Differentiating \frac{dx}{dy} = (\frac{dy}{dx})^{-1} with Respect to x

[John O' Meara]

Given that \frac{dx}{dy} = (\frac{dy}{dx})^{-1}\\, differentiate throughout with respect to x and show that \frac{d^2x}{dy^2} = \frac{- \frac{d^2y}{dx^2}}{(\frac{dy}{dx})^3}\\.

An attempt: \frac{d^2 x}{dx dy} = \frac{d (\frac{dy}{dx})^{-1}}{dx} \\.

I need help to get me started. Thanks for the help.

 

[Kummer]

Given that \frac{dx}{dy} = (\frac{dy}{dx})^{-1}\\, differentiate throughout with respect to x and show that \frac{d^2x}{dy^2} = \frac{- \frac{d^2y}{dx^2}}{(\frac{dy}{dx})^3}\\.

An attempt: \frac{d^2 x}{dx dy} = \frac{d (\frac{dy}{dx})^{-1}}{dx} \\.

I need help to get me started. Thanks for the help.

Sorry, I am not familar with differentials.

Let f:(a,b)\to \mathbb{R} be a one-to-one differentiable function. Since differentiable functions are continuous it means f is an increasing function or decreasing function. Also it has an inverse function f^{-1} which we shall write as g for convienace.

Thus, g(f(x))=x by definition of inverse function. Since f is differentiable on (a,b) it follows from a inverse derivative theorem that g is differentiable on \mbox{Im}(f). And so g(f(x)) is differentiable on (a,b). Thus, by the chain rule we have upon differentiating both sides, f'(x)g'(f(x))=1. Now since f(x) is a strict monotonic function as explained above it means f'(x)>0 \mbox{ or }f'(x)<0 \mbox{ on }(a,b). The important point is that f'(x)\not = 0 at any point on (a,b). Thus, we have that g'(f(x))=\frac{1}{f'(x)}. Now we cannot continue to go further without knowing that f is twice differentiable. Assume that it is then by the chain rule again we have, f'(x)g''(f(x)) = \frac{-f''(x)}{[f'(x)]^2}. Since f'(x)\not = 0 as explained we finally arive at, g''(f(x)) = \frac{-f''(x)}{[f'(x)]^3}.

Now in differential notation this means, I believe,
\frac{d^2 x}{dy^2} = -\frac{\frac{d^2 y}{dx^2}}{\left( \frac{dy}{dx} \right)^3}.
Q.E.D.