Problem with derivatives and implicit derivation.

Question

The problem is the next

Proof that the identity $\left(\displaystyle\frac{dy}{dx}\right)\left( \displaystyle\frac{dx}{dy}\right)=1$ is equivalent with the next relation:

$$\left(\displaystyle\frac{dx}{dy} \right)^2\displaystyle\frac{d^3 y}{dx^3}+3\displaystyle\frac{d^2y}{dx^2}\displaystyle\frac{d^2x}{dy^2}+\left(\displaystyle\frac{dy}{dx} \right)^{2}\displaystyle\frac{d^3x}{dy^3}=0$$

First, I derivated the first identity respect to $x$, but, I'm so confussed with the variables. In this way, I can see that $\displaystyle\frac{dy}{dx}=y'(x)$ and $\displaystyle\frac{dx}{dy}=x(y)$, i.e., the function $y$ depends on $x$ and the function $x$ depends on $y$. Then, we have the next:

The derivative of $y'(x)x'(y)=1$ respect to $x$ is, if we rewrite the expression to:

$y'(x)=\displaystyle\frac{1}{x'(y)}$

$y''(x)=\displaystyle\frac{-x''(y)y'}{(x'(y))^{2}}$

But, my derivatives are correct? It is the correct via for solve the problem?

Answer 1

Hint: Suppose \begin{align} y = f(x) \ \ \text{ and } \ \ x= g(y) \end{align} then it follows we have the implicit function \begin{align} y = f(g(y)). \end{align} Hence taking the third derivative with respect to $y$ yields \begin{align} 0=&\ \frac{d^3}{dy^3}f(g(y)) = \frac{d^2}{dy^2}\left[\frac{df}{dx}\frac{dg}{dy}\right]= \frac{d}{dy}\left[\frac{d^2f}{dx^2}\left(\frac{dg}{dy}\right)^2+\frac{df}{dx}\frac{d^2g}{dy^2}\right]\\ =&\ \frac{d^3f}{dx^3}\left(\frac{dg}{dy}\right)^3+3\frac{d^2f}{dx^2}\frac{dg}{dy}\frac{d^2g}{dy^2}+\frac{df}{dx}\frac{d^3g}{dy^3} \end{align}