1. **State the problem:** (a) Differentiate the function $$f(x) = \arcsin x + 2\sqrt{1 - x^2}$$ for $$x \in [-1,1]$$. (b) Find the coordinates of the point on the graph of $$y = f(x)$$ where the gradient of the tangent is zero. 2. **Recall formulas and rules:** - Derivative of $$\arcsin x$$ is $$\frac{1}{\sqrt{1 - x^2}}$$. - Derivative of $$\sqrt{1 - x^2}$$ is $$\frac{d}{dx} (1 - x^2)^{1/2} = \frac{1}{2}(1 - x^2)^{-1/2} \cdot (-2x) = \frac{-x}{\sqrt{1 - x^2}}$$. - Use sum rule: $$\frac{d}{dx}[u + v] = u' + v'$$. 3. **Differentiate $$f(x)$$:** $$f'(x) = \frac{d}{dx} \arcsin x + 2 \cdot \frac{d}{dx} \sqrt{1 - x^2} = \frac{1}{\sqrt{1 - x^2}} + 2 \cdot \left( \frac{-x}{\sqrt{1 - x^2}} \right) = \frac{1}{\sqrt{1 - x^2}} - \frac{2x}{\sqrt{1 - x^2}} = \frac{1 - 2x}{\sqrt{1 - x^2}}$$ 4. **Find where gradient is zero:** Set $$f'(x) = 0$$: $$\frac{1 - 2x}{\sqrt{1 - x^2}} = 0 \implies 1 - 2x = 0 \implies x = \frac{1}{2}$$ 5. **Find corresponding $$y$$ coordinate:** Calculate $$f\left( \frac{1}{2} \right)$$: $$f\left( \frac{1}{2} \right) = \arcsin \left( \frac{1}{2} \right) + 2 \sqrt{1 - \left( \frac{1}{2} \right)^2} = \frac{\pi}{6} + 2 \sqrt{1 - \frac{1}{4}} = \frac{\pi}{6} + 2 \sqrt{\frac{3}{4}} = \frac{\pi}{6} + 2 \cdot \frac{\sqrt{3}}{2} = \frac{\pi}{6} + \sqrt{3}$$ **Final answers:** (a) $$f'(x) = \frac{1 - 2x}{\sqrt{1 - x^2}}$$ (b) The point where the gradient is zero is $$\left( \frac{1}{2}, \frac{\pi}{6} + \sqrt{3} \right)$$.