Calculating total derivative from partial derivatives I have a univariate function of a multivariate function:
$$F(g(x,y,z)),$$
$$g(x,y,z)=2x+2y+z.$$
I know $F'(x)$, $F'(y)$ and $F'(z)$. I need to know $F'(g)$. How can I do that? I'm confused, but it seems like the chain rule does not work directly here, nor the total derivative as a sum of partial derivatives.
 A: I assume by "$F'(x)$" that you mean that you have a formula for $\frac{d}{dx}\left(F\circ g\right): \mathbb{R}^3\to\mathbb{R}$ (your notation is not standard; most people would read $F'(x)$ as "take the derivative of $F$ with respect to its only argument, then plug in $x$" in which case you have nothing you need to do).
If so, you can use the chain rule:
$$\frac{d(F\circ g)}{dx}\left(x,y,z\right) = \frac{\partial F}{\partial g}\left[g(x,y,z)\right]\frac{dg}{dx}(x),$$ which gives
$$\frac{1}{2}\frac{d(F\circ g)}{dx}\left(x,y,z\right) = F'(2x+2y+z),$$
and in particular, setting $x=0,y=0,z=u$,
$$F'(u) = \frac{1}{2}\frac{d(F\circ g)}{dx}\left(0,0,u\right).$$
A: By the chain rule we have
$$\,dF(g)=F'(g)\,dg$$
On the other hand by total derivative rule we get
$$\,dg=g_x\,dx+g_y\,dy+g_z\,dz$$
where $g_x:=\partial g/\partial x$. Therefore
$$\,dF(g)=F'(g)(g_x\,dx+g_y\,dy+g_z\,dz)$$
This then implies for instance 
$$\frac{\,dF(g)}{\,dx}=F'(g)(g_x+g_y\frac{dy}{dx}+g_z\frac{dz}{dx})$$
In particular if $x,y,z$ are independent i.e. $0=dy/dx=dz/dx=...$ one obtains
$$\frac{\,dF(g)}{\,dx}=F'(g)g_x, \frac{\,dF(g)}{\,dy}=F'(g)g_y, \frac{\,dF(g)}{\,dz}=F'(g)g_z$$
