# Proving an equation of f and f^(-1) derivatives

1. Jan 25, 2012

### karkas

1. The problem statement, all variables and given/known data

In one of my class's tests I've come across the following equation:

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

2. Relevant equations

Considering that $$\frac{dy}{dx}=\frac{1}{\frac{dx}{dy}}$$ how does one prove this statement?

3. The attempt at a solution

I've tried substituting like this:

$$\frac{d^2y}{dx^2}=\frac{d}{dx}\frac{dy}{dx}= \frac{d}{dx}\left(\frac{1}{\frac{dx}{dy}}\right)= \frac{ d \left( \frac{1}{u} \right)}{du}\frac{du}{dx}=-\left(\frac{1}{\frac{dx}{dy}}\right)^2 \frac{d\left(\frac{1}{\frac{dx}{dy}}\right)}{dx}$$

but i don't see how this can continue to finalize the proof...

2. Jan 25, 2012

### tiny-tim

hi karkas!

use the chain rule to replace that d/dx at the end by a d/dy

3. Jan 25, 2012

### karkas

How would that give me the needed second derivative of the inverse function??? I can't see it :(

4. Jan 25, 2012

### tiny-tim

hint: what is d/dy of 1/(dx/dy) ?

5. Jan 25, 2012

### karkas

Its (-d^2x/dy^2)/(dx/dy)^2 isn't it?

6. Jan 25, 2012

### tiny-tim

then doesn't that solve the question?

7. Jan 25, 2012

### karkas

oh man I should really learn to improve my writing...

Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook