Suppose that $g:(0,+\infty)\times(0,+\infty)\rightarrow \mathbb{R}$ is a function of $x$ and $y(x)$, where $x\in (0,+\infty)$ and $y:(0,+\infty)\rightarrow (0,+\infty)$ and y is a linear and decreasing function of $x$. If we want to calculate the partial derivative of $g(x,y(x))$ with respect to $x$, then do i need to use the chain rule?
Namely, what is the $\frac{\mathrm dg(x,y(x))}{\mathrm dx}$ in it's general formula if we know only the above information and ton the explicit formula of $g$?
As $y$ is a linear and decreasing function of $x$, we can put $$ y(x) = a - bx, $$ where $a$ and $b$ are some constants such that $b > 0$.
But as both $x \in (0, +\infty)$ and $y \in (0, +\infty)$ so we must also have $$ a - bx > 0 \mbox{ for all } x > 0. $$ and so we must have $$ 0 < x < \frac{a}{b}. $$
So the requirement that that $x$ and $y$ both vary over the entire range $(0, +\infty)$ cannot be met.
However, barring the above-mentioned unachievable condition, we can put $$ g\big(x, y(x) \big) = g(x, a-bx)= h(x), $$ and then $$ h^\prime(x) = \frac{\partial g}{\partial x} + \frac{\partial g}{\partial y}y^\prime(x) = \frac{\partial g}{\partial x} + \frac{\partial g}{\partial y} (-b) = \frac{\partial g}{\partial x} -b \frac{\partial g}{\partial y}. $$