Partial derivative of a function with another function inside it? What is
$\cfrac {\partial f(x, y, g(x))} {\partial x}$ expanded out?
I want to say $\cfrac {\partial f(x, y, g(x))} {\partial g(x)} \times \cfrac {\partial g(x)} {\partial x}$ but I don't think that's quite right.

Let consider: 


*

*a function $f:\mathbb{R}^3\rightarrow\mathbb{R}:(x,y,z) \mapsto f(x,y,z)$ ;

*a function $g:\mathbb{R}\rightarrow\mathbb{R}:x\rightarrow g(x)$ ;

*a function $h:\mathbb{R}^2\rightarrow\mathbb{R}:(x,y) \mapsto f(x,y,g(x))=: h(x,y)$.


The derivative of $h(\cdot,\cdot)$ with respect to $x$ evaluated in $(\bar{x},\bar{y})$ is given by
\begin{equation}
\frac{dh}{dx}(\bar{x},\bar{y}) = \frac{df}{dx}(\bar{x},\bar{y},g(\bar{x})) = \frac{\partial f}{\partial x}(\bar{x},\bar{y},g(\bar{x})) + \frac{\partial f}{\partial z}(\bar{x},\bar{y},g(\bar{x})) \, \frac{\partial g}{\partial x}(\bar{x}) .
\end{equation}
The point is to make a difference between the name of the arguments of a function and the point at which you evaluate the function (or its derivative). Usually one drops the $\bar{ \cdot}$ to simplify the notation.
