To differentiate the function:

\[
f(x) = \frac{1}{x \sqrt{x+1}}
\]

we'll use the quotient rule, which states that if \( f(x) = \frac{g(x)}{h(x)} \), then

\[
f'(x) = \frac{g'(x) h(x) - g(x) h'(x)}{[h(x)]^2}.
\]

Let \( g(x) = 1 \) and \( h(x) = x \sqrt{x+1} \). Then:

1. **Find \( g'(x) \):**  
   \[
   g'(x) = 0.
   \]

2. **Find \( h'(x) \):**  
   First, we differentiate \( h(x) = x \sqrt{x+1} \) using the product rule, \( (uv)' = u'v + uv' \), where \( u = x \) and \( v = \sqrt{x+1} \):
   \[
   u' = 1 \quad \text{and} \quad v' = \frac{1}{2\sqrt{x+1}}.
   \]
   Thus,
   \[
   h'(x) = (1)(\sqrt{x+1}) + (x)\left(\frac{1}{2\sqrt{x+1}}\right) = \sqrt{x+1} + \frac{x}{2\sqrt{x+1}}.
   \]

3. **Apply the quotient rule:**
   \[
   f'(x) = \frac{(0)(x \sqrt{x+1}) - (1)(\sqrt{x+1} + \frac{x}{2\sqrt{x+1}})}{(x \sqrt{x+1})^2}.
   \]
   Simplifying, we get:
   \[
   f'(x) = -\frac{\sqrt{x+1} + \frac{x}{2\sqrt{x+1}}}{x^2 (x+1)}.
   \]
   Combining the terms in the numerator:
   \[
   f'(x) = -\frac{2(x+1) + x}{2x^2 (x+1)\sqrt{x+1}} = -\frac{3x + 2}{2x^2 (x+1)\sqrt{x+1}}.
   \]

The derivative of the function is:

\[
f'(x) = -\frac{3x + 2}{2x^2 (x+1)\sqrt{x+1}}.
\]

Would you like further details or have any questions? Here are some related problems to consider:

1. Find the derivative of \( \frac{1}{\sqrt{x}(x+1)} \).
2. Differentiate \( \frac{x^2}{\sqrt{x+1}} \).
3. Find the derivative of \( \frac{1}{x^2 \sqrt{x+1}} \).
4. Differentiate \( \frac{1}{x \sqrt{x-1}} \).
5. Find the derivative of \( \frac{\ln(x)}{x \sqrt{x+1}} \).

**Tip:** When differentiating functions involving products or quotients of multiple terms, consider breaking them down using rules like the product or quotient rule for easier calculations.

Learn how to differentiate the function 1/(x*sqrt(x+1)) using calculus techniques like the quotient and product rules. This detailed solution includes key steps and formulas, suitable for advanced high school or college-level calculus students.

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