Chain rule in multivarible calculus

Question

I recently was reading spivak calculus on manifold and I've arrived to chain rule he wrote that

\begin{gather}F(x)=f(g_{1}(x),...,g_{m}(x))\end{gather}

then he took the derivative which lead to

\begin{gather} \frac{d F(x)}{dx}=\sum_{j=1}^{m}\frac{\partial f}{\partial g_{j}}.\frac{dg_{j} }{dx}\end{gather}

let's talk about the case that $x$ is a scalar and $g_{1}(x)=x$ and $g_{2}(x)=x^2$

\begin{gather} \frac{d F(x)}{dx}=\frac{\partial f}{\partial x}.\frac{d x}{dx}+\frac{\partial f}{\partial (x^2)}.2x \end{gather}

doesn't that mean that we've considered $f$ as being function of $2$ variables ,because we took the partial derivatives w.r.t variables inside it ?

Answer 1

$f$ is a real-valued function of $m$ real variables $x_1, \ldots, x_n$. We can form the partial derivatives $\dfrac{\partial f}{\partial x_{j}}$ with respect to these variables which are again functions of the $m$ real variables $(x_1, \ldots, x_n)$. The naming of the input variable is irrelevant, for $m = 2$ one often writes $f(x,y)$ instead of $f(x_1,x_2)$ and gets the two partial derivatives $\dfrac{\partial f}{\partial x}, \dfrac{\partial f}{\partial y}$. Similarly, for $m = 3$ one often writes $f(x,y,z)$ instead of $f(x_1,x_2,x_3)$ and gets the three partial derivatives $\dfrac{\partial f}{\partial x}, \dfrac{\partial f}{\partial y}, \dfrac{\partial f}{\partial z}$.

$F$ is the composition of $f$ and the function $g(x) = (g_1(x),\ldots, g_m(x)) \in \mathbb R^m$ of the real variable $x$ which has $m$ coordinate functions $g_j$. Thus the composition $F = f \circ g$ is a real-valued function of one real variable. We can also say that $F(x) = f(x_1,\ldots,x_m)$ with $x_j = g_j(x)$.

In my opinion one should not write $\dfrac{\partial f}{\partial g_{j}}$ for the partial derivative of $f$ with respect to ist $j$-th input variable because this notation suggests that there exist partial derivatives of $f$ with respect to functions $g_j$ - which is not the case. However, that is just a matter of taste and if you correctly understand what it means it will cause no problem to write $\dfrac{\partial f}{\partial g_{j}}$ instead of $\dfrac{\partial f}{\partial x_{j}}$.

I suggest to write $$\frac{d F}{dx}=\sum_{j=1}^{m}\frac{\partial f}{\partial x_{j}}\cdot\frac{dg_{j} }{dx} . \tag{1}$$ At the point $x$ we get $$\frac{d F}{dx}(x)=\sum_{j=1}^{m}\frac{\partial f}{\partial x_{j}}(g(x))\cdot\frac{dg_{j} }{dx}(x) . \tag{2}$$ In your example we have $m = 2$ and $g_1(x) = x, g_2(x) = x^2$. This gives $$\frac{d F}{dx}(x)=\frac{\partial f}{\partial x_{1}}(x,x^2) + \frac{\partial f}{\partial x_{2}}(x , x^2) \cdot 2 x . \tag{3}$$

Let us emphasize that $f(x_1,x_2)$ is a function of two variables. However, $F(x) = f(x,x^2)$ is a function of one variable which allows to form the usual derivative $\dfrac{d F}{dx}$. It can be expressed via $(3)$ in terms of the two partial derivatives $\dfrac{\partial f}{\partial x_{1}}, \dfrac{\partial f}{\partial x_{2}}$. If we write $f(x,y)$ instead of $f(x_1,x_2)$. formula $(3)$ takes the form $$\frac{d F}{dx}(x)=\frac{\partial f}{\partial x}(x ,x^2) + \frac{\partial f}{\partial y}(x , x^2) \cdot 2 x . \tag{4}$$

Let us consider the example $f(x,y) = x +y$. We have $\dfrac{\partial f}{\partial x}(x,y) = 1$ and $\dfrac{\partial f}{\partial y}(x,y) = 1$. With $g_1(x) = x$ and $g_2(x) = x^2$ we get $F(x) = f(x,x^2) = x + x^2$, thus $\dfrac{d F}{dx}(x) = 1 + 2x$. On the RHS of $(4)$ we have $$\frac{\partial f}{\partial x}(x,x^2))\cdot\frac{dg_{1} }{dx}(x) + \frac{\partial f}{\partial y}(x,x^2)\cdot\frac{dg_{2} }{dx}(x) = 1 \cdot 1 + 1 \cdot 2x = 1 + 2x .$$