Question: Find the derivative of $f (y) = \frac{16 \cdot \sqrt[4]{y}}{\ln(y)}$
1. **State the problem:** Find the derivative of the function $$f(y) = \frac{16 \cdot y^{\frac{1}{4}}}{\ln(y)}$$ with respect to $y$.
2. **Recall the formula:** To differentiate a quotient $$\frac{u(y)}{v(y)}$$, use the quotient rule:
$$f'(y) = \frac{u'(y) \cdot v(y) - u(y) \cdot v'(y)}{(v(y))^2}$$
where $u(y) = 16 \cdot y^{\frac{1}{4}}$ and $v(y) = \ln(y)$.
3. **Find $u'(y)$:**
$$u(y) = 16 \cdot y^{\frac{1}{4}}$$
Using the power rule:
$$u'(y) = 16 \cdot \frac{1}{4} y^{\frac{1}{4} - 1} = 16 \cdot \frac{1}{4} y^{-\frac{3}{4}} = 4 y^{-\frac{3}{4}}$$
4. **Find $v'(y)$:**
$$v(y) = \ln(y)$$
Derivative of natural log:
$$v'(y) = \frac{1}{y}$$
5. **Apply the quotient rule:**
$$f'(y) = \frac{4 y^{-\frac{3}{4}} \cdot \ln(y) - 16 y^{\frac{1}{4}} \cdot \frac{1}{y}}{(\ln(y))^2}$$
6. **Simplify the numerator:**
Rewrite $16 y^{\frac{1}{4}} \cdot \frac{1}{y}$ as $16 y^{\frac{1}{4} - 1} = 16 y^{-\frac{3}{4}}$.
So numerator becomes:
$$4 y^{-\frac{3}{4}} \ln(y) - 16 y^{-\frac{3}{4}} = y^{-\frac{3}{4}} (4 \ln(y) - 16)$$
7. **Final derivative:**
$$f'(y) = \frac{y^{-\frac{3}{4}} (4 \ln(y) - 16)}{(\ln(y))^2}$$
This is the derivative of the given function $f(y)$.
**Answer:**
$$\boxed{f'(y) = \frac{y^{-\frac{3}{4}} (4 \ln(y) - 16)}{(\ln(y))^2}}$$