Question: The function h is defined by h(x) = 2\sqrt{x} + x^3. Find h'(1).
1. **State the problem:** We need to find the derivative of the function $$h(x) = 2\sqrt{x} + x^3$$ and then evaluate it at $$x=1$$.
2. **Recall the derivative rules:**
- The derivative of $$\sqrt{x}$$ is $$\frac{1}{2\sqrt{x}}$$.
- The derivative of $$x^n$$ is $$nx^{n-1}$$.
3. **Rewrite the function for easier differentiation:**
$$h(x) = 2x^{\frac{1}{2}} + x^3$$
4. **Differentiate each term:**
$$h'(x) = 2 \cdot \frac{1}{2} x^{\frac{1}{2} - 1} + 3x^{3-1}$$
5. **Simplify the derivative:**
$$h'(x) = \cancel{2} \cdot \frac{1}{\cancel{2}} x^{-\frac{1}{2}} + 3x^2 = x^{-\frac{1}{2}} + 3x^2$$
6. **Evaluate the derivative at $$x=1$$:**
$$h'(1) = 1^{-\frac{1}{2}} + 3 \cdot 1^2 = 1 + 3 = 4$$
**Final answer:** $$h'(1) = 4$$