Chapter 5: Integrals - Practice Exercises - Page 309: 79

$-\sqrt {1+t^2}$

Step 1. Based on the Fundamental Theorem of Calculus, if we define $F(x)=\int_a^x f(t)dt$; then, we have $F'(x)=f(x)$ Step 2. In the case of the exercise, we have $y=\int_x^1\sqrt {1+t^2}dt=-\int_1^x\sqrt {1+t^2}dt=\int_1^x(-\sqrt {1+t^2})dt$ Thus, we can see that $f(t)=-\sqrt {1+t^2}$ and $y'=f(x)=-\sqrt {1+t^2}$