## Calculus: Early Transcendentals 8th Edition

$\dfrac{d}{dt}[u(t) \times v(t)]=u'(t)v(t)+u(t)v'(t)$
Our aim is to prove: $\dfrac{d}{dt}[u(t) \times v(t)]=u'(t)v(t)+u(t)v'(t)$ ...(1) Suppose $u(t)=u_1(t)i+u_2(t)j+u_3(t)k$ and $v(t)=v_1(t)i+v_2(t)j+v_3(t)k$ Take the left side of the equation (1). $\dfrac{d}{dt}[u(t) \times v(t)]=\dfrac{d}{dt}[\lt u_1(t),u_2(t),u_3(t) \gt \times \lt v_1(t),v_2(t),v_3(t) \gt]$ $=\dfrac{d}{dt}[\begin {vmatrix} i&j&k\\u_1(t)&u_2(t)&u_3(t)\\v_1(t)&v_2(t)&v_3(t) \end{vmatrix}]$ $=[(u_2'(t)v_3(t)+u_2(t)v_3'(t))-(u_3'(t)v_2(t)+u_3(t)v_2'(t)]i+[(u_3'(t)v_1(t)+u_3(t)v_1'(t))-(u_1'(t)v_3(t)+u_1(t)v_3'(t)]j+[(u_1'(t)v_2(t)+u_1(t)v_2'(t))-(u_1'(t)v_2(t)+u_2'(t)v_1(t)]k$ Take the right side of the equation (1). $u'(t)v(t)+u(t)v'(t)=\lt u_1'(t),u_2'(t),u_3'(t) \gt \lt v_1(t),v_2(t),v_3(t) \gt +\lt u_1(t),u_2(t),u_3(t) \gt \lt v_1'(t),v_2'(t),v_3'(t) \gt$ $=[(u_2'(t)v_3(t)+u_2(t)v_3'(t))-(u_3'(t)v_2(t)+u_3(t)v_2'(t)]i+[(u_3'(t)v_1(t)+u_3(t)v_1'(t))-(u_1'(t)v_3(t)+u_1(t)v_3'(t)]j+[(u_1'(t)v_2(t)+u_1(t)v_2'(t))-(u_1'(t)v_2(t)+u_2'(t)v_1(t)]k$ Hence, $\dfrac{d}{dt}[u(t) \times v(t)]=u'(t)v(t)+u(t)v'(t)$