Implicit Differentiation usage of $\frac {dy}{dx}$

Question

Why, after differentiating $y$ on one side of the equation, is $dy/dx$ added?
As clarification an example I will provide an example:

Implicitly differentiate $y^2 = x$.

You get $$2y\frac {dy}{dx} = 1$$ as one of the first steps in differentiation. Why is the $dy/dx$ added after $y^2$ is differentiated?

Thanks!

Answer 1

$$y^2=(y(x))^2=x\\ \dfrac{d(y^2)}{dx}=1\\ \implies \dfrac{d(y^2)}{dx}=\dfrac{d(y^2)}{dy}\dfrac{dy}{dx}\text{ (chain rule )}=1\\ \implies 2y\dfrac{dy}{dx}=1$$

Answer 2

The idea behind implicit differentiation is that while $y$ isn't (necessarily) a function of $x$ we treat it as it it was.

So say that $y = f(x)$. Then you want to find $f'(x)$ in the following $$f(x)^2 = x.$$ So you take the derivative on both sides and use the chain rule to find the derivative of $f(x)^2$: $$\begin{align} \frac{d}{dx} f(x)^2 &= \frac{d}{dx} x & \Rightarrow \\ 2f(x)f'(x) &= 1 \end{align} $$ Now then we just replace $f(x)$ by $y$ and get $$ 2y\frac{dy}{dx} = 1. $$ So to answer you question about where the $\frac{dy}{dx}$ comes from, you can think of it as the derivative of the inner function $y$.