confusion with the notation of partial derivatives Let $f\in C^2(\mathbb{R}^n)$ and $g:\mathbb{R}\to\mathbb{R}$. I have a confusion with the notation of partial derivatives. Are these derivatives correct?

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*$(g(f))'=g'(f)\cdot\sum_{i=1}^{n}f_{x_i}$


*$\dfrac{\partial}{\partial x_i}g(f(x))=\dfrac{\partial g}{\partial f}\dfrac{\partial f}{\partial x_i}$


*$\dfrac{\partial g(f)}{\partial x_i}(x)=g'(f(x))\cdot f_{x_i}$


*$\nabla (g(f))=g'(f)\nabla f$


*$\dfrac{\partial g}{\partial f}(f)=\dfrac{\partial g}{\partial f}\cdot\dfrac{\partial f}{\partial x}$


*$D g(f)=g'(f)Df$
 A: 

*

*$(g(f))'=g'(f)\cdot\sum_{i=1}^{n}f_{x_i}$

The prime denotes ordinary differentiation, not partial differentiation. Ordinary differentiation is defined when there is only one variable to differentiate, and that is not the case for $g(f) = g(f(x_1,x_2,\dots, x_n))$, so there is no "$(g(f))'$".


*

*$\dfrac{\partial}{\partial x_i}g(f(x))=\dfrac{\partial g}{\partial f}\dfrac{\partial f}{\partial x_i}$

This is technically acceptable, but it is not preferred. A partial derivative is "partial" because it depends on more than just the variable of differentiation. It also depends on the variables that are being held constant. For example, you can provide the plane with the ordinary $(x,y)$ coordinates. But if you define $t = y - x$, then $(x, t)$ also provides a coordinate system for the plane. If $f(p)$ is a function on the points $p$ in the plane, then $f$ can be considered to be a function of $(x,y)$, say $f(x,y) = x + y$ or a function of $(x,t)$, in which case the same function would be $f(x,t) = x + (t + x) = 2x + t$. When considered a function of $(x,y)$ we have $\frac{\partial f}{\partial x} = 1$. When considered a function of $(x,t)$ we have $\frac{\partial f}{\partial x} = 2$. So $\frac{\partial f}{\partial x}$ depends not just on $x$, but also on what other variables you are holding constant while you differentiate with respect to $x$. This is why it gets
its own name and a separate symbolism than ordinary integration. They are used to help us keep in mind this dependence on variables not explicitly included in the notation.
But $g$ is not a function of multiple variables. It is a function of a single variable, and thus its derivative is not partial, but an ordinary derivative, so it is denoted with a $d$, not a $\partial$. $g\circ f$ is $g(f(x_1,\dots,x_n))$, so it is a function of the same coordinates as $f$. So the correct notation is
$$\dfrac{\partial}{\partial x_i}g(f(x))=\dfrac{d g}{d f}\dfrac{\partial f}{\partial x_i}$$
This also mixes up a couple of notational conventions, so it looks a little strange, but it is acceptable.


*

*$\dfrac{\partial g(f)}{\partial x_i}(x)=g'(f(x))\cdot f_{x_i}$

That is fine. Though generally people just denote the multiplication by juxtaposition, like in the previous example. Using a $\cdot$ or $\times$ is only done when there is some ambiguity in what is meant when it isn't used.


*

*$\nabla (g(f))=g'(f)\nabla f$

That's fine. One has to understand that $\nabla f$ is a vector, so $g'(f)\nabla f$ is scalar multiplication of a vector.


*

*$\dfrac{\partial g}{\partial f}(f)=\dfrac{\partial g}{\partial f}\cdot\dfrac{\partial f}{\partial x}$

No. Not at all. Once again we have the derivative of $g$ denoted by $\partial$, which while not technically incorrect, pretends to a problem that does not exist for $g$. But the $\dfrac{\partial f}{\partial x}$ makes no sense at all. $f$ is not a function of $x$, it is a function of $x_1, \dots , x_n$. If you call $x = (x_1, \dots, x_n)$, I have seen $\dfrac{\partial f}{\partial x}$ used to mean $\nabla f$, but it is not a common notation, so you'd have to introduce it before using it.
But finally, you are essentially saying $g'(f) = g'(f)\nabla f$ which makes no sense at all. The left side is a scalar, while the right side is that same scalar times a vector.


*

*$Dg(f)=g'(f)Df$

This is just $\nabla (g(f))=g'(f)\nabla f$ in a different notation.
