To solve the given problem, we use Part 1 of the Fundamental Theorem of Calculus (FTC Part 1), which states:

$\frac{d}{dx} \left( \int_{a}^{g(x)} f(t) \, dt \right) = f(g(x)) \cdot g'(x)$

Here:

• $f(t)$ is the integrand function,
• $g(x)$ is the upper limit of the integral,
• $a$ is a constant lower limit.

Given:

$h(x) = \int_{2}^{1/x} \arctan(3t) \, dt$

We need to find $h'(x)$.

Steps:

1. Recognize the form of the integral:

   • The lower limit $2$ is constant.
   • The upper limit is $g(x) = \frac{1}{x}$.

2. Apply FTC Part 1: To differentiate $h(x)$, apply the rule: $\frac{d}{dx} \left( \int_{a}^{g(x)} f(t) \, dt \right) = f(g(x)) \cdot g'(x).$ Here:

   • $f(t) = \arctan(3t)$,
   • $g(x) = \frac{1}{x}$,
   • $g'(x) = -\frac{1}{x^2}$.

3. Substitute into the formula: Replace $t$ with $g(x) = \frac{1}{x}$ in $f(t)$: $f\left( g(x) \right) = \arctan\left( 3 \cdot \frac{1}{x} \right) = \arctan\left( \frac{3}{x} \right).$

   Multiply by $g'(x) = -\frac{1}{x^2}$: $h'(x) = \arctan\left( \frac{3}{x} \right) \cdot \left( -\frac{1}{x^2} \right).$

4. Simplify: $h'(x) = -\frac{1}{x^2} \cdot \arctan\left( \frac{3}{x} \right).$

Final Answer:

$h'(x) = -\frac{1}{x^2} \arctan\left( \frac{3}{x} \right).$

Let me know if you would like a step-by-step explanation of any part or further details!

Related Questions:

1. What is the Fundamental Theorem of Calculus, and how is it applied?
2. How do you handle integrals with variable limits?
3. What is the derivative of an arctangent function, $\arctan(u)$?
4. How does the chain rule apply in the FTC Part 1?
5. How can this method be generalized to other integrals with different limits?

Tip: When differentiating definite integrals with variable limits, always account for the derivative of the upper (or lower) limit using the chain rule!