Answer
$\dfrac{d}{dt}[u(t)+v(t)]=u'(t)+v'(t)$
Work Step by Step
Need to prove $\dfrac{d}{dt}[u(t)+v(t)]=u'(t)+v'(t)$
Let us consider $u(t)=u_1(t)i+u_2(t)j+u_3(t)k$ and
$v(t)=v_1(t)i+v_2(t)j+v_3(t)k$
$\dfrac{d}{dt}[u(t)+v(t)]=\dfrac{d}{dt}[(u_1(t)i+u_2(t)j+u_3(t)k)+(v_1(t)i+v_2(t)j+v_3(t)k)]$
or, $=\dfrac{d}{dt}[(u_1(t)+v_1(t))i+(u_2(t)+v_2(t))j+(u_3(t)+v_3(t))k$
or, $=(u_1'(t)+v_1'(t))i+(u_2'(t)+v_2'(t))j+(u_3'(t)+v_3'(t))k$
or, $=(u_1'(t)i+u_2'(t)j+u_3'(t)k)+(v_1'(t)i+v_2'(t)j+v_3'(t)k)$
or, $=u'(t)+v'(t)$
Hence, the result.
$\dfrac{d}{dt}[u(t)+v(t)]=u'(t)+v'(t)$
