To analyze the function $y = \ln(\ln(x))$, let's go through a few key aspects:

1. Domain:

The function $\ln(x)$ is only defined for $x > 0$. Additionally, for $y = \ln(\ln(x))$ to be defined, the inner function $\ln(x)$ must also be greater than 0:

$\ln(x) > 0 \implies x > 1$

Thus, the domain of $y = \ln(\ln(x))$ is $x > 1$.

2. Range:

Since the natural logarithm function $\ln(x)$ can take on any positive value for $x > 1$, the function $\ln(\ln(x))$ can take any real value. Hence, the range of $y = \ln(\ln(x))$ is:

$(-\infty, +\infty)$

3. Derivative:

To find the derivative of $y = \ln(\ln(x))$, we use the chain rule:

[ \frac{dy}{dx} = \frac{d}{dx} [\ln(\ln(x))] = \frac{1}{\ln(x)} \cdot \frac{d}{dx} [\ln(x)] = \frac{1}{\ln(x)} \cdot \frac{1}{x} = \frac{1}{x \ln(x)} ]

So, the derivative of the function is:

$\frac{dy}{dx} = \frac{1}{x \ln(x)}$

4. Critical Points:

To find critical points, we set the derivative to zero:

$\frac{1}{x \ln(x)} = 0$

Since the expression $\frac{1}{x \ln(x)}$ is never zero for $x > 1$, there are no critical points where the derivative is zero. However, we need to check the behavior at the boundary of the domain, $x \to 1^+$.

5. Behavior at the Boundary:

As $x \to 1^+$, we have:

$\ln(x) \to 0^+ \implies \ln(\ln(x)) \to -\infty$

Thus, the function decreases to negative infinity as $x$ approaches 1 from the right.

Conclusion:

• Domain: $x > 1$
• Range: $(-\infty, +\infty)$
• Derivative: $\frac{dy}{dx} = \frac{1}{x \ln(x)}$

Would you like more details or have any questions? Here are some related questions you might be interested in:

1. What is the second derivative of $y = \ln(\ln(x))$?
2. How do you find the concavity and inflection points of $y = \ln(\ln(x))$?
3. How would you plot the function $y = \ln(\ln(x))$?
4. How does the function behave as $x \to \infty$?
5. How can you apply logarithmic differentiation to more complex functions?

Tip: Remember that logarithmic functions have unique properties when it comes to growth rates and asymptotic behavior!