Derivative Quotient 346F9A

1. **State the problem:** We need to find the derivative of the function $$f(y) = \frac{16 \cdot y^{1/4}}{\ln(y)}.$$ 2. **Recall the formula:** To differentiate a quotient $$\frac{u(y)}{v(y)}$$, use the quotient rule: $$$f'(y) = \frac{u'(y) \cdot v(y) - u(y) \cdot v'(y)}{(v(y))^2}.$$ 3. **Identify parts:** Here, $$u(y) = 16 \cdot y^{1/4}, \quad v(y) = \ln(y).$$ 4. **Find derivatives:** - Derivative of $u(y)$: $$u'(y) = 16 \cdot \frac{1}{4} y^{1/4 - 1} = 16 \cdot \frac{1}{4} y^{-3/4} = 4 y^{-3/4}.$$ - Derivative of $v(y)$: $$v'(y) = \frac{1}{y}.$$ 5. **Apply quotient rule:** $$f'(y) = \frac{4 y^{-3/4} \cdot \ln(y) - 16 y^{1/4} \cdot \frac{1}{y}}{(\ln(y))^2}.$$ 6. **Simplify numerator:** Rewrite $16 y^{1/4} \cdot \frac{1}{y}$ as $16 y^{1/4 - 1} = 16 y^{-3/4}$. So numerator becomes: $$4 y^{-3/4} \ln(y) - 16 y^{-3/4} = y^{-3/4} (4 \ln(y) - 16).$$ 7. **Final derivative:** $$f'(y) = \frac{y^{-3/4} (4 \ln(y) - 16)}{(\ln(y))^2} = y^{-3/4} \cdot \frac{4 \ln(y) - 16}{(\ln(y))^2}.$$ This matches the given answer: $$f'(y) = \frac{3}{4} y^{-3/4} \cdot \frac{4 \ln(y) - 16}{(\ln(y))^2}$$ **Note:** The coefficient $\frac{3}{4}$ in the user's answer seems to be a typo; the correct coefficient from the derivative of $y^{1/4}$ is $\frac{1}{4}$ multiplied by 16, which is 4, as shown above.