Total derivative of vector function

Question

Let's asume that I have a (vector) function $f \space (s,t,u,v):\mathbb{R}^4 \rightarrow \mathbb{R}^n.$ I would like to calculate: $\frac{d}{dt}\biggr|_{t=t_0} f(t,t,t,t).$

Intuitively:

This should just be given by:

$$\frac{d}{dt}\biggr|_{t=t_0} f(t,t,t,t)=\frac{\partial}{\partial s}\biggr|_{(s,t,u,v)=(t_0,t_0,t_0,t_0)}f(s,t,u,v)+\frac{\partial}{\partial t}\biggr|_{(s,t,u,v)=(t_0,t_0,t_0,t_0)}f(s,t,u,v)+\frac{\partial}{\partial u}\biggr|_{(s,t,u,v)=(t_0,t_0,t_0,t_0)}f(s,t,u,v)+\frac{\partial}{\partial v}\biggr|_{(s,t,u,v)=(t_0,t_0,t_0,t_0)}f(s,t,u,v).$$

Am I right? Can someone provide some explanation. I know this is probably a stupid question, but I'm not a mathematician and have trouble finding like a theorem or something regarding this, couldn't find something useful here on math.SE either .It seems like this just the total deriative of the function, but in this case Wikipedia just provides the formula for a scalar function.

Answer 1

Generally, in appropriate conditions, when $f$ have $n$ variables $z=f(y_1, \cdots, y_n)$ and we take its composition with functions $y_i=\psi_i(x_1, \cdots, x_m)$, then well known chain rule is:

$$\frac{\partial f \circ \psi}{\partial x_k}= \frac{\partial f}{\partial y_1}\frac{\partial \psi_1}{\partial x_k}+\frac{\partial f}{\partial y_2}\frac{\partial \psi_2}{\partial x_k}+ \cdots+\frac{\partial f}{\partial y_n}\frac{\partial \psi_n}{\partial x_k}$$

In your case $m=1, n=4$ and $\forall k=\overline{1,4}$ we have $\psi_k(t)=t$, which gives $\psi_k^{'}=1$. So your formula is right and can be written as: $$\frac{df}{dt}= \sum_{i=1}^{4}f_{y_i}^{'}=f_{y_1}^{'}+f_{y_2}^{'}+f_{y_3}^{'}+f_{y_4}^{'}$$

Answer 2

From this part of the same Wikipedia page, you can consider a more general version as (with a relabling of the variables) \begin{equation} \frac{d}{dt}\bigg\rvert_{t_0} f(t,x_1(t),x_2(t),x_3(t)) = \left( \frac{\partial}{\partial t} + \sum_{i=1}^3 \frac{d x_i}{dt} \frac{\partial}{\partial x_i}\right) f \end{equation}

In your case, $\frac{d x_i}{d t} = 1$ and this reduces to the formula you produced.