Answer
Proof of the Chain Rule for Gradients:
$\nabla \left( {F\left( {f\left( {x,y,z} \right)} \right)} \right) = F'\left( {f\left( {x,y,z} \right)} \right)\nabla f$
Work Step by Step
Let $f\left( {x,y,z} \right)$ and $g\left( {x,y,z} \right)$ be differentiable and $c$ a constant. If $F\left( t \right)$ is a differentiable function of one variable, then according to Theorem 1, the Chain Rule for Gradients is given by
$\nabla \left( {F\left( {f\left( {x,y,z} \right)} \right)} \right) = F'\left( {f\left( {x,y,z} \right)} \right)\nabla f$
Proof.
Using the definition of gradient we get
$\nabla \left( {F\left( {f\left( {x,y,z} \right)} \right)} \right) = \left( {\frac{{dF}}{{df}}\frac{{\partial f}}{{\partial x}},\frac{{dF}}{{df}}\frac{{\partial f}}{{\partial y}},\frac{{dF}}{{df}}\frac{{\partial f}}{{\partial z}}} \right)$
Since $F$ is a function of one variable, we can write $\frac{{dF}}{{df}} = F'\left( {f\left( {x,y,z} \right)} \right)$. Thus,
$\nabla \left( {F\left( {f\left( {x,y,z} \right)} \right)} \right) = \left( {F'\left( {f\left( {x,y,z} \right)} \right)\frac{{\partial f}}{{\partial x}},F'\left( {f\left( {x,y,z} \right)} \right)\frac{{\partial f}}{{\partial y}},F'\left( {f\left( {x,y,z} \right)} \right)\frac{{\partial f}}{{\partial z}}} \right)$
$ = F'\left( {f\left( {x,y,z} \right)} \right)\left( {\frac{{\partial f}}{{\partial x}},\frac{{\partial f}}{{\partial y}},\frac{{\partial f}}{{\partial z}}} \right)$
Since $\nabla f = \left( {\frac{{\partial f}}{{\partial x}},\frac{{\partial f}}{{\partial y}},\frac{{\partial f}}{{\partial z}}} \right)$, hence
$\nabla \left( {F\left( {f\left( {x,y,z} \right)} \right)} \right) = F'\left( {f\left( {x,y,z} \right)} \right)\nabla f$
