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How can I prove the gradient satisfies the following properties of the derivative?

(a) For functions $f(x,y)$, $g(x,y)$ of $2$ variables, we have that $$\nabla{(fg)}(x,y)=f(x,y)\nabla{g(x,y)}+g(x,y)\nabla{f(x,y)}$$

(b) For a function $u(x,y)$ of $2$ variables and for an integer $n \geq 1$, we have that $$\nabla{u^{n}(x,y)=nu^{n-1}(x,y)\nabla{u(x,y)}}$$

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    @user251257 well I know that the gradient of f(x,y) is (the partial derivative of f respect to x, partial derivative respect to y), but I'm really not sure where to go from there – 2017-02-13

1 Answers 1

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$fg=B\circ (f,g)$, where $B$ is the bilinear map $(x,y)\mapsto xy$. $B$ is continuous, so $C^{\infty}$, and we have:

$$D(fg)(x) = DB(f(x),g(x)) \circ (Df(x),Dg(x))$$

i.e.

$$D(fg)(x)h = DB(f(x),g(x))(Df(x)h,Dg(x)h) \\ = B(f(x), Dg(x)h) + B(Df(x)h,g(x)) = f(x) Dg(x) h + g(x)Df(x)h = [f(x)Dg(x) + g(x)Df(x)] h$$

Thus $D(fg)(x) = f(x)Dg(x) + g(x)Df(x)$. Recall that $Df(x) = \nabla f(x) ^T$.

For the second one, use induction.