# I Chain rule and change of variables again

1. Jan 11, 2017

### jonjacson

d2y/dx2

And we want to consider x as function of y instead of y as function of x.

I understand this equality:

dy/dx = 1/ (dx/dy)

But for the second order this equality is provided:

d2y/dx2 =- d2x/dy2 / (dx/dy)3

Does anybody understand where is it coming from? The cubic term in the denominator looks quite strange, I don't know how to understand that equation.

ANy help is welcome.

2. Jan 11, 2017

### Staff: Mentor

Write your second equation above as $dy/dx = (dx/dy)^{-1}$
Now take the derivative with respect to x of both sides:
$d^2y/dx^2 = (-1) (dx/dy)^{-2} \cdot d/dx(dx/dy)$
On the right side above, I'm using the chain rule.
Is that enough of a start?

3. Jan 11, 2017

### jonjacson

It looks like a great start but I have a question.

If we have x as function of y, dx/dy will be a function of y too, WHat is the meaning of deriving this respect to x?
X is the independent variable now, Does it make sense to differenciate a function respect itself?

Maybe I am confused and I should go to bed.

4. Jan 11, 2017

### Staff: Mentor

You can differentiate it with respect to x (not derive it).
No, if x is a function of y, x is the dependent variable, and y is the independent variable.
In the last part of what I wrote I have $d/dx(dx/dy)$. That's the same as $\frac d {dy} \left(\frac {dx}{dy} \right)\cdot \frac{dy}{dx}$, with the chain rule being applied once more.