If $f=f(x(s),y(t))$, then $\frac{\partial f}{\partial t}=0$ is wrong? enter link description here
You may see through the link that someone pointed out that if$$f=f(x(t),y(t))$$then$$\cfrac{\partial f}{\partial t}=0$$So now i want to challenge this by giving an example:
Let $f=f(x,y)$, where $x=x(t)$,$y=y(s)$. Because $f$ is independent of $t$, so we have $$\frac{\partial f}{\partial t}=0\qquad(1)$$
Now, given that $x=x(t)=2t^2$, $y=y(s)=3s$, then f is given by:$$f=f(x,y)=xy\qquad(2)$$
Substitute $x=x(t)=2t^2$, $y=y(s)=3s$ into (2), we have$$f=6t^2s\qquad(3)$$It follows that $$\frac{\partial f}{\partial t}=12ts\qquad(4)$$Equation(1) and Equation (4) are not equal, what is wrong?

$t$ affects $f$. in other words $f$ and $t$ is dependent since $f$=$f$($x$($s$),$y$($t$)) hope this helps! :)
