Please assume this is my initial exposure to calculus and derivatives.
I am having difficulty making the connection between the application of the chain rule to explicit differentiation and that of implicit differentiation. Everything I’ve learned so far about differentiation has been based on explicitly defined functions and limits. Applying the chain rule to explicit functions makes sense to me, as I am just recognizing composite functions within an original function. But applying the chain rule to the dependent variable in an implicit function doesn't feel the same to me. I know it works, but it just doesn't feel right. Take for example the equation $x\sqrt{y}=1$. I understand that $y$ is a function of $x$, but it is the very function $x\sqrt{y}=1$ that defines the relationship between x and y. This is different than how I have applied the chain rule when performing explicit differentiation. Take for example the function $y=x\sin(2x^2+2x+1)$. Recognizing that $2x^2+2x+1$ is a composite function, and applying the chain rule, makes sense to me. But for some reason, I cannot wrap my head around making that same association to $y$ in the function $x\sqrt{y}=1$.
Perhaps I am not completely understanding the chain rule. What am I missing? Thanks for any help you can provide.
Consider: $g(x)=x\sqrt{y(x)}$. Then: $$g'(x)=(x\sqrt{y(x)})'=1\cdot \sqrt{y(x)}+x\cdot \frac{1}{2\sqrt{y(x)}}\cdot y'(x)=0 \Rightarrow \\ y'(x)=\frac{-\sqrt{y(x)}\cdot 2\sqrt{y(x)}}{x}=-\frac{2y(x)}{x}.$$
Addendum: You are right that the balance must be preserved when $\frac{d}{dx}$ is applied: $$\frac{d}{dx}\left(x\sqrt{y}\right)=\frac{d}{dx}(1) \Rightarrow \sqrt{y}\frac{d}{dx}(x)+x\frac{d}{dx}(\sqrt{y})=0 \Rightarrow \cdots.$$ Also the differential of a function is $dy=y'dx$. Hence: $$y=x \Rightarrow dy=dx \Rightarrow y'dx=dx \Rightarrow y'=1 \\ y^2=x^2 \Rightarrow dy^2=dx^2 \Rightarrow 2ydy=2xdx \Rightarrow 2y\cdot y'dx=2xdx \Rightarrow y'=\frac{x}{y}=\frac{x}{x}=1.$$