How do I prove $\operatorname{div} (f \nabla g) = f\Delta g + \nabla f \cdot\nabla g$?

Question

Let $f\in C^1$ and $g\in C^2$ be scalar functions. How do I prove the identity

$$\operatorname{div} (f \nabla g) = f\Delta g + \nabla f \cdot\nabla g$$

?

Answer 1

It's all just a matter of plugging things into the definitions. Recall the following definitions:

$$\operatorname{div}(F)=\sum_{j=1}^n\partial_jF_j,$$

$$\nabla G=(\partial_1 G,\dots,\partial_n G),$$

$$\Delta G=\sum_{j=1}^n\partial_j^2G,$$

$$\langle x,y \rangle = \sum_{j=1}^n x_jy_j.$$

The left hand side then becomes

$$\operatorname{div}(f\cdot\nabla g)=\operatorname{div}(f\cdot\partial_1g,\dots,f\cdot\partial_ng)=\sum_{j=1}^n\partial_j(f\cdot\partial_jg)=\sum_{j=1}^n\bigl(\partial_jf\cdot\partial_jg+f\cdot\partial_j^2g\bigr).$$

One the other hand we can compute the right hand side as

$$f\cdot\Delta g+\langle \nabla f,\nabla g\rangle=\sum_{j=1}^nf\cdot\partial_j^2g+\sum_{j=1}^n\partial_jf\cdot\partial_jg=\sum_{j=1}^n\bigl(\partial_jf\cdot\partial_jg+f\cdot\partial_j^2g\bigr).$$

Now both computations gave us the same thing, and so $\operatorname{div}(f\cdot\nabla g)=f\cdot\Delta g+\langle \nabla f,\nabla g\rangle$, as was desired.