Question: If f(x) = (3x² + 1)\sqrt{5} + x, then f'(x) =
1. **State the problem:** Find the derivative of the function $$f(x) = (3x^2 + 1)\sqrt{5} + x$$.
2. **Recall the rules:**
- The derivative of a sum is the sum of the derivatives.
- The derivative of $$x$$ is 1.
- $$\sqrt{5}$$ is a constant.
- The derivative of $$3x^2 + 1$$ is $$6x$$.
3. **Apply the derivative:**
$$f'(x) = \frac{d}{dx} \left((3x^2 + 1)\sqrt{5}\right) + \frac{d}{dx}(x)$$
4. Since $$\sqrt{5}$$ is constant, use constant multiple rule:
$$f'(x) = \sqrt{5} \cdot \frac{d}{dx}(3x^2 + 1) + 1$$
5. Compute derivative inside:
$$\frac{d}{dx}(3x^2 + 1) = 6x$$
6. Substitute back:
$$f'(x) = \sqrt{5} \cdot 6x + 1 = 6x\sqrt{5} + 1$$
7. Check answer choices: none exactly match $$6x\sqrt{5} + 1$$ but option b is $$6x\sqrt{5} + x$$ which is close but has $$x$$ instead of 1.
Since the original function has $$+ x$$, derivative of $$x$$ is 1, so correct derivative is $$6x\sqrt{5} + 1$$.
**Final answer:** $$f'(x) = 6x\sqrt{5} + 1$$
**Note:** None of the given options exactly match this derivative, but the closest is option b if the $$x$$ is a typo for 1.