The problem here is to calculate the derivative of the function $f(x) = \sqrt{6\pi} \cdot \int_{1}^{x} \frac{dr}{\sin(r^2)}$ at the point $x = \frac{\pi}{6}$.

To solve this, let’s proceed step-by-step:

1. Identify the Integral Form: The function $f(x)$ is defined in terms of an integral with a variable upper limit of $x$, which suggests we can use the Fundamental Theorem of Calculus to find the derivative.

2. Apply the Fundamental Theorem of Calculus: According to the Fundamental Theorem of Calculus, if $F(x) = \int_{a}^{x} g(r) \, dr$, then $F'(x) = g(x)$. In this case: $f'(x) = \sqrt{6\pi} \cdot \frac{1}{\sin(x^2)}$

3. Evaluate at $x = \frac{\pi}{6}$: Substitute $x = \frac{\pi}{6}$ into $f'(x)$: $f'\left(\frac{\pi}{6}\right) = \sqrt{6\pi} \cdot \frac{1}{\sin\left(\left(\frac{\pi}{6}\right)^2\right)}$ Simplify inside the sine function: $f'\left(\frac{\pi}{6}\right) = \sqrt{6\pi} \cdot \frac{1}{\sin\left(\frac{\pi^2}{36}\right)}$

Therefore, the derivative of $f(x)$ at $x = \frac{\pi}{6}$ is: $f'\left(\frac{\pi}{6}\right) = \frac{\sqrt{6\pi}}{\sin\left(\frac{\pi^2}{36}\right)}$

Would you like more details on any step, or have any questions?

Here are some additional questions to explore:

1. What is the Fundamental Theorem of Calculus and how does it apply to integrals with variable limits?
2. How does the derivative of a function defined as an integral compare to the derivative of standard functions?
3. What are some common methods to evaluate derivatives at specific points?
4. How can trigonometric simplifications aid in solving integrals or derivatives?
5. What are potential applications of this derivative in real-world problems?

Tip: When dealing with derivatives of integrals, always check if the Fundamental Theorem of Calculus can simplify the process.