrt{f(x)}$: $\frac{d}{dx} \sqrt{f(x)} = \frac{f'(x)}{2 \sqrt{f(x)}}$ 3. **Apply the product rule to the first term:** Let $u = x$, $v = \arcsin(x)$. $$u' = 1, \quad v' = \frac{1}{\sqrt{1 - x^2}}$$ So, $$\frac{d}{dx}[x \arcsin(x)] = 1 \cdot \arcsin(x) + x \cdot \frac{1}{\sqrt{1 - x^2}} = \arcsin(x) + \frac{x}{\sqrt{1 - x^2}}$$ 4. **Differentiate the second term:** $$y_2 = \sqrt{1 - x^2} = (1 - x^2)^{1/2}$$ Using the chain rule: $$\frac{d}{dx} y_2 = \frac{1}{2} (1 - x^2)^{-1/2} \cdot (-2x) = \frac{-x}{\sqrt{1 - x^2}}$$ 5. **Combine the derivatives:** $$y' = \arcsin(x) + \frac{x}{\sqrt{1 - x^2}} - \frac{x}{\sqrt{1 - x^2}} = \arcsin(x)$$ 6. **Final answer:** $$\boxed{y' = \arcsin(x)}$$ The derivative simplifies nicely because the terms involving $\frac{x}{\sqrt{1 - x^2}}$ cancel out.