Q:

Taking partial derivatives of function with dependent variables

Consider the following function:
$$
f(x(t), y(t), t) = x(t)^2 + y(t)^3 + t^4
$$
I am not sure about how the (partial) derivatives should be carried out.
My understanding is that:
$$
\frac{d}{dt} f(x(t), y(t), t) = \frac{\partial}{\partial t} f(x(t), y(t), t) = 2x(t) x'(t) + 3y(t)^2 y'(t) + 4t^3
$$
And that:
$$
\frac{\partial}{\partial x(t)} f(x(t), y(t), t) = 2x(t) \\
\frac{\partial}{\partial y(t)} f(x(t), y(t), t) = 3y(t)^2 y'(t)
$$
I was wondering if this is correct, or where I went wrong.

A:

You can calculate it in two different ways and lead to the same answer (courtesy to Newton who invented calculus!) as following$${df\left(x(t),y(t),t\right)\over dt}=2xx'+3y'y^2+4t^3$$also $f(x,y,t)=x^2+y^3+t^4$ so we have $$df(x,y,t)={\partial f\over\partial x}dx+{\partial f\over\partial y}dy+{\partial f\over\partial t}dt=2xdx+3y^2dy+4t^3dt$$if $x$ and $y$ are both functions of $t$ we may proceed further and obtain$${df\over dt}=2x{dx\over dt}+3y^2{dy\over dt}+4t^3=2xx'+3y^2y'+4t^3$$as we concluded before.

