Chapter 14 - Calculus of Vector-Valued Functions - 14.2 Calculus of Vector-Valued Functions - Exercises - Page 720: 10

$$ b'(t) =\langle 3e^{3t-4} ,-e^{6-t}, -(t+1)^{-2} \rangle.$$

Since $ b(t)=\langle e^{3t-4},e^{6-t}, (t+1)^{-1} \rangle $, then by using the chain rule, the derivative $ b'(t)$ is given by $$ b'(t)= \langle e^{3t-4} (3t-4)',e^{6-t}(6-t)', -(t+1)^{-2} \rangle\\ =\langle 3e^{3t-4} ,-e^{6-t}, -(t+1)^{-2} \rangle.$$