Differentiate Square Root 423Daf

1. **State the problem:** Differentiate the function $$y=(2+\sqrt{x})^2$$ with respect to $$x$$. 2. **Recall the formula:** Use the chain rule for differentiation. If $$y = [u(x)]^2$$, then $$\frac{dy}{dx} = 2u(x) \cdot \frac{du}{dx}$$. 3. **Identify the inner function:** Here, $$u(x) = 2 + \sqrt{x} = 2 + x^{1/2}$$. 4. **Differentiate the inner function:** $$\frac{du}{dx} = 0 + \frac{1}{2}x^{-1/2} = \frac{1}{2\sqrt{x}}$$. 5. **Apply the chain rule:** $$\frac{dy}{dx} = 2(2 + \sqrt{x}) \cdot \frac{1}{2\sqrt{x}}$$ 6. **Simplify:** $$\frac{dy}{dx} = \cancel{2} (2 + \sqrt{x}) \cdot \frac{1}{\cancel{2}\sqrt{x}} = \frac{2 + \sqrt{x}}{\sqrt{x}}$$ 7. **Rewrite the derivative:** $$\frac{dy}{dx} = \frac{2}{\sqrt{x}} + \frac{\sqrt{x}}{\sqrt{x}} = \frac{2}{\sqrt{x}} + 1$$ **Final answer:** $$\boxed{\frac{dy}{dx} = \frac{2}{\sqrt{x}} + 1}$$