Implicit differentiation. Is y a function of x? What is going on?

Question

I am a bit confused at the concept of implicit differentiation. Say we have this question:

So I guess the implication here is that y is a function of x?

Is that that why $\frac{dy}{dx}$ of $\sin{y}$ is cos(y)$\frac{dy}{dx}$? I'm a bit confused.

I guess this stems from my confusion about the difference between $\frac{dy}{dx}$ vs $\frac{d}{dx}$

When we differentiate a simple equation like $y = 2x$ and say $\frac{dy}{dx} = 2$, the implication here is that y is a function of x obviously which is why we can take the derivative of y with respect to x. Is the same thing going on when taking the derivative of $\sin{y} = x$? I feel like I'm missing something or that I'm glossing over an important concept here.

Answer 1

Yes, $y$ is a function of $x$. Strictly speaking, we are taking $\sin (y(x)) = x$, and differentiating with respect to $x$ using the chain rule to obtain $$y'(x) \cos(y(x)) = 1$$ whereupon the conclusion follows as written.

This explicit writing of $y$ as $y(x)$ is something that is often not taught at all, I'm afraid, but whenever you define $y = ax+b$, for instance, this is shorthand for defining a function $y(x) = ax+b$.

Answer 2

The idea is that you have some values $x^*,y^*$ such that $\sin y^* = x^*$ and furthermore, $\sin' y^* \neq 0$.

I am using the $ ^*$ so I can distinguish values from functions.

If this is true, then the inverse function theorem says that there is a local differentiable inverse $y$ (to $\sin$) such that (for $z$ close to $y^*$) $y(\sin z ) = z$.

In particular, $y( \sin y^*) = y(x^*) = y^*$.

If we differentiate, we get $y'( \sin z ) \sin' z = y'( \sin z ) \cos z = 1$ and so, setting $z=y^*$ we get $y'(\sin y^*) = y'(x^*) = {1 \over \cos y^*}$.