Derivative Product Root 1C8Cee

1. **State the problem:** We need to find the derivative $f'(x)$ of the function $$f(x) = (3x^2 + 1)\sqrt{5 + x}.$$ 2. **Rewrite the function:** Note that $$\sqrt{5 + x} = (5 + x)^{\frac{1}{2}}.$$ So, $$f(x) = (3x^2 + 1)(5 + x)^{\frac{1}{2}}.$$ 3. **Use the product rule:** The derivative of a product $u(x)v(x)$ is $$f'(x) = u'(x)v(x) + u(x)v'(x).$$ Here, let $$u(x) = 3x^2 + 1$$ and $$v(x) = (5 + x)^{\frac{1}{2}}.$$ 4. **Find $u'(x)$:** $$u'(x) = \frac{d}{dx}(3x^2 + 1) = 6x.$$ 5. **Find $v'(x)$:** Use the chain rule: $$v'(x) = \frac{1}{2}(5 + x)^{-\frac{1}{2}} \cdot \frac{d}{dx}(5 + x) = \frac{1}{2}(5 + x)^{-\frac{1}{2}} \cdot 1 = \frac{1}{2}(5 + x)^{-\frac{1}{2}}.$$ 6. **Apply the product rule:** $$f'(x) = u'(x)v(x) + u(x)v'(x) = 6x(5 + x)^{\frac{1}{2}} + (3x^2 + 1) \cdot \frac{1}{2}(5 + x)^{-\frac{1}{2}}.$$ 7. **Simplify the expression:** To combine terms, write both with a common denominator: $$f'(x) = \frac{6x(5 + x)^{\frac{1}{2}} \cdot (5 + x)^{\frac{1}{2}}}{(5 + x)^{\frac{1}{2}}} + \frac{(3x^2 + 1)}{2(5 + x)^{\frac{1}{2}}} = \frac{6x(5 + x) + \frac{1}{2}(3x^2 + 1)}{(5 + x)^{\frac{1}{2}}}.$$ 8. **Calculate numerator:** $$6x(5 + x) = 6x \cdot 5 + 6x \cdot x = 30x + 6x^2.$$ So numerator is $$30x + 6x^2 + \frac{3x^2 + 1}{2} = \frac{2(30x + 6x^2) + 3x^2 + 1}{2} = \frac{60x + 12x^2 + 3x^2 + 1}{2} = \frac{60x + 15x^2 + 1}{2}.$$ 9. **Final derivative:** $$f'(x) = \frac{15x^2 + 60x + 1}{2(5 + x)^{\frac{1}{2}}}.$$ **Answer:** $$\boxed{f'(x) = \frac{15x^2 + 60x + 1}{2\sqrt{5 + x}}}.$$