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To find the derivative of the function \( f(x) = \frac{x}{\sqrt{x+1}} \), we can use the Quotient Rule, which states that if \( f(x) = \frac{u(x)}{v(x)} \), then \( f'(x) = \frac{u'v - uv'}{v^2} \).

In this case, we define:
- \( u(x) = x \)
- \( v(x) = \sqrt{x+1} \)

### Step 1: Find \( u' \) and \( v' \)
1. \( u(x) = x \) so \( u' = 1 \).
2. \( v(x) = \sqrt{x+1} = (x+1)^{1/2} \). To differentiate, we use the chain rule:
   \[
   v' = \frac{1}{2}(x+1)^{-1/2} \cdot 1 = \frac{1}{2\sqrt{x+1}}.
   \]

### Step 2: Apply the Quotient Rule
Now, substitute \( u \), \( u' \), \( v \), and \( v' \) into the Quotient Rule formula:
\[
f'(x) = \frac{u'v - uv'}{v^2}.
\]

Substitute the values:
\[
f'(x) = \frac{(1) \cdot \sqrt{x+1} - x \cdot \frac{1}{2\sqrt{x+1}}}{(\sqrt{x+1})^2}.
\]

### Step 3: Simplify the Expression
1. The denominator simplifies as follows:
   \[
   (\sqrt{x+1})^2 = x + 1.
   \]

2. For the numerator:
   \[
   \sqrt{x+1} - \frac{x}{2\sqrt{x+1}} = \frac{2(x+1) - x}{2\sqrt{x+1}} = \frac{x+2}{2\sqrt{x+1}}.
   \]

So the derivative is:
\[
f'(x) = \frac{x+2}{2(x+1)^{3/2}}.
\]

Would you like further details or clarification on any steps?

### Related Questions

1. How can we verify this derivative using a different method?
2. What is the second derivative of \( f(x) \)?
3. How does this function behave as \( x \) approaches infinity?
4. Can we find the critical points of \( f(x) \)?
5. What is the behavior of \( f(x) \) as \( x \) approaches -1?

### Tip
When using the Quotient Rule, it helps to simplify each derivative term separately before combining them.

Find the derivative of f(x) = x / sqrt(x + 1).

Find the derivative of f(x) = x / sqrt(x + 1) using the Quotient Rule. This solution provides a step-by-step process, covering differentiation techniques including the Quotient Rule and Chain Rule. Suitable for students in Grades 11-12.

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