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We are given the following functions:

\[
f'(x) = \frac{1}{1 + x^2}
\]
and
\[
g(x) = \tan{x}
\]

We need to compute \((f \circ g)'(x)\), which means we are looking for the derivative of the composite function \( f(g(x)) \). By the chain rule:

\[
(f \circ g)'(x) = f'(g(x)) \cdot g'(x)
\]

### Step 1: Compute \( g'(x) \)
Since \( g(x) = \tan{x} \), the derivative is:

\[
g'(x) = \sec^2{x}
\]

### Step 2: Substitute into the chain rule
Now, substitute \( f'(x) \) and \( g(x) \) into the chain rule expression:

\[
(f \circ g)'(x) = f'(g(x)) \cdot g'(x)
\]

We know \( f'(x) = \frac{1}{1 + x^2} \), so:

\[
f'(g(x)) = \frac{1}{1 + (\tan{x})^2}
\]

Using the identity \( 1 + \tan^2{x} = \sec^2{x} \), we get:

\[
f'(g(x)) = \frac{1}{\sec^2{x}}
\]

Thus:

\[
(f \circ g)'(x) = \frac{1}{\sec^2{x}} \cdot \sec^2{x} = 1
\]

### Final Answer:
The correct answer is \( \boxed{1} \), which corresponds to option \( b \).

Would you like more details on any part of this solution?

Here are 5 related questions you might find interesting:
1. How does the chain rule apply to compositions of three functions?
2. What is the derivative of \( \cot{x} \)?
3. Can the chain rule be applied to implicit differentiation problems?
4. How do trigonometric identities simplify derivative calculations?
5. What are the applications of composite function derivatives in physics?

**Tip:** Always remember key trigonometric identities like \( 1 + \tan^2{x} = \sec^2{x} \) as they often simplify calculations in calculus.

If f'(x) = 1/(1 + x^2), g(x) = tan(x), then (f ∘ g)'(x) = … ?

Learn how to apply the chain rule to compute the derivative of a composite function involving trigonometric functions. In this problem, f'(x) = 1/(1 + x^2) and g(x) = tan(x). We use trigonometric identities and the chain rule to find (f ∘ g)'(x).

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