consider the function
$$f = f(x(t),y(t))$$
I know that its total derivative wrt t is
$$\frac {df}{dt} = \frac {\partial f} {\partial x} \frac {dx}{dt} + \frac {\partial f}{\partial y} \frac {dy}{dt}$$
and that the total derivative wrt x is $$ \frac {df} {dx} = \frac {\partial f} {\partial x} \frac {dx}{dx} + \frac {\partial f} {\partial y} \frac {dy} {dx}$$
However I am not fully familiar with the notation and the forms in which it takes during more extreme conditions, such as the following, could anyone fill the blanks in for me?
1)$$f = f(x(t,w),y(t,w))$$ $$\frac {df}{dt} = ?$$ 2) $$f = f(x(g(t,w)),y(g(t,w)))$$ $$\frac {df}{dt} = ?$$ 3) $$f = f(x(g(t,w),z),y(g(t,w),z))$$ $$\frac {df}{dt} = ?$$ 4) $$f = f(x(g(t)),y(g(t)))$$ $$\frac {df}{dt} = ?$$
how would I go about writing these properly? Any answer will help illustrate the semantics in much greater detail
we suppose that $f(x,y)$ is differentiable, so by definition of the partial derivatives $\frac{\partial f}{\partial x}$ and $\frac{\partial f}{\partial y}$ : $$f(x+h,y) = f(x,y) + h \frac{\partial f(x,y)}{\partial x} + h \epsilon_1(h)$$ $$f(x,y+h) = f(x,y) + h \frac{\partial f(x,y)}{\partial y} + h \epsilon_2(h)$$ where $\epsilon_i(h)$ are residual functions such that $\lim_{h \to 0} \epsilon_i(h) = 0$
now if $\frac{\partial f(x,y)}{\partial x}$ and $\frac{\partial f(x,y)}{\partial y}$ are continuous, i.e. $f(x,y)$ is continuously differentiable :
$$f(x+h,y+C h) = f(x,y) + h \frac{\partial f(x,y)}{\partial x} + C h \frac{\partial f(x,y)}{\partial y} + h\epsilon(h)$$ (it is a good exercice to prove it)
finally, because we suppose that $x(t)$ and $y(t)$ are differentiable : $x(t+h) = x(t) + h x'(t) + h\epsilon_3(h)$ and $y(t+h) = y(t) + h y'(t) + h\epsilon_4(h)$ hence :
$$f(x(t+h),y(t+h)) = f(x(t)+h x'(t),y(t)+h y'(t)) + h \epsilon_5(h) = f(x(t),y(t)) + h x'(t) \frac{\partial f(x,y)}{\partial x} + h y'(t) \frac{\partial f(x,y)}{\partial y} + h \epsilon_6(h)$$
which proves that :
$$\frac{\partial f(x(t),y(t))}{\partial t} = \lim_{h \to 0} \frac{f(x(t+h),y(t+h))-f(x(t),y(t))}{h} = x'(t) \frac{\partial f(x(t),y(t))}{\partial x} + y'(t) \frac{\partial f(x(t),y(t))}{\partial y}$$
now adapt all this to the simpler notations above, and to compute the nested derivatives of your first question.