The derivative of \( \arctan\left(\frac{1}{x}\right) \) can be found using the chain rule. Here's how we proceed step by step:

1. Let \( y = \arctan\left(\frac{1}{x}\right) \).

2. The derivative of \( \arctan(u) \) with respect to \( u \) is \( \frac{1}{1+u^2} \), so:
   \[
   \frac{d}{dx} \arctan\left( \frac{1}{x} \right) = \frac{1}{1 + \left(\frac{1}{x}\right)^2} \cdot \frac{d}{dx}\left(\frac{1}{x}\right)
   \]

3. Simplify the expression inside the arctangent's derivative:
   \[
   \frac{1}{1 + \left(\frac{1}{x}\right)^2} = \frac{1}{1 + \frac{1}{x^2}} = \frac{x^2}{x^2 + 1}
   \]

4. Now, find the derivative of \( \frac{1}{x} \), which is \( -\frac{1}{x^2} \).

5. Multiply the results:
   \[
   \frac{d}{dx} \arctan\left( \frac{1}{x} \right) = \frac{x^2}{x^2 + 1} \cdot \left(-\frac{1}{x^2}\right)
   \]

6. Simplify:
   \[
   \frac{d}{dx} \arctan\left( \frac{1}{x} \right) = -\frac{1}{x^2 + 1}
   \]

Thus, the derivative of \( \arctan\left(\frac{1}{x}\right) \) is:
\[
\boxed{-\frac{1}{x^2 + 1}}
\]

Would you like further details on the steps? Here are some related questions:

1. What is the derivative of \( \arctan(x) \)?
2. How does the chain rule work with composite functions?
3. Can you apply the chain rule to more complex trigonometric functions?
4. What is the second derivative of \( \arctan\left(\frac{1}{x}\right) \)?
5. How can you apply implicit differentiation to inverse functions?

**Tip**: When differentiating functions with nested parts (like \( \arctan \) with an inner function), always break it down using the chain rule carefully!

Find the derivative of arctan(1/x) using the chain rule in calculus. This problem breaks down the step-by-step solution, including applying the derivative of arctan(u) and simplifying expressions. Ideal for college-level calculus students.

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