To determine the correct statements in this problem, let’s go through each option one by one. The function provided is:

$y = \ln(2x + 3)$

Step 1: Find the Derivative $y'$

Using the chain rule, we can find the derivative of $y = \ln(2x + 3)$:

$y' = \frac{d}{dx} \ln(2x + 3) = \frac{1}{2x + 3} \cdot 2 = \frac{2}{2x + 3}$

Step 2: Evaluate $y'$ at Given Points

1. When $x_0 = 1$: $y'(1) = \frac{2}{2 \cdot 1 + 3} = \frac{2}{5}$

2. When $x_0 = 0$: $y'(0) = \frac{2}{2 \cdot 0 + 3} = \frac{2}{3}$

Step 3: Analyze Each Statement

1. Statement 1: $y'$ at $x_0 = 1$ is $\frac{2}{5}$.
   True, as we calculated $y'(1) = \frac{2}{5}$.

2. Statement 2: $y'$ at $x_0 = 0$ is $\frac{1}{3}$.
   False, since $y'(0) = \frac{2}{3}$, not $\frac{1}{3}$.

3. Statement 3: $y'$ at $x_0 = 1$ is $\frac{1}{5}$.
   False, as $y'(1) = \frac{2}{5}$, not $\frac{1}{5}$.

4. Statement 4: $y'$ at $x_0 = 0$ is $\frac{2}{3}$.
   True, since $y'(0) = \frac{2}{3}$.

5. Statement 5: Approximation at $x_0 = 0$ in the form $y = \frac{2}{3} \Delta x + \alpha(\Delta x)$, where $\alpha(\Delta x)$ is a higher-order term.
   True, as this correctly represents the linear approximation with $y' = \frac{2}{3}$ at $x_0 = 0$.

6. Statement 6: Approximation at $x_0 = 0$ in the form $y = \frac{2}{3} + \alpha(\Delta x)$.
   False, since the constant $\frac{2}{3}$ represents $y'(0) \cdot \Delta x$, not the function itself at $x_0 = 0$.

Conclusion

The correct statements are 1, 4, and 5.

Would you like further details on any of these calculations, or have any additional questions?

Related Questions to Explore

1. How is the chain rule applied in finding derivatives of logarithmic functions?
2. What is the process of linear approximation, and why is it useful?
3. How can Taylor series expansions help in approximating functions around a point?
4. Why do we use higher-order terms like $\alpha(\Delta x)$ in approximations?
5. How does differentiating composite functions differ from differentiating simpler functions?

Tip

For logarithmic derivatives, remember that $\frac{d}{dx} \ln(f(x)) = \frac{f'(x)}{f(x)}$ – a useful shortcut for quick calculations.