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.