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To find the derivative of the function \( y = \ln \sqrt{x + 1} \), we can follow these steps:

1. **Rewrite the Expression**: 
   Rewrite \( y = \ln \sqrt{x + 1} \) using exponent rules. Since \( \sqrt{x + 1} = (x + 1)^{1/2} \), we can write:
   \[
   y = \ln \left( (x + 1)^{1/2} \right)
   \]
   By using the logarithmic property \( \ln(a^b) = b \cdot \ln(a) \), we get:
   \[
   y = \frac{1}{2} \ln(x + 1)
   \]

2. **Differentiate with Respect to \( x \)**:
   Now, we differentiate \( y = \frac{1}{2} \ln(x + 1) \) with respect to \( x \):
   \[
   \frac{dy}{dx} = \frac{1}{2} \cdot \frac{1}{x + 1}
   \]

3. **Simplify the Result**:
   This simplifies to:
   \[
   \frac{dy}{dx} = \frac{1}{2(x + 1)}
   \]

### Final Answer:
\[
\frac{dy}{dx} = \frac{1}{2(x + 1)}
\]

Would you like a more detailed breakdown of each step, or do you have any questions?

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### Additional Related Questions:
1. How would you differentiate \( y = \ln(x^2 + 1) \)?
2. What is the derivative of \( y = \ln(2x + 3) \)?
3. How would you find the derivative of \( y = \ln(x) \cdot \sqrt{x + 1} \)?
4. What are the derivative rules for logarithmic and square root functions?
5. How can you apply the chain rule when differentiating \( y = \ln \sqrt{x + 1} \)?

### Tip:
When differentiating functions involving both logarithms and roots, rewrite the function using properties of logarithms and exponents first to make differentiation easier.

Find the derivative of y = ln(√(x + 1)).

Find the derivative of y = ln(√(x + 1)) using logarithmic differentiation and the chain rule. This solution provides a step-by-step approach, covering properties of logarithms and square roots, ideal for calculus students in Grades 11-12.

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