https://file.aiquickdraw.com/mathbot/file/1726767619155jpj9mgis.jpg

To solve the problem in the image, we need to find the slope of the tangent line to the curve given by:

\[
y = \int_{2}^{\sec(x)} \left( \sqrt{\frac{2 - 1}{t}} + 35 \right) dt
\]

at \( x = \frac{\pi}{3} \).

The slope of the tangent line at a given point is the derivative \( \frac{dy}{dx} \). So, let's differentiate the integral function using the **Fundamental Theorem of Calculus** combined with the **chain rule**.

### Step-by-step approach:

1. **Differentiate the integral**:  
   The integral has a variable upper limit of integration \( \sec(x) \), and we apply the Fundamental Theorem of Calculus to differentiate:
   \[
   \frac{d}{dx} \left[ \int_{2}^{\sec(x)} f(t) \, dt \right] = f(\sec(x)) \cdot \frac{d}{dx} [\sec(x)]
   \]
   where \( f(t) = \sqrt{\frac{2 - 1}{t}} + 35 = \frac{1}{\sqrt{t}} + 35 \).

2. **Evaluate at \( x = \frac{\pi}{3} \)**:
   - \( \sec\left(\frac{\pi}{3}\right) = 2 \), since \( \sec(\theta) = \frac{1}{\cos(\theta)} \) and \( \cos\left(\frac{\pi}{3}\right) = \frac{1}{2} \).
   - Now, apply this result and calculate the slope.

I’ll calculate the slope for you.The slope of the tangent line to the curve at \( x = \frac{\pi}{3} \) is:

\[
\frac{\sqrt{3} \left( 2\sqrt{2} + 140 \right)}{2}
\]

Would you like more details or clarifications on this result?

Here are five related questions for further exploration:
1. What is the Fundamental Theorem of Calculus and how does it apply to this problem?
2. How does the chain rule factor into differentiating integrals with variable limits?
3. What is the secant function and how is it evaluated at common angles?
4. Can we approximate the numerical value of the slope at \( x = \frac{\pi}{3} \)?
5. How does the tangent line at a point relate to the derivative of the function?

**Tip**: Always verify the limits of integration and the argument of trigonometric functions to avoid small mistakes.

The slope of the tangent line to the curve y = ∫[2 to sec(x)] (√(2−1)/t + 35) dt at x = π/3 is?

Find the slope of the tangent line to the curve y = ∫[2 to sec(x)] (√(2−1)/t + 35) dt at x = π/3 using the Fundamental Theorem of Calculus and the chain rule. This solution applies advanced calculus techniques, including differentiation of an integral with a variable limit.

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