# derivative problem

$t = {n^2} + 2n\$/extract_itex] $dt = 2ndn + 2dn\\$ $ddt = 2d{n^2}\\$ ${d^3}t = 0$ i know how dt, ${d^3}t$ is calculated but have no idea regarding ddt,..... what does it even mean • June 23rd 2013, 08:42 AM Plato Re: derivative problem Quote: Originally Posted by mpx86 $t = {n^2} + 2n\\$ $dt = 2ndn + 2dn\\$ $ddt = 2d{n^2}\\$ ${d^3}t = 0$ i know how dt, ${d^3}t$ is calculated but have no idea regarding ddt,..... what does it even mean Your notation is awful. $\\t=n^2+2n\\ \frac{dt}{dn}=2n+2\text{ first derivative} \\\frac{d^2t}{dn^2}=2\text{ second derivative}\\\frac{d^3t}{dn^3}=0\text{ third derivative}$ • June 23rd 2013, 09:01 AM mpx86 Re: derivative problem Quote: Originally Posted by Plato Your notation is awful. $\\t=n^2+2n\\ \frac{dt}{dn}=2n+2\text{ first derivative} \\\frac{d^2t}{dn^2}=2\text{ second derivative}\\\frac{d^3t}{dn^3}=0\text{ third derivative}$ i know this ...... but i saw $\[{\rm{ }}d(dt) = 2d{n^2}$$ in a book and m curious to know how the author solved it...