If $\frac{dy}{dx}\frac{dx}{dy} = 1$, does $\frac{d^2 y}{dx^2} \frac{d^2 x}{dy^2} = 1$? I know $\frac{dy}{dx}\frac{dx}{dy} = 1$ because the chain rule says $1 = \frac{dy}{dy} = \frac{dy}{dx}\frac{dx}{dy}$.  But does $\frac{d^2 y}{dx^2} \frac{d^2 x}{dy^2} = 1$?  Or would that be too good to be true?
 A: What is true is that (assuming always that $y$ is a one-to-one function of $x$ on some interval)
$$ \eqalign{\dfrac{d^2 x}{dy^2} &= \dfrac{d}{dy} \dfrac{dx}{dy} = \dfrac{dx}{dy} \dfrac{d}{dx} \left(\dfrac{dy}{dx}\right)^{-1}\cr
&= - \left(\dfrac{dy}{dx}\right)^{-3} \dfrac{d^2 y}{dx^2} }$$
In order to have $\dfrac{d^2 x}{dy^2} \dfrac{d^2 y}{dx^2} = 1$, you'd need
$$ \left(\dfrac {d^2 y}{dx^2}\right)^2 = - \left( \dfrac{dy}{dx}\right)^3$$
The solutions of this differential equation include 
$$ y = \dfrac{4}{a+x} + b$$
for constants $a,b$.
A: If you want an example where the derivatives don't vanish, consider $y=x^3$ (so $x=y^{1/3}$).  Then:
$$
\frac{dy}{dx}=3x^2
$$ 
and
$$
\frac{dx}{dy}=\frac{1}{3}y^{-2/3}=\frac{1}{3}x^{-2}
$$
and
$$
\frac{dy}{dx}\frac{dx}{dy}=3x^2\cdot\frac{1}{3x^2}=1.
$$
On the other hand,
$$
\frac{d^2y}{dx^2}=6x
$$
and
$$
\frac{d^2x}{dy^2}=-\frac{2}{9}y^{-5/3}=-\frac{2}{9x^5}.
$$
However, 
$$
\frac{d^2y}{dx^2}\frac{d^2x}{dy^2}=6x\cdot\left(-\frac{2}{9x^5}\right)=-\frac{4}{3x^4}\not=1.
$$
A: Consider $y=x$. Then $\frac{dy}{dx} = 1$ and $\frac{dx}{dy} = 1$. However, $\frac{d^2y}{dx^2} = 0 = \frac{d^2x}{dy^2}$.
A: No it doesn't. Keep in mind that derivatives are operators, not variables. Still, that's a good thought process. For a counter-example, see above with $y=x$.
