Chain rule for multivariable functions
With a single variable function the chain rule tells us that [math]\displaystyle{ [f(g(x))]' = g'(x)f'(g(x)) }[/math]. For multivariable functions the idea is the same, it's still a product of derivatives. Both functions have to be differentiable for the chain rule to work. Now some textbooks have a different approach here. We have essentially two cases to treat: one is [math]\displaystyle{ f(g(t),h(t)) }[/math]; the other is [math]\displaystyle{ f(g(a,b),h(t,s)) }[/math]. One of the textbooks that I follow go for a general form [math]\displaystyle{ f(\gamma(t)) }[/math], where [math]\displaystyle{ \gamma(t) }[/math] is a (vector valued) function of n variables.
I'm going to begin with the easiest case [math]\displaystyle{ f(\gamma(t)) }[/math], where [math]\displaystyle{ \gamma(t) = (x(t), y(t)) }[/math] is a vector function or a (differentiable) curve. [math]\displaystyle{ x(t) }[/math] and [math]\displaystyle{ y(t) }[/math] are both differentiable. Before moving on to calculations, notice that any change in [math]\displaystyle{ t }[/math] is going to change the value of [math]\displaystyle{ f }[/math]. Which means that [math]\displaystyle{ f }[/math] depends, indirectly, on [math]\displaystyle{ t }[/math].
An increment [math]\displaystyle{ \Delta t }[/math] is going to produce the increments [math]\displaystyle{ \Delta x }[/math] and [math]\displaystyle{ \Delta y }[/math]. As such:
[math]\displaystyle{ \Delta f = f(x + \Delta x, y + \Delta y) - f(x,y) = \frac{\partial f}{\partial x}\Delta x + \frac{\partial f}{\partial y}\Delta y + \eta r }[/math].
where [math]\displaystyle{ \eta \to 0 }[/math] with [math]\displaystyle{ r = \sqrt{\Delta x^2 + \Delta y^2} \to 0 }[/math], because [math]\displaystyle{ f }[/math] is differentiable. The last term is about error when we use a linear approximation. Now since we are differentiating in respect to [math]\displaystyle{ t }[/math]:
[math]\displaystyle{ \frac{\Delta f}{\Delta t} = \frac{\partial f}{\Delta x} \cdot \frac{\Delta x}{\Delta t} + \frac{\Delta f}{\Delta y} \cdot \frac{\Delta y}{\Delta t} \pm \sqrt{\left(\frac{\Delta x}{\Delta t}\right)^2 + \eta\left(\frac{\Delta y}{\Delta t}\right)^2} }[/math]
where the sign of the last term is positive if [math]\displaystyle{ \Delta t \gt 0 }[/math] and negative if [math]\displaystyle{ \Delta t \lt 0 }[/math]. When we take the limit, [math]\displaystyle{ \Delta t \to 0 }[/math], [math]\displaystyle{ \eta \to 0 }[/math] and the last term goes away. The resulting expression is:
[math]\displaystyle{ \frac{d}{dt}f(x(t),y(t)) = \frac{\partial f}{\partial x} \frac{\partial x}{\partial t} + \frac{\partial f}{\partial y} \frac{\partial y}{\partial t} }[/math]
