Suppose that $x = g(t)$ and $y = h(t)$ are differentiable functions of $t$ and $z = f(x,y)$
Then, $z$ is a differentiable function of $t$ and $\mathrm dz/\mathrm dt = (\partial z/\partial x)(\mathrm dx/\mathrm dt) + (\partial z/\partial y)(\mathrm dy/\mathrm dt)$
The above is from a calc lecture, and I ask: why can we add the $x$ and $y$ components? And likewise if $x$ and $y$ (and thus $z$, a function of $x$ and $y$) depended on multiple variables--why can we add them?
My preference is to define partials in the following way: if $z = f(x, y)$, then $dz = \partial_x z + \partial_y z$. In other words, $dz$ is the result of adding together the changes in $z$ which result from allowing $x$ to vary and the changes in $z$ which result from allowing $y$ to vary.
A better notation, in my opinion, for $\frac{\partial z}{\partial x}$ is actually $\frac{\partial_x z}{dx}$. So, you can derive this as follows:
$$ dz = \partial_x z + \partial_y z $$
Divide both sides by $dt$:
$$\frac{dz}{dt} = \frac{\partial_x z}{dt} + \frac{\partial_y z}{dt}$$
Now, $\frac{dx}{dx} = \frac{dy}{dy} = 1$, so we can freely multiply by this if we want. So, we can say:
$$ \frac{dz}{dt} = \frac{\partial_x z}{dt}\frac{dx}{dx} + \frac{\partial_y z}{dt}\frac{dy}{dy}$$
Now, using basic fraction rules, we can rearrange this as:
$$ \frac{dz}{dt} = \frac{\partial_x z}{dx}\frac{dx}{dt} + \frac{\partial_y z}{dy}\frac{dy}{dt} $$
Rewriting this in "typical" partial derivative notation gives:
$$ \frac{dz}{dt} = \frac{\partial z}{\partial x}\frac{dx}{dt} + \frac{\partial z}{\partial y}\frac{dy}{dt} $$
\partial. – David K Mar 04 '22 at 02:20