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.