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Let's solve the derivative of the given function step by step. The function is:

**g(x) = (√x + 1) / √x**

This is a quotient, so we use the **quotient rule** for derivatives:

\[
\left( \frac{u}{v} \right)' = \frac{u'v - uv'}{v^2}
\]

Here:
- \( u = \sqrt{x} + 1 \)
- \( v = \sqrt{x} \)

### Step 1: Compute \( u' \) and \( v' \)
- \( u = \sqrt{x} + 1 \implies u' = \frac{1}{2\sqrt{x}} \) (derivative of \( \sqrt{x} \))
- \( v = \sqrt{x} \implies v' = \frac{1}{2\sqrt{x}} \)

### Step 2: Apply the quotient rule
Substitute into the quotient rule formula:

\[
g'(x) = \frac{u'v - uv'}{v^2}
\]

Substitute \( u, u', v, v' \):
\[
g'(x) = \frac{\left(\frac{1}{2\sqrt{x}}\right)(\sqrt{x}) - (\sqrt{x} + 1)\left(\frac{1}{2\sqrt{x}}\right)}{(\sqrt{x})^2}
\]

### Step 3: Simplify the numerator
- The first term: \( \frac{1}{2\sqrt{x}} \cdot \sqrt{x} = \frac{1}{2} \)
- The second term: \( (\sqrt{x} + 1) \cdot \frac{1}{2\sqrt{x}} = \frac{\sqrt{x}}{2\sqrt{x}} + \frac{1}{2\sqrt{x}} = \frac{1}{2} + \frac{1}{2\sqrt{x}} \)

Combine these:

\[
g'(x) = \frac{\frac{1}{2} - \left(\frac{1}{2} + \frac{1}{2\sqrt{x}}\right)}{x}
\]

Simplify further:
\[
g'(x) = \frac{\frac{1}{2} - \frac{1}{2} - \frac{1}{2\sqrt{x}}}{x}
\]

\[
g'(x) = \frac{-\frac{1}{2\sqrt{x}}}{x}
\]

### Step 4: Final simplification
Simplify the fraction:

\[
g'(x) = \frac{-1}{2x\sqrt{x}}
\]

### Final Answer:
\[
g'(x) = -\frac{1}{2x\sqrt{x}}
\]

Let me know if you want further details or clarifications! 

### Related Questions:
1. How do you calculate the derivative of square root functions?
2. What is the general quotient rule formula and when is it used?
3. How do you simplify complex fractions like in this example?
4. Can you apply the chain rule to a similar problem?
5. How does the derivative of a quotient behave when the denominator is constant?

**Tip:** Always simplify both numerator and denominator separately before combining for better clarity.

Learn how to calculate the derivative of g(x) = (√x + 1)/√x using the quotient rule. This step-by-step solution covers core calculus concepts like derivatives and simplifies the function for better understanding. Ideal for Grades 11-12.

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