Question: Find the derivative.
y = (\sqrt{x} - 4)(x^2 + 17);
1. **State the problem:** We need to find the derivative of the function $$y = (\sqrt{x} - 4)(x^2 + 17)$$ with respect to $$x$$.
2. **Recall the formula:** To differentiate a product of two functions, use the product rule:
$$\frac{d}{dx}[u \cdot v] = u'v + uv'$$
where $$u = \sqrt{x} - 4$$ and $$v = x^2 + 17$$.
3. **Find the derivatives of each part:**
- $$u = \sqrt{x} - 4 = x^{\frac{1}{2}} - 4$$
- $$u' = \frac{1}{2}x^{-\frac{1}{2}} - 0 = \frac{1}{2\sqrt{x}}$$
- $$v = x^2 + 17$$
- $$v' = 2x$$
4. **Apply the product rule:**
$$y' = u'v + uv' = \left(\frac{1}{2\sqrt{x}}\right)(x^2 + 17) + (\sqrt{x} - 4)(2x)$$
5. **Simplify each term:**
- First term: $$\frac{x^2 + 17}{2\sqrt{x}}$$
- Second term: $$2x\sqrt{x} - 8x = 2x^{\frac{3}{2}} - 8x$$
6. **Combine the terms:**
$$y' = \frac{x^2 + 17}{2\sqrt{x}} + 2x^{\frac{3}{2}} - 8x$$
7. **Optional: Express all terms with powers of $$x$$:**
- $$\frac{x^2}{2x^{\frac{1}{2}}} = \frac{x^{2}}{2x^{\frac{1}{2}}} = \frac{x^{\frac{3}{2}}}{2}$$
- $$\frac{17}{2\sqrt{x}} = \frac{17}{2x^{\frac{1}{2}}}$$
So,
$$y' = \frac{x^{\frac{3}{2}}}{2} + \frac{17}{2x^{\frac{1}{2}}} + 2x^{\frac{3}{2}} - 8x$$
8. **Combine like terms:**
$$\frac{x^{\frac{3}{2}}}{2} + 2x^{\frac{3}{2}} = \frac{1}{2}x^{\frac{3}{2}} + \frac{4}{2}x^{\frac{3}{2}} = \frac{5}{2}x^{\frac{3}{2}}$$
Final derivative:
$$y' = \frac{5}{2}x^{\frac{3}{2}} + \frac{17}{2x^{\frac{1}{2}}} - 8x$$