We are tasked with finding the derivative of the function:

$y = \sqrt{\frac{x - 1}{x^6 + 2}}$

using logarithmic differentiation.

Step 1: Rewrite the function

First, rewrite the square root as a fractional exponent for easier manipulation: $y = \left( \frac{x - 1}{x^6 + 2} \right)^{1/2}$

Step 2: Take the natural logarithm of both sides

Apply the natural logarithm to both sides: $\ln(y) = \ln\left( \left( \frac{x - 1}{x^6 + 2} \right)^{1/2} \right)$

Using the logarithmic property $\ln(a^b) = b \ln(a)$, we can simplify this to: $\ln(y) = \frac{1}{2} \ln\left( \frac{x - 1}{x^6 + 2} \right)$

Step 3: Apply logarithmic properties

Now use the property $\ln\left(\frac{a}{b}\right) = \ln(a) - \ln(b)$: $\ln(y) = \frac{1}{2} \left( \ln(x - 1) - \ln(x^6 + 2) \right)$

Step 4: Differentiate both sides

Differentiate both sides with respect to $x$. Remember, when differentiating $\ln(y)$, you need to apply the chain rule: $\frac{d}{dx}[\ln(y)] = \frac{1}{y} \cdot y'$ On the right-hand side, differentiate each term: $\frac{1}{2} \cdot \frac{d}{dx} \left( \ln(x - 1) - \ln(x^6 + 2) \right)$ The derivative of $\ln(x - 1)$ is: $\frac{d}{dx}[\ln(x - 1)] = \frac{1}{x - 1}$ The derivative of $\ln(x^6 + 2)$ using the chain rule is: $\frac{d}{dx}[\ln(x^6 + 2)] = \frac{6x^5}{x^6 + 2}$

Thus, the derivative of the right-hand side becomes: $\frac{1}{2} \left( \frac{1}{x - 1} - \frac{6x^5}{x^6 + 2} \right)$

Step 5: Solve for $y'$

Now multiply both sides by $y$ to solve for $y'$: $y' = y \cdot \frac{1}{2} \left( \frac{1}{x - 1} - \frac{6x^5}{x^6 + 2} \right)$ Substitute back $y = \sqrt{\frac{x - 1}{x^6 + 2}}$: $y' = \sqrt{\frac{x - 1}{x^6 + 2}} \cdot \frac{1}{2} \left( \frac{1}{x - 1} - \frac{6x^5}{x^6 + 2} \right)$

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

Would you like me to provide further details or clarification on any step? Here are 5 related questions to deepen your understanding:

1. How can you simplify a function's derivative when using logarithmic differentiation?
2. What are the benefits of using logarithmic differentiation for complex functions?
3. How does the chain rule apply when differentiating logarithmic functions?
4. What are common mistakes to avoid when applying logarithmic differentiation?
5. How would this derivative change if the power of the square root was different?

Tip: When dealing with products, quotients, or powers of complicated expressions, logarithmic differentiation can simplify the process by turning multiplication into addition and division into subtraction.