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$$y(t)=x(t)+g(t)$$

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....

user277455
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    $g$ iand $x$ are not variables. They are functions of the variable $t$. – Julián Aguirre Oct 06 '15 at 08:59
  • You answered your question in the last line. "I am confused about the x(t), because it's dependent on t." Oh, yes, it is dependent on $t$. So you cannot quite differentiate with respect to the functions. $g$ is a function. You can differentiate with respect to $t$ here. – Aditya Agarwal Oct 06 '15 at 09:18
  • I think it depends on whether you would take the partial or full derivative, see this question. Since $g$ is a function of $t$, if $g$ changes, it would be very likely that $t$ would have to change as well and therefore $x$ and $y$ also. – Kwin van der Veen Oct 06 '15 at 11:23

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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*}

Frits Veerman
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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. $$d(y_g(g))/dg = d(x(t(g))/dg + 1$$

lesnik
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