It is correct to state that $\partial f=\frac{\mathrm d}{\mathrm dt} f$?

Question

I have the custom to write the total derivative of $f$ just as $\partial f$. Here $\partial f$ is the linear function $\partial f(v)$ such that

$$\lim_{h\to 0}\frac{\|f(v+h)-f(v)-\partial f(v)h\|}{\|h\|}=0\tag1$$

for a vector $v$ (this is the definition that I know about total derivative of a function).

Let a function $f$ defined through the three variables $t,x,y$, such that $x$ and $y$ depends on $t$. Also suppose that the partial derivatives of $f$ are continuous, hence we have that

$$\partial f(t,x,y)=\partial_t f(t,x,y)+\partial_x f(t,x,y)+\partial_y f(t,x,y)\tag2$$

My question is: it is the same $\partial f$ and $\frac{\mathrm d}{\mathrm dt }f$? If not, what is the difference? Thank you in advance.

Answer 1

The total derivative $\frac{df}{dt}$is only defined for a function of one variable. If we have a function $f(x_{1},...,x_{n})$ of many variables, then we can define a total derivative of $f$ only if each of the variables $x_{1},...,x_{n}$ depends on some parameter $t$, say $x_{i} = x_{i}(t)$. In this case,

$$\frac{d}{dt} f(x_{1}(t),...,x_{n}(t)) = \frac{\partial f}{\partial x_{1}}\dot{x_{1}} + \ldots + \frac{\partial f}{\partial x_{n}}\dot{x_{n}}$$

What you call $\partial f$ is a linear map $\mathbb{R}^{n} \to \mathbb{R}$. It can be identified with the row vector
$$\partial f = \left(\frac{\partial f}{\partial x_{1}}, \ldots ,\frac{\partial f}{\partial x_{n}}\right)$$

In the case that the coordinates $x_{i}$ are considered to all depend on some parameter $t$, we obtain the vector $\dot{\mathbf{x}} = \left(\dot{x_{1}},\ldots \dot{x_{n}}\right) \in \mathbb{R}^{n}$. Then, by the chain rule, we have $$\frac{d}{dt} f(x_{1}(t),...,x_{n}(t)) = \partial f (\dot{\mathbf{x}})$$ Does that make sense?

edit: response to comments
In my opinion, the way to avoid being confused about this sort of thing is to be absolutely clear about exactly what function you are talking about. Suppose I have $f: \mathbb {R}^n \to \mathbb {R}^m$ (the most general case). There are two kinds of derivative we can define:

1. What I just call "the derivative": the unique linear map $A_{x}:\mathbb {R}^n \to \mathbb {R}^m$ (if one exists) such that $$\frac {f(x+h)-f (x)-A_{x}(h)}{||h||} \to 0$$ as $h \to 0$, where $h $ is a vector in $\mathbb {R}^n $
2. The partial derivative with respect to argument $i $, which is a vector in $\mathbb {R}^m $, defined by only changing the $i^{th} $ argument: $$\lim_{t \to 0} \frac {f (x+t\mathbf {e_{i}})}{t}$$ where $\mathbf{e}_{i}=(0,...,1,...,0) $ is the $i^{th} $ basis vector in $\mathbb {R}^n $, and $t$ is a real number.

I don't like the Leibniz notation in this context because in some sense the independent variables "don't exist" and the operation of differentiating a function only depends on the function. For example when I wrote $\frac {df}{dt} $ earlier, what that really means is we chose some curve $\gamma(t) $ in $\mathbb {R}^n $, and differentiated the composition $f \circ \gamma $

Answer 2

Yes, you are totally correct. The two symbols bear the same meaning.

But this equation that you wrote above,

$$\partial f(t,x,y)=\partial_t f(t,x,y)+\partial_x f(t,x,y)+\partial_y f(t,x,y)$$

is not exactly correct. It should be revised, since the total derivative is with respect to $t$.

$$\partial f(t,x,y)=\partial_t f(t,x,y)+\partial_x f(t,x,y) \dot x+\partial_y f(t,x,y) \dot y$$