Differentiation - Chain Rule

[PhyAmateur]

In one physics problem if $$r^2= \lambda^2(1+\frac{m}{2\lambda})^2$$

what is $dr^2 ?$

Should I find $dr$ starting from $r= \lambda(1+\frac{m}{2\lambda})$ first and then square or find $dr^2$ starting from r^2? I know this is a basic question in differentiation using chain rule but it seems I am stuck over this..

 

[SteamKing]

A little more context would be helpful here.

Are you trying to find (dr)² or dr². I think there's a difference.

 

[PhyAmateur]

It is written as $$dr^2$$ I was just wondering if I should derive once and then after finding the answer, derive twice.. What do you say?

 

[SteamKing]

This notation is ambiguous to me. I can't advise you further without more information about this problem where you found it.

 

[Fredrik]

I would interpret $dr^2$ as $(dr)^2$, not as $d(r^2)$. The straightforward way is to solve for r, compute $dr=\frac{dr}{d\lambda}d\lambda$, and then square the result.

 

[PhyAmateur]

It is the same $$dr^2$$ found in the 3 sphere metric for example..

 

[PeroK]

It is the same $$dr^2$$ found in the 3 sphere metric for example..

$r^2 = λ^2(1 + \frac{m}{λ} + \frac{m^2}{4λ^2}) = λ^2 + mλ + \frac{m^2}{4}$

$r = λ(1 + \frac{m}{2λ}) = λ + \frac{m}{2}$

$\frac{d(r^2)}{dλ} = 2λ + m$

$d(r^2) = (2λ + m)dλ \ \ (A)$

$\frac{dr}{dλ} = 1$

$dr = dλ$

$(dr)^2 = (dλ)^2 \ \ (B)$

A or B, take your pick.