Find the derivative of the function.
y = 	3
√x + 6

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.

Discover how to find the derivative of the function y = 3/sqrt(x) + 6 using differentiation rules. This problem involves applying the power rule and understanding how constants affect derivatives. Suitable for advanced high school and college students.

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