Let $f(t, dummyVar)=t$ and $t=x+y$, now we want to calculate $\frac{\partial f}{\partial t}$
here are two solutions which yield two different answers, why? First solution: from the equation, it is obvious that:
$\frac{\partial f}{\partial t} = 1$
from the chain rule we could write below equation:
$\frac{\partial f}{\partial t} = \frac{\partial f}{\partial x}*\frac{\partial x}{\partial t}+\frac{\partial f}{\partial y}*\frac{\partial y}{\partial t}$
and we know that: $f(x,y)=x+y$
so:
$\frac{\partial f}{\partial t} = 2$
I am now so confused. where is my mistake?
here is the visualization:
Your mistake is that $f$ is a function of the variable $t$, so $\frac{\partial f}{\partial x}$ does not really make sense, and similarly for $y$.
In much the same manner, how does $x$ depend on $t$, in order for $\frac{\partial x}{\partial t}$ to make sense?
The chain rule speaks about the composition of functions. There are two functions in your question: $f$, which depends on $t$, and an unnamed function which I'll call $g$, given by $g(x,y) = x+y$.
The function $f$ depends only on $t$ so $\frac{df}{dt}$ can be determined via usual single-variable calculus methods. Conversely, if you consider $h:=f\circ g$, now we have a function that depends on $x$ and $y$. We may apply the chain rule to calculate $\frac{\partial h}{\partial x}$ and $\frac{\partial h}{\partial y}$, but $\frac{\partial h}{\partial t} = 0$ because $h$ does not depend on $t$.