Chapter 14 - Calculus of Vector-Valued Functions - 14.2 Calculus of Vector-Valued Functions - Exercises - Page 720: 8

$$ r'(t)=\left\langle -1,\frac{2}{\sqrt{t}},0\right \rangle.$$

Since $ r(t)=\langle 7-t,4\sqrt{t}, 8 \rangle $, then the derivative $ r'(t)$ is given by $$ r'(t)=\left\langle -1,\frac{4}{2\sqrt{t}},0\right \rangle =\left\langle -1,\frac{2}{\sqrt{t}},0\right \rangle.$$