The given problem involves finding the derivative of the function $y$ defined as:

$y = \int_{\sin x}^x \sqrt{1 + t^2} \, dt$

Solution:

The Fundamental Theorem of Calculus, Part 1, helps us evaluate derivatives of definite integrals with variable limits of integration. Here's how to proceed:

Step 1: Identify variable limits and split contributions

The integral has both an upper and a lower limit as functions of $x$. To differentiate $y$, we use the rule:

$\frac{d}{dx} \int_{u(x)}^{v(x)} f(t) \, dt = f(v(x)) \cdot v'(x) - f(u(x)) \cdot u'(x).$

Here:

• $f(t) = \sqrt{1 + t^2}$,
• Upper limit $v(x) = x$, so $v'(x) = 1$,
• Lower limit $u(x) = \sin x$, so $u'(x) = \cos x$.

Step 2: Apply the rule

$\frac{dy}{dx} = f(v(x)) \cdot v'(x) - f(u(x)) \cdot u'(x),$ $\frac{dy}{dx} = \sqrt{1 + x^2} \cdot 1 - \sqrt{1 + (\sin x)^2} \cdot \cos x.$

Final Answer:

$\frac{dy}{dx} = \sqrt{1 + x^2} - \cos x \cdot \sqrt{1 + \sin^2 x}.$

Would you like a deeper explanation of any step or more details?

Related Questions:

1. What is the Fundamental Theorem of Calculus and how is it applied in general?
2. How would the solution change if the limits of integration were constants instead of functions of $x$?
3. Why is the derivative of the integral's lower limit term negative in this case?
4. Can you explain the geometric interpretation of the Fundamental Theorem of Calculus?
5. How does one compute derivatives of integrals with two variable limits?

Tip:

When applying the Fundamental Theorem of Calculus to functions with variable limits, always treat the limits as functions of $x$ and use the chain rule to differentiate them.