Intuitive explanation for $f(x,y) = g(nx+y) \implies \frac{\partial f}{\partial x} = n \frac{\partial f}{\partial y}$

Question

Consider the question I asked earlier. This tells that if $f(x,y) = g(nx+y)$ for $f,g : \mathbb{R}^2 \to \mathbb{R}$ where $f$ and $g$ are both differentiable, then $\frac{\partial f}{\partial x} = n \frac{\partial f}{\partial y}$.

The proof goes like this (to avoid reading the tagged question): $\frac{\partial f(x,y)}{\partial x} = \frac{dg(nx+y)}{dx} = n \frac{dg(nx+y)}{d(nx+y)} = n \frac{dg(nx+y)}{dy} = n \frac{\partial f(x,y)}{\partial y}$.

I am looking for an intuitive explanation for this. While it is (certainly) justified, the proof still feels less satisfactory.

Answer 1

An intuitive explanation can be the following, based on what a partial derivative is.

If you evaluate the partial derivative of $f(x,y)$ w.r.t $x$ you are approximating how much $f$ changes when $x$ changes by a unit value, everything else (in this case the other variable y) unchanged. Same for $y$.

Now change $x$ by 1. You see that the argument of $g$ than changes by $n$. If you change $y$ by 1 the argument of $g$ instead changes only by $1$. Since the overall change of $f$ at linear order is $g'$ (common in both cases) multiplied by the change of the argument, you expect a change of $f$ much larger when you change $x$ by a unit value, instead of $y$.

( If you are not convinced of considering changes by unit values replace 1 with a small number $\epsilon$. The argument is the same, I.e. changing x by a small amount changes the argument of g n times more than a change of y by the same amount. This larger change of the argument than propagates to a larger change of f )

Answer 2

Intuitively, this just says that $f(x,y)$ is constant on the family of level curves $$nx+y=c~({\rm of~level~}g(c)).$$ Since gradient vector is orthogonal to the level curves, one has $$\nabla f=k<n,1>~{\rm for~some}~k,$$ hence $$\frac{\partial f}{\partial x}=n\frac{\partial f}{\partial y}.$$