[b]7.-[/b][math]div\left(fF\right)=fdivF+F\cdot\bigtriangledown f[/math][br][br]Demostración.[br][br]Sea[br][br] [math]div\left(fF\right)=\left(\frac{\partial\left(fF_1\right)}{\partial x},\frac{\partial\left(fF_2\right)}{\partial y},\frac{\partial\left(fF_3\right)}{\partial z}\right)[/math][br][br]Entonces:[br][br] [math]=f\left(\frac{\partial\left(F_1\right)}{\partial x}\right)+F_1\left(\frac{\partial\left(f\right)}{\partial x}\right)+f\left(\frac{\partial\left(F_2\right)}{\partial y}\right)+F_2\left(\frac{\partial\left(f\right)}{\partial y}\right)+f\left(\frac{\partial\left(F_3\right)}{\partial z}\right)+F_3\left(\frac{\partial\left(f\right)}{\partial z}\right)[/math][br][br]Factorizamos y reorganizamos valores iguales:[br][br] [math]=f\left(\frac{\partial\left(F_1\right)}{\partial x}+\frac{\partial\left(F_2\right)}{\partial y}+\frac{\partial\left(F_3\right)}{\partial z}\right)+\left(F_1\frac{\partial\left(f\right)}{\partial x}+F_2\frac{\partial\left(f\right)}{\partial x}+F_3\frac{\partial\left(f\right)}{\partial x}\right)[/math][br][br]En consecuencia:[br][br] [math]=fdivF+F\cdot\bigtriangledown f[/math][br][br]Por lo tanto, hemos demostrado que [math]div\left(fF\right)=fdivF+F\cdot\bigtriangledown f[/math].[br]