1. The problem is to find the derivative of the function $$f(x) = x \arctan(\sqrt{x})$$ and verify if the given expressions for $$f'(x)$$ are correct and equivalent. 2. We use the product rule for derivatives: $$\frac{d}{dx}[u(x)v(x)] = u'(x)v(x) + u(x)v'(x)$$ where $$u(x) = x$$ and $$v(x) = \arctan(\sqrt{x})$$. 3. Compute $$u'(x)$$: $$u'(x) = \frac{d}{dx}[x] = 1$$. 4. Compute $$v'(x)$$ using the chain rule: $$v(x) = \arctan(\sqrt{x})$$ Let $$g(x) = \sqrt{x} = x^{1/2}$$, then $$v'(x) = \frac{1}{1 + (g(x))^2} \cdot g'(x) = \frac{1}{1 + x} \cdot \frac{1}{2\sqrt{x}} = \frac{1}{2\sqrt{x}(1+x)}$$. 5. Apply the product rule: $$f'(x) = u'(x)v(x) + u(x)v'(x) = 1 \cdot \arctan(\sqrt{x}) + x \cdot \frac{1}{2\sqrt{x}(1+x)} = \arctan(\sqrt{x}) + \frac{x}{2\sqrt{x}(1+x)}$$. 6. Simplify the fraction: $$\frac{x}{2\sqrt{x}(1+x)} = \frac{\sqrt{x} \cdot \sqrt{x}}{2\sqrt{x}(1+x)} = \frac{\sqrt{x}}{2(1+x)}$$. 7. Therefore, the derivative is: $$f'(x) = \arctan(\sqrt{x}) + \frac{\sqrt{x}}{2(1+x)}$$. 8. The two expressions given: - $$\frac{x}{(1+x)(2\sqrt{x})} + \arctan(\sqrt{x})$$ - $$\frac{\sqrt{x}}{2(1+x)} + \arctan(\sqrt{x})$$ are equivalent because $$\frac{x}{(1+x)(2\sqrt{x})} = \frac{\sqrt{x}}{2(1+x)}$$. Hence, the answer is correct and the two forms are equivalent.