The problem asks to determine $f'(x)$ and $f'(1)$, where the function $f(x)$ is defined as follows:

$f(x) = \int_1^{\sqrt{x^2}} \frac{2x^2}{t} \, dt.$

Step 1: Differentiating $f(x)$ using the Fundamental Theorem of Calculus

The function $f(x)$ is defined as a definite integral with a variable upper limit. To differentiate it with respect to $x$, we apply the Leibniz rule for differentiation under the integral sign, particularly when the upper limit depends on $x$.

General Rule:

If $F(x) = \int_{a}^{g(x)} h(t) \, dt$, then:

$F'(x) = h(g(x)) \cdot g'(x).$

Step 2: Application to $f(x)$

In this case:

• $h(t) = \frac{2x^2}{t}$,
• The lower limit $1$ is constant (its derivative is $0$),
• The upper limit $g(x) = \sqrt{x^2} = |x|$.

Since $x \in \{1, \infty\}$ (i.e., $x \geq 1$), we have $|x| = x$. Thus $g(x) = x$, and its derivative is:

$g'(x) = \frac{d}{dx} x = 1.$

Using the Leibniz rule:

$f'(x) = \frac{2x^2}{g(x)} \cdot g'(x).$

Substitute $g(x) = x$ and $g'(x) = 1$:

$f'(x) = \frac{2x^2}{x} \cdot 1.$

Simplify:

$f'(x) = 2x.$

Step 3: Evaluating $f'(1)$

To find $f'(1)$, substitute $x = 1$ into $f'(x)$:

$f'(1) = 2(1) = 2.$

Final Answers:

1. $f'(x) = 2x$,
2. $f'(1) = 2$.

Do you need further clarification or a detailed explanation of any step?

Here are some follow-up questions to deepen understanding:

1. What is the Fundamental Theorem of Calculus, and how is it applied here?
2. How does the Leibniz rule generalize differentiation of definite integrals?
3. Why does $\sqrt{x^2}$ simplify to $x$ for $x \geq 1$?
4. How would the result change if the lower limit were also variable?
5. Can we confirm $f(x)$ is differentiable on its domain $\{1, \infty\}$?

Tip: The Fundamental Theorem of Calculus simplifies evaluating derivatives of integrals with variable limits.