Taking partial derivatives of function with dependent variables

Question

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.

Answer 1

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.

Answer 2

your calculation is correct. You have $$z=x^2 +y^3 +t^4$$ where $x$ and $y$ are functions of $t$.

Thus you get, $$ \frac {dz}{dt} = 2x\frac {dx}{dt} +3y^2 \frac {dy}{dt} +4t^3.$$