## Thomas' Calculus 13th Edition

$\dfrac{d(u \pm v)}{dt}=\dfrac{d}{dt}u \pm \dfrac{d}{dt}v$
Let us consider $u=\lt u_x(t), u_y(t), u_z(t) \gt$ and $v=\lt v_x(t), v_y(t), v_z(t) \gt$ Need to use product rule such as:$\dfrac{d(u \pm v)}{dt}=\lt \dfrac{d}{dt}(u_xt \pm v_xt), \dfrac{d}{dt}(u_yt \pm v_yt),\dfrac{d}{dt}(u_zt \pm v_zt) \\ \implies \dfrac{d(u \pm v)}{dt}=\lt \dfrac{d}{dt}(u_xt) , \dfrac{d}{dt}(u_y t),\dfrac{d}{dt}(u_z t)\gt \pm \lt \dfrac{d}{dt}(v_xt) , \dfrac{d}{dt}(v_y t),\dfrac{d}{dt}(v_z t)\gt = \dfrac{d}{dt}(u) \pm \dfrac{d}{dt}(v)$