To find the derivative of $y = (x - 3)(x^4 + 3)$ using logarithmic differentiation, follow these steps:

1. Take the natural logarithm of both sides of the equation:

   $\ln y = \ln((x - 3)(x^4 + 3))$

2. Apply the properties of logarithms (in particular, $\ln(ab) = \ln a + \ln b$):

   $\ln y = \ln(x - 3) + \ln(x^4 + 3)$

3. Differentiate both sides with respect to $x$. Using implicit differentiation on the left and the chain rule on the right:

   $\frac{1}{y} \frac{dy}{dx} = \frac{1}{x - 3} \cdot 1 + \frac{1}{x^4 + 3} \cdot 4x^3$

4. Simplify the expression on the right side:

   $\frac{1}{y} \frac{dy}{dx} = \frac{1}{x - 3} + \frac{4x^3}{x^4 + 3}$

5. Multiply both sides by $y$ to solve for $\frac{dy}{dx}$. Since $y = (x - 3)(x^4 + 3)$, substitute back for $y$:

   $\frac{dy}{dx} = \left( \frac{1}{x - 3} + \frac{4x^3}{x^4 + 3} \right)(x - 3)(x^4 + 3)$

6. Further simplification (optional, depending on the form desired):

   $\frac{dy}{dx} = (x - 3)(x^4 + 3) \left( \frac{1}{x - 3} + \frac{4x^3}{x^4 + 3} \right)$

   $\frac{dy}{dx} = (x^4 + 3) + 4x^3(x - 3)$

Thus, the derivative of $y = (x - 3)(x^4 + 3)$ is:

$\frac{dy}{dx} = (x^4 + 3) + 4x^3(x - 3)$

---

Would you like further details on any step?

---

1. How does logarithmic differentiation simplify finding derivatives of products?
2. How do you apply the chain rule in logarithmic differentiation?
3. What are the applications of logarithmic differentiation in calculus?
4. Can you use this method for functions with more than two factors?
5. How does implicit differentiation work in the context of logarithmic differentiation?

Tip: Logarithmic differentiation is especially useful for functions with products or quotients of terms raised to powers, making differentiation simpler.