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We are tasked with finding \( f'(1) \) for the function \( f(x) = (1 + x^2) \cdot \tan^{-1}(x) \), and then rounding the answer to two decimal places.

Let's begin by applying the **product rule** to differentiate \( f(x) = (1 + x^2) \cdot \tan^{-1}(x) \). The product rule is:

\[
f'(x) = u'(x) \cdot v(x) + u(x) \cdot v'(x)
\]

Here, let:
- \( u(x) = 1 + x^2 \)
- \( v(x) = \tan^{-1}(x) \)

Now, let's find the derivatives of \( u(x) \) and \( v(x) \):

1. **Derivative of \( u(x) = 1 + x^2 \)**:
   \[
   u'(x) = 2x
   \]

2. **Derivative of \( v(x) = \tan^{-1}(x) \)**:
   \[
   v'(x) = \frac{1}{1 + x^2}
   \]

Now we can apply the product rule:
\[
f'(x) = (2x) \cdot \tan^{-1}(x) + (1 + x^2) \cdot \frac{1}{1 + x^2}
\]

Simplifying the second term:
\[
f'(x) = 2x \cdot \tan^{-1}(x) + 1
\]

### Step 2: Evaluate \( f'(1) \)
Now, substitute \( x = 1 \) into the expression for \( f'(x) \):

\[
f'(1) = 2(1) \cdot \tan^{-1}(1) + 1
\]

Since \( \tan^{-1}(1) = \frac{\pi}{4} \) (or approximately \( 0.7854 \)):

\[
f'(1) = 2 \cdot 0.7854 + 1 = 1.5708 + 1 = 2.5708
\]

### Step 3: Rounding
Finally, rounding \( 2.5708 \) to two decimal places:

\[
f'(1) \approx 2.57
\]

Thus, \( f'(1) \approx 2.57 \).

Would you like any further details or have questions?

Here are five questions that expand on this topic:
1. How do you apply the product rule to more complex functions?
2. What is the geometric interpretation of the inverse tangent function \( \tan^{-1}(x) \)?
3. How does knowing the value of \( \tan^{-1}(1) \) help in trigonometric differentiation problems?
4. How would the solution change if \( f(x) \) were defined with a different trigonometric function?
5. What techniques are useful for differentiating functions that combine trigonometric and polynomial terms?

**Tip**: Always simplify your derivatives as much as possible before substituting values.

Find f'(1) for f(x) = (1 + x^2) * tan^(-1)(x). Round your answer to two decimal places.

Learn how to find the derivative f'(1) for the function f(x) = (1 + x^2) * tan^(-1)(x). This solution uses the product rule and involves differentiating trigonometric and polynomial terms. Suitable for undergraduate calculus students.

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