To find the derivative of the function \( f(x) = \sqrt{\frac{x}{x+1}} \), we can proceed as follows:

### Step 1: Rewrite the function in a more convenient form
Express the square root as a power:
\[
f(x) = \left(\frac{x}{x+1}\right)^{\frac{1}{2}}
\]

### Step 2: Apply the chain rule
Let \( u(x) = \frac{x}{x+1} \). Then, \( f(x) = u(x)^{\frac{1}{2}} \). According to the chain rule:
\[
f'(x) = \frac{1}{2} \cdot u(x)^{-\frac{1}{2}} \cdot u'(x)
\]
This simplifies to:
\[
f'(x) = \frac{1}{2} \cdot \frac{1}{\sqrt{u(x)}} \cdot u'(x)
\]

### Step 3: Differentiate \( u(x) \)
Now, we need to find the derivative of \( u(x) = \frac{x}{x+1} \). Use the quotient rule:
\[
u'(x) = \frac{(x+1) \cdot 1 - x \cdot 1}{(x+1)^2} = \frac{1}{(x+1)^2}
\]

### Step 4: Substitute \( u'(x) \) back into the derivative of \( f(x) \)
\[
f'(x) = \frac{1}{2} \cdot \frac{1}{\sqrt{\frac{x}{x+1}}} \cdot \frac{1}{(x+1)^2}
\]

### Step 5: Simplify the expression
Simplify the expression by dealing with the square root:
\[
f'(x) = \frac{1}{2} \cdot \frac{\sqrt{x+1}}{\sqrt{x}} \cdot \frac{1}{(x+1)^2} = \frac{1}{2} \cdot \frac{\sqrt{x+1}}{\sqrt{x} \cdot (x+1)^2}
\]

This gives the final derivative:
\[
f'(x) = \frac{\sqrt{x+1}}{2x^{\frac{1}{2}}(x+1)^{\frac{5}{2}}}
\]

This is the derivative of \( f(x) = \sqrt{\frac{x}{x+1}} \).

Would you like a more detailed explanation or have any questions?

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**Related questions:**

1. How do you apply the quotient rule in differentiation?
2. What are some common derivatives involving square roots?
3. Can the derivative of \( \sqrt{\frac{x}{x+1}} \) be simplified further?
4. How does the chain rule work in calculus?
5. What is the derivative of \( \sqrt{x^2 + 1} \)?

**Tip:** Always rewrite expressions in a more convenient form before differentiating to simplify the process.

Learn how to find the derivative of the square root function f(x) = sqrt(x/(x+1)) step-by-step. This problem involves calculus, differentiation, and the application of the chain rule and quotient rule in calculus. Suitable for college-level students and anyone studying calculus.

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