John O' Meara said:
Given that [tex]\frac{dx}{dy} = (\frac{dy}{dx})^{-1}\\[/tex], differentiate throughout with respect to x and show that [tex]\frac{d^2x}{dy^2} = \frac{- \frac{d^2y}{dx^2}}{(\frac{dy}{dx})^3}\\[/tex].
An attempt: [tex]\frac{d^2 x}{dx dy} = \frac{d (\frac{dy}{dx})^{-1}}{dx} \\[/tex].
I need help to get me started. Thanks for the help.
Sorry, I am not familar with differentials.
Let [tex]f:(a,b)\to \mathbb{R}[/tex] be a one-to-one differentiable function. Since differentiable functions are continuous it means [tex]f[/tex] is an increasing function or decreasing function. Also it has an inverse function [tex]f^{-1}[/tex] which we shall write as [tex]g[/tex] for convienace.
Thus, [tex]g(f(x))=x[/tex] by definition of inverse function. Since [tex]f[/tex] is differentiable on [tex](a,b)[/tex] it follows from a inverse derivative theorem that [tex]g[/tex] is differentiable on [tex]\mbox{Im}(f)[/tex]. And so [tex]g(f(x))[/tex] is differentiable on [tex](a,b)[/tex]. Thus, by the chain rule we have upon differentiating both sides, [tex]f'(x)g'(f(x))=1[/tex]. Now since [tex]f(x)[/tex] is a strict monotonic function as explained above it means [tex]f'(x)>0 \mbox{ or }f'(x)<0 \mbox{ on }(a,b)[/tex]. The important point is that [tex]f'(x)\not = 0[/tex] at any point on [tex](a,b)[/tex]. Thus, we have that [tex]g'(f(x))=\frac{1}{f'(x)}[/tex]. Now we cannot continue to go further without knowing that [tex]f[/tex] is twice differentiable. Assume that it is then by the chain rule again we have, [tex]f'(x)g''(f(x)) = \frac{-f''(x)}{[f'(x)]^2}[/tex]. Since [tex]f'(x)\not = 0[/tex] as explained we finally arive at, [tex]g''(f(x)) = \frac{-f''(x)}{[f'(x)]^3}[/tex].
Now in differential notation this means, I believe,
[tex]\frac{d^2 x}{dy^2} = -\frac{\frac{d^2 y}{dx^2}}{\left( \frac{dy}{dx} \right)^3}[/tex].
Q.E.D.