# Why $\frac{dx}{dt}$ not $\frac{\partial x}{\partial t}$ in $\frac{d}{dt}\big(f(t,x(t),y(t)\big)$?

The definition for "total derivative" given at wikipedia here gives the formula for $f(t,x(t),y(t))$ as: $$\frac{d}{dt}\big(f(t,x(t),y(t)\big)=\frac{\partial f}{\partial t} \frac{dt}{dt}+ \frac{\partial f}{\partial x} \frac{dx}{dt} + \frac{\partial f}{\partial y} \frac{dy}{dt}$$

why does this formula not use only partial derivatives? (i.e. what is the point of the $\frac{dx}{dt}$ term (and similar for y and t)?

Alternatively, in this post with a proof of this total derivative formula, why is it that $$D_a(g) = \displaystyle \begin{pmatrix}dx/dt\\dy/dt\end{pmatrix}$$? since $D_a(g)$ is the jacobian, which should be a matrix of partial derivatives?

My guess is that it is because, since $x$ and $y$ are only a function of $t$, the total and partial derivatives are the same thing. but in general where $x$ and $y$ are not only a function of $t$, this logic does not hold

Thanks

Yes, your guess is basically correct - when $x$ is a function only of $t$, $\frac{dx}{dt}$ and $\frac{\partial x}{\partial t}$ are the same, basically by definition.
To elaborate a little more, though: It wouldn't be incorrect to use $\frac{\partial x}{\partial t}$ here, just misleading. The notation $\frac{\partial x}{\partial t}$ says "the derivative of $x$ with respect to $t$", but it heavily implies "...where $x$ is also a function of other variables". Similarly, if I said "some bachelors are unmarried", I would be saying something that is technically true, but suggesting the additional comment "some bachelors are not".
• if $x$ IS a function of other variables, then do I use $\frac{dx}{dt}$ in the formula or $\frac{\partial x}{\partial t}$ (This is where I think I am misunderstanding something). For example, what if instead of $f(t,x(t),y(t)$ we had $f(t,x(t,z), y(t))$, where $z$ may or may not be a function of $t$? (I apologize if my notation is wrong here). and if I DO use $\frac{dx}{dt}$, the isn't the proof at answer i liked wrong? – user106860 Apr 26 '18 at 0:22
• Oh! For that, you'll want to use the total derivative $\frac{dx}{dt}$ still, so that $\frac{dz}{dt}$ will get incorporated into it. – Reese Apr 26 '18 at 0:24
Here, $x$ and $y$ (and incidentally, $t$) all appear to be functions of $t$ only.