Question

PROVE that d/du(AxB) = Ax(dB/du) + (dA/du)xB where A,B are differentiated functions of u.

Answer 1

Question #22405 Prove that $d / du(AB) = A(dB / du) + (dA / du)B$ where A,B are differentiated functions of u.

Solution. Really, $\frac{AB(u + \Delta u) - AB(u)}{\Delta u} = \frac{A(u + \Delta u)(B(u + \Delta u) - B(u)) + B(u)(A(u + \Delta u) - A}{\Delta u}$.

$A(u + \Delta u)\frac{B(u + \Delta u) - B(u)}{\Delta u} + B(u)\frac{A(u + \Delta u) - A(u)}{\Delta u}$. Passing to the limit as $\Delta u \to 0$ and using the fact that $A$ is continuous, due to it is differentiated one gets the desired equality.