We are tasked with using logarithmic differentiation to find the derivative of the function:

$y = (x^2 + 2)^2 (x^4 + 4)^4$

Step 1: Take the natural logarithm of both sides

Start by applying $\ln$ to both sides of the equation to simplify the product and exponents:

$\ln y = \ln \left[ (x^2 + 2)^2 (x^4 + 4)^4 \right]$

Using the logarithmic property $\ln(ab) = \ln a + \ln b$, we get:

$\ln y = \ln (x^2 + 2)^2 + \ln (x^4 + 4)^4$

Now, apply the property $\ln(a^b) = b \ln a$:

$\ln y = 2 \ln (x^2 + 2) + 4 \ln (x^4 + 4)$

Step 2: Differentiate implicitly with respect to $x$

Differentiate both sides. Recall that the derivative of $\ln y$ is $\frac{1}{y} \frac{dy}{dx}$, and use the chain rule:

$\frac{1}{y} \frac{dy}{dx} = 2 \cdot \frac{1}{x^2 + 2} \cdot \frac{d}{dx}(x^2 + 2) + 4 \cdot \frac{1}{x^4 + 4} \cdot \frac{d}{dx}(x^4 + 4)$

First, compute the derivatives of the inner terms:

1. $\frac{d}{dx}(x^2 + 2) = 2x$
2. $\frac{d}{dx}(x^4 + 4) = 4x^3$

Substitute these into the equation:

$\frac{1}{y} \frac{dy}{dx} = 2 \cdot \frac{2x}{x^2 + 2} + 4 \cdot \frac{4x^3}{x^4 + 4}$

Simplify the terms:

$\frac{1}{y} \frac{dy}{dx} = \frac{4x}{x^2 + 2} + \frac{16x^3}{x^4 + 4}$

Step 3: Solve for $\frac{dy}{dx}$

Multiply through by $y$ to isolate $\frac{dy}{dx}$. Recall that $y = (x^2 + 2)^2 (x^4 + 4)^4$:

$\frac{dy}{dx} = y \left( \frac{4x}{x^2 + 2} + \frac{16x^3}{x^4 + 4} \right)$

Substitute back $y = (x^2 + 2)^2 (x^4 + 4)^4$:

$\frac{dy}{dx} = (x^2 + 2)^2 (x^4 + 4)^4 \left( \frac{4x}{x^2 + 2} + \frac{16x^3}{x^4 + 4} \right)$

Final Answer:

$\frac{dy}{dx} = (x^2 + 2) (x^4 + 4)^4 \cdot 4x + (x^2 + 2)^2 (x^4 + 4)^3 \cdot 16x^3$

Would you like me to simplify or explain a particular step further?

Follow-Up Questions:

1. What are the key advantages of using logarithmic differentiation in complex products?
2. Can you explain why the chain rule is necessary in logarithmic differentiation?
3. How would the solution change if the powers were variables instead of constants?
4. Could you verify the solution using a computational tool or alternate differentiation methods?
5. How does logarithmic differentiation simplify derivatives for functions involving multiple terms?

Tip:

When dealing with products of functions raised to powers, logarithmic differentiation is often the easiest way to simplify the problem before finding the derivative.