Question: Find the derivative of $f(y) = 9\tan(y) - \frac{\sqrt{y}}{8} - \frac{6}{y^2\sqrt{y}}$. Enclose arguments of functions, numerators, and denominators in parentheses. For example, $\sin(2x)$ or $\frac{(a - b)}{(1 + n)}$. $f'(y) = 9 \sec^2(y) - \frac{1}{16\sqrt{y}} + \frac{15}{y^3\sqrt{y}}$
1. **State the problem:** We want to find the derivative of the function $$f(y) = 9\tan(y) - \frac{\sqrt{y}}{8} - \frac{6}{y^2\sqrt{y}}.$$
2. **Recall derivative formulas:**
- The derivative of $\tan(y)$ is $\sec^2(y)$.
- The derivative of $\sqrt{y} = y^{1/2}$ is $\frac{1}{2}y^{-1/2}$.
- For powers of $y$, use the power rule: $\frac{d}{dy} y^n = n y^{n-1}$.
3. **Rewrite the function for easier differentiation:**
$$f(y) = 9\tan(y) - \frac{1}{8} y^{1/2} - 6 y^{-2} y^{-1/2} = 9\tan(y) - \frac{1}{8} y^{1/2} - 6 y^{-\frac{5}{2}}.$$
4. **Differentiate each term:**
- Derivative of $9\tan(y)$ is $$9 \sec^2(y).$$
- Derivative of $-\frac{1}{8} y^{1/2}$ is $$-\frac{1}{8} \cdot \frac{1}{2} y^{-1/2} = -\frac{1}{16} y^{-1/2} = -\frac{1}{16 \sqrt{y}}.$$
- Derivative of $-6 y^{-5/2}$ is $$-6 \cdot \left(-\frac{5}{2}\right) y^{-7/2} = 15 y^{-7/2} = \frac{15}{y^{7/2}} = \frac{15}{y^3 \sqrt{y}}.$$
5. **Combine all derivatives:**
$$f'(y) = 9 \sec^2(y) - \frac{1}{16 \sqrt{y}} + \frac{15}{y^3 \sqrt{y}}.$$
6. **Final answer:**
$$\boxed{f'(y) = 9 \sec^2(y) - \frac{1}{16 \sqrt{y}} + \frac{15}{y^3 \sqrt{y}}}.$$