Derivative of implicit fuction

Question

I want the proof of implicit fuction derivative. I don't know why I should calculate derivative of all monomials towards x for finding $y'$ (derivative of $y$) with respect to x at equations such as $y^2=x$ or $x^5+4xy^3-5=2$? Why is the derivative of $y^2$ with respect to $x$ equal to $2y.dx/dy$ in the equation $y^2=x$?

Answer 1

It follows from the chain rule. Think of $y$ as a function of $x$. The equation $y^2 = x$ is actually $y(x)^2 = x$.

One way of writing the chain rule is: $$\frac{d}{dx} f(g(x)) = f'(g(x)) g'(x).$$

So in your question, $$\frac{d}{dx} y(x)^2 = 2 y(x) y'(x) = 2 y \frac{dy}{dx}.$$

Answer 2

Consider $y^2=x$. Now for implicit derivative, we use the chain rule. Get an idea of the chain rule here. Since $y=y(x)$, and we are interested in finding derivative wrt x. But there is a term involving $y^2$. So we first differentiate $y^2$ and then since $y$ is dependent on $x$, we differentiate wrt x. So this gives
$2y\times\dfrac{dy}{dx}=1\implies \dfrac{dy}{dx}=\dfrac{1}{2y} \quad \square$