Undergrad Chain rule and change of variables again

[jonjacson]

We start with:

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.

 

[Mark44]

We start with:

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.

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?

 

[jonjacson]

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?

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.

 

[Mark44]

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?

You can differentiate it with respect to x (not derive it).

X is the independent variable now

No, if x is a function of y, x is the dependent variable, and y is the independent variable.

, Does it make sense to differenciate a function respect itself?

Maybe I am confused and I should go to bed.

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.