MHB Proving vector calculus identities w/ tensor notation

[skate_nerd]

I have an vector calculus identity to prove and I need to use vector notation to do it. The identity is $$\vec{\nabla}(fg)=f\vec{\nabla}{g}+g\vec{\nabla}{f}$$ I tried starting with the left side by writing $\vec{\nabla}(fg)=\nabla_j(fg)$. Now I look and that and it really looks like there is nowhere I can go from there. Is there something I am unaware of that you can do with those scalar functions \(f\) and \(g\)? Or would it be a better idea to start this proof with the right side of the identity?

 

[I like Serena]

I have an vector calculus identity to prove and I need to use vector notation to do it. The identity is $$\vec{\nabla}(fg)=f\vec{\nabla}{g}+g\vec{\nabla}{f}$$ I tried starting with the left side by writing $\vec{\nabla}(fg)=\nabla_j(fg)$. Now I look and that and it really looks like there is nowhere I can go from there. Is there something I am unaware of that you can do with those scalar functions \(f\) and \(g\)? Or would it be a better idea to start this proof with the right side of the identity?

Hi skatenerd! :)

Let's take a look at the first component which is the x component.
$$\nabla_1(fg) = \frac{\partial}{\partial x}(f \cdot g)$$
Can you apply the product rule to that?Btw, I have moved your thread to the sub forum Calculus, which covers this topic.

 

[skate_nerd]

I guess I was expecting it to be more complicated than that...haha thank you

 

[Deveno]

I am fond of the "big D" notation, which seems easier to read to me.

In this notation, we have:

$\nabla(fg) = (D_1(fg),D_2(fg),D_3(fg))$

$= (fD_1g + gD_1f,fD_2g + gD_2f,fD_3g + gD_3f)$

$= (fD_1g,fD_2g,fD_3g) + (gD_1f,gD_2f,gD_3f)$

$=f(D_1g,D_2g,D_3g) + g(D_1f,D_2f,D_3f) = f\nabla g + g\nabla f$