The problem you provided is:

$y = \ln(10x - 1)$

To find $y'$, we take the derivative of $y$ with respect to $x$:

Step-by-step solution:

1. The derivative of $\ln(u)$ is: $\frac{d}{dx}[\ln(u)] = \frac{1}{u} \cdot \frac{du}{dx}.$

2. Let $u = 10x - 1$. Then, $\frac{du}{dx} = 10$.

3. Using the chain rule: $y' = \frac{1}{10x - 1} \cdot 10.$

4. Simplify: $y' = \frac{10}{10x - 1}.$

Final Answer:

$y' = \frac{10}{10x - 1}.$

Would you like me to elaborate further or solve similar problems?

Here are 5 related questions to explore:

1. How do you differentiate other logarithmic functions such as $\ln(ax + b)$?
2. What happens when you differentiate $\ln(u)$ with $u$ being a more complex function?
3. How do logarithmic derivatives apply in real-world problems like population growth?
4. Can you solve for critical points of $y = \ln(10x - 1)$?
5. How does the natural logarithm relate to exponential functions?

Tip: Always use the chain rule carefully when the argument of a function (like $\ln$) is more than just $x$.