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.
 A: You have a composition:
$$(x,y)\longmapsto(x,y,g(x))\longmapsto f(x,y,g(x)).$$
Give a name to the first function, $h:{\Bbb R}^2\longrightarrow{\Bbb R}^3$ for example.
The second function is simply $f:{\Bbb R}^3\longrightarrow{\Bbb R}$.
Apply the chain rule.
A: $d_{x}f(x, y, g(x)) + \frac{dg}{dx} d_{z}f(x, y, g(x))$
f = f(x, y, z) and g(x) = f(x, y, g(x)). When x moves by dx, you are evaluating f to a new point where x AND z have changed.
(x, y, g(x)) -> (x+dx, y, g(x+dx)) ~ (x+dx, y, g(x) + g'(x)dx)
So when x moves a little bit, we will see a change in x and z.
Generally speaking, f(x+dx, y+dy, z+dz) ~ $f(x, y, z) + dx ∂_{x}f(x, y, z) + dy ∂_{y}f(x, y, z) + dz ∂_{z}f(x, y, z)$
So you replace dz by g'(x)dx and you will see the answer
A: 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.
