To find the derivative of the function $y = \frac{3}{\sqrt{x}} + 6$, we need to apply the rules of differentiation.

The function can be rewritten as:

$y = 3x^{-\frac{1}{2}} + 6$

Now, we'll differentiate term by term:

1. The derivative of $3x^{-\frac{1}{2}}$ with respect to $x$ is: $\frac{d}{dx} \left( 3x^{-\frac{1}{2}} \right) = 3 \cdot \left(-\frac{1}{2}\right) \cdot x^{-\frac{1}{2} - 1} = -\frac{3}{2}x^{-\frac{3}{2}}$

2. The derivative of the constant $6$ is: $\frac{d}{dx}(6) = 0$

So, the derivative of the function $y$ is:

$\frac{dy}{dx} = -\frac{3}{2}x^{-\frac{3}{2}}$

This can also be expressed as:

$\frac{dy}{dx} = -\frac{3}{2\sqrt{x^3}}$

Would you like any additional details or have any questions? Here are five related questions you might consider:

1. What is the second derivative of $y = \frac{3}{\sqrt{x}} + 6$?
2. How does the derivative behave as $x$ approaches zero?
3. What is the slope of the function at a specific point, say $x = 4$?
4. How would you integrate $-\frac{3}{2\sqrt{x^3}}$ to find the original function?
5. Can you explore the concavity of the function $y = \frac{3}{\sqrt{x}} + 6$ using the second derivative?

Tip: Always remember that constants like 6 in differentiation disappear because the rate of change of a constant is zero.