To solve for $y'$ using logarithmic differentiation, let's go through the following steps:

Given: $y = \frac{\sin^2(x) \cdot \tan^6(x)}{(x^2 + 3)^2}$

Step 1: Take the Natural Logarithm of Both Sides

Applying the natural logarithm to both sides, we get: $\ln(y) = \ln\left(\frac{\sin^2(x) \cdot \tan^6(x)}{(x^2 + 3)^2}\right)$

Using logarithmic properties (specifically, $\ln \frac{a}{b} = \ln a - \ln b$ and $\ln(ab) = \ln a + \ln b$), this becomes: $\ln(y) = \ln(\sin^2(x)) + \ln(\tan^6(x)) - \ln((x^2 + 3)^2)$

Now we can simplify further by bringing down the exponents: $\ln(y) = 2 \ln(\sin(x)) + 6 \ln(\tan(x)) - 2 \ln(x^2 + 3)$

Step 2: Differentiate Both Sides with Respect to $x$

Now, differentiate implicitly with respect to $x$:

$\frac{1}{y} \cdot y' = 2 \cdot \frac{\cos(x)}{\sin(x)} + 6 \cdot \frac{1}{\tan(x)} \cdot \sec^2(x) - 2 \cdot \frac{2x}{x^2 + 3}$

This simplifies to: $\frac{y'}{y} = 2 \cot(x) + 6 \sec^2(x) - \frac{4x}{x^2 + 3}$

Step 3: Solve for $y'$

Now, multiply both sides by $y$ to isolate $y'$: $y' = y \cdot \left(2 \cot(x) + 6 \sec^2(x) - \frac{4x}{x^2 + 3}\right)$

Finally, substitute back the expression for $y$: $y' = \frac{\sin^2(x) \cdot \tan^6(x)}{(x^2 + 3)^2} \cdot \left(2 \cot(x) + 6 \sec^2(x) - \frac{4x}{x^2 + 3}\right)$

This is the derivative $y'$ of the function.

Would you like more details on any part of this process or have any questions?

Here are five related questions for further practice:

1. How would the process change if the exponent of $\sin(x)$ was 3 instead of 2?
2. Can you use logarithmic differentiation on a product involving exponential and trigonometric functions?
3. How do you simplify expressions involving $\cot(x)$ and $\sec^2(x)$ after differentiation?
4. What if we applied logarithmic differentiation to only the numerator? Would that be valid?
5. How does the derivative of a function change if it’s under a square root?

Tip: When using logarithmic differentiation, always simplify by expanding the log expression first before differentiating—it makes the process more straightforward.