Could someone explain chain rule when it comes to implicit differentiation?

Question

so basically I want to know why when we have something like:

$$v(x) = x - y + 1$$ If we take the derivative with respect to x, it yields:

$$v'(x) = 1 - \frac{dy}{dx}$$

Now I still don't understand why when it comes to implicit differentiation, we need to tag a $y'$ or $\frac{dy}{dx}$ after every time we take the derivative of a y term.

Thanks

Answer 1

Basically when we are "taking differentiation with respect to $x$", we mean that (intuitively) "when $x$ has a small change, how will the function change".

Now $y$ can be a function of $x$, e.g. $y=x^2$ or $y=e^{e^x}$. So a small change in $x$ will cause a change (called $\frac{dy}{dx}$) in $y$.

For example, if you are dealing with $\frac{d}{dx}y^2$, you need to do it as "if I change slightly $y$, we will get $2y$ as the change ($\frac{d}{dy}y^2=2y$). But since we are doing it with respect to $x$ so we need to multiply the term by 'when we change $x$ a bit, how $y$ will be changed [$\frac{dy}{dx}=y'$]. Hence the answer is $2y\frac{dy}{dx}$."

Of course it is only intuitively speaking, and the formal proof of chain rules etc. can be found online easily. I am trying to clarify why the "correction" term $\frac{dy}{dx}$ is necessary.

Answer 2

$y$ is a function of $x$ so when $x$ varies, it is going to cause $y$ to vary along with it.

If you want to treat $y$ as a variable that is independent from $x$ that is called partial differentiation, and it would give the same result as setting $\frac {dy}{dx} = 0$