Differentiation with respect to several variables $$y(t)=x(t)+g(t)$$
say we take derivative with respect to g, does it become $$dy/dg=1$$ or $$dy/dg=dx/dg +1$$
I am confused about the x(t), because it's dependent on t, and so is g(t), so it would seem that changing g should affect x....
 A: I think the best way to look at this problem is to split it up in the following way. First, we introduce the function $Y(a,b) = a + b$. Second, we are given two time-dependent quantities $x(t)$ and $g(t)$. We can use these to find
\begin{equation*}
y(t) := Y(x(t),g(t)) = x(t)+g(t).
\end{equation*}
Now we can as ourselves what ''$\frac{d y}{d g}$'' could mean. Intuitively, this tells us how much $y$ changes when we change $g$ a little. That $g$ itself in turn depends on $t$ is then not really important -- this only specifies how $g$ might change as a function of $t$.
The most straightforward answer to your question needs an assumption: namely, that we want to see how $y$ changes when we change $g$ a little, while keeping $x$ fixed. Then, this is equivalent to taking the partial derivative of $y$ with respect to $g$, which is the same as the partial derivative of $Y$ with respect to $b$, i.e.
\begin{equation*}
\frac{\partial y}{\partial g} = \frac{\partial Y}{\partial b} = 1.
\end{equation*}
A: Short answer: $$dy/dg=dx/dg +1$$
But I guess some clarifications are required.
Try to understand what exactly you are trying to calculate. It's not a matter of symbols manipulations, it's a matter of understanding. What does it mean $dy/dg$? $y$ as a function does not even have $g$ as an argument. Generally speaking it may not be even possible to calculate $dy/dg$ in this case. It may turn out that it's possible to change some other arguments in a way that $y$ changes, but $g$ does not. To be able to calculate $dy/dg$ you must have some additional constraints, which would allow you to obtain a relationship $y(g)$. After that you differentiate this function of a single argument and get the answer.
I need another example here. Suppose we have a particle moving on a plane and some function $f(x, y)$. What would be $df/dx$? It's not possible to answer this question. For one particle, moving parallel to $x$ axe we may get some non-zero answer. For another, which moves along a curve where $f(x,y)=const$ we would have 0 as an answer. We need to know some additional information, not contained in $f(x, y)$, to be able to transform $f(x, y)$ in a form $f(x)$. In this we need to know how exactly particle moves.
Back to your example. You are lucky, you can get $y$ as a function of single argument $g$ because you know how both of them depend on time. Suppose you can reverse $g(t)$ and get $t(g)$. Then you can find out how $y$ depend on $g$.
$$y_g(g) = y(t(g)) = x(t(g)) + g(t(g)) = x(t(g)) + g$$
(Here $y_g$ is a function which returns value of $y$ given value of $g$. I created a special symbol for it because this function is very different from $y(t)$. They both return value of $y$, but they manipulate the argument very differently to get the answer.)
And now you can differentiate this function of a single argument.
$$d(y_g(g))/dg = d(x(t(g))/dg + 1$$
