How should I understand the notation $dy/dx$ in implicit function differentiation?

Question

Let's say for the function $x^2+y^2=25$.

Taking the differential of $x^2$ and $y^2$, we have $2xdx+2ydy=0$

Then we end up with ${dy \over dx}=-{{x \over y}}$

My question is when talking about derivatives $dy \over dx$ should be taken as a whole, but here we end with the same notation, should I perceive ${dy \over dx}=-{{x \over y}}$ as y's derivative with respect to x as well?

Answer 1

Yes. It is the derivative of $y$ w.r.t. $x$. In fact you could have done it this way. Take the derivative w.r.t. $x$ on both sides. Then $$2x+2y\frac{dy}{dx}=0$$ $$\frac{dy}{dx}=-\frac{x}{y}$$

Answer 2

Yes, $\frac{dy}{dx}$ means the same thing in this case as everywhere else. It is the ratio of the instantaneous changes in $y$ compared to $x$.

The important thing to note is that with implicit differentiation, the derivative will often end up being a function of both $x$ and $y$. The reason for this is that, for implicit functions, there are often more than one point for any given $x$ value, so you need both $x$ and $y$ to identify the point whose slope you are looking at.

So, $\frac{dy}{dx} = -\frac{x}{y}$ means that you need both $x$ and $y$ to figure out what the slope is at that point. With explicit differentiation, you usually can figure it out with just the $x$ value.