VC3.32

From Exampleproblems

By definition, $\nabla g=(\frac{\partial g}{\partial x})i+(\frac{\partial g}{\partial y})j+(\frac{\partial g}{\partial z})k\,$

Hence $f\nabla g=[f\frac{\partial g}{\partial x}]i+[f\frac{\partial g}{\partial y}]j+[f\frac{\partial g}{\partial z}]k\,$

and so by definition of divergence, we get

$\mathrm{div}(f\nabla g)=\frac{\partial}{\partial x}[f\frac{\partial g}{\partial x}]+\frac{\partial}{\partial y}[f\frac{\partial g}{\partial y}]+\frac{\partial}{\partial z}[f\frac{\partial g}{\partial z}]\,$

= $\frac{\partial f}{\partial x}\frac{\partial g}{\partial x}+f\frac{\partial^2 g}{\partial x^2}+\frac{\partial f}{\partial y}\frac{\partial g}{\partial y}+f\frac{\partial^2 g}{\partial y^2}+\frac{\partial f}{\partial z}\frac{\partial g}{\partial z}+f\frac{\partial^2 g}{\partial z^2}\,$

= $f[\frac{\partial^2 g}{\partial x^2}+\frac{\partial^2 g}{\partial y^2}+\frac{\partial^2 g}{\partial z^2}]+[\frac{\partial f}{\partial x}\frac{\partial g}{\partial x}+\frac{\partial f}{\partial y}\frac{\partial g}{\partial y}+\frac{\partial f}{\partial z}\frac{\partial g}{\partial z}]\,$

= $f[\frac{\partial^2 }{\partial x^2}+\frac{\partial^2 }{\partial y^2}+\frac{\partial^2 }{\partial z^2}]g+[\frac{\partial f}{\partial x}i+\frac{\partial f}{\partial y}j+\frac{\partial f}{\partial z}k]\cdot[\frac{\partial g}{\partial x}i+\frac{\partial g}{\partial y}j+\frac{\partial g}{\partial z}k]\,$