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?

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.