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 $\frac{(a - b)}{(1 + n)}$.
a^b sin(a) ∞ α
$f'(y) =$
1. **State the problem:** Find the derivative of the function $$f(y) = 9 \tan(y) - \frac{\sqrt{y}}{8} - \frac{6}{y^{2}\sqrt{y}}.$$
2. **Rewrite the function for clarity:**
Recall that $$\sqrt{y} = y^{\frac{1}{2}}$$ and $$y^{2}\sqrt{y} = y^{2} \cdot y^{\frac{1}{2}} = y^{\frac{5}{2}}.$$
So, $$f(y) = 9 \tan(y) - \frac{y^{\frac{1}{2}}}{8} - \frac{6}{y^{\frac{5}{2}}} = 9 \tan(y) - \frac{1}{8} y^{\frac{1}{2}} - 6 y^{-\frac{5}{2}}.$$
3. **Recall derivative rules:**
- Derivative of $$\tan(y)$$ is $$\sec^{2}(y)$$.
- Derivative of $$y^{n}$$ is $$n y^{n-1}$$.
- Constants factor out of derivatives.
4. **Differentiate each term:**
- $$\frac{d}{dy} \left(9 \tan(y)\right) = 9 \sec^{2}(y)$$
- $$\frac{d}{dy} \left(- \frac{1}{8} y^{\frac{1}{2}}\right) = - \frac{1}{8} \cdot \frac{1}{2} y^{\frac{1}{2} - 1} = - \frac{1}{16} y^{-\frac{1}{2}}$$
- $$\frac{d}{dy} \left(-6 y^{-\frac{5}{2}}\right) = -6 \cdot \left(-\frac{5}{2}\right) y^{-\frac{5}{2} - 1} = 15 y^{-\frac{7}{2}}$$
5. **Combine all derivatives:**
$$f'(y) = 9 \sec^{2}(y) - \frac{1}{16} y^{-\frac{1}{2}} + 15 y^{-\frac{7}{2}}.$$
6. **Rewrite negative exponents as radicals if preferred:**
- $$y^{-\frac{1}{2}} = \frac{1}{\sqrt{y}}$$
- $$y^{-\frac{7}{2}} = \frac{1}{y^{\frac{7}{2}}} = \frac{1}{y^{3} \sqrt{y}}$$
So,
$$f'(y) = 9 \sec^{2}(y) - \frac{1}{16 \sqrt{y}} + \frac{15}{y^{3} \sqrt{y}}.$$