To solve the given problems, we need to find the derivatives $\frac{dy}{dx}$ for the functions $y$ defined by the integrals in the exercises. These problems leverage the Fundamental Theorem of Calculus, which states that:

$\frac{d}{dx} \int_a^x f(t) \, dt = f(x),$ where $f(t)$ is continuous over the interval of integration.

45. $y = \int_0^x \sqrt{1 + t^2} \, dt$

Using the Fundamental Theorem of Calculus:

$\frac{dy}{dx} = \sqrt{1 + x^2}.$

46. $y = \int_1^x \frac{1}{t} \, dt, \quad x > 0$

Again, applying the Fundamental Theorem of Calculus:

$\frac{dy}{dx} = \frac{1}{x}.$

Final Answers:

1. For $\#45$, $\frac{dy}{dx} = \sqrt{1 + x^2}$.
2. For $\#46$, $\frac{dy}{dx} = \frac{1}{x}$, valid for $x > 0$.

Would you like a deeper explanation of any step? Let me know!

Follow-Up Questions:

1. What is the Fundamental Theorem of Calculus, and why is it useful here?
2. How can we interpret these derivatives geometrically?
3. What are some practical applications of these integral functions?
4. How does the continuity of $f(t)$ affect the solution?
5. Can this process be extended to variable limits of integration?

Tip: Remember that the Fundamental Theorem of Calculus links derivatives and integrals, which is crucial in solving many problems in calculus!