# Simple Differential problem

1. Aug 8, 2007

### 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.

2. Aug 8, 2007

### Kummer

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 continous 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.