# derivative problem

• Jun 23rd 2013, 08:34 AM
mpx86
derivative problem
$\displaystyle t = {n^2} + 2n\\$
$\displaystyle dt = 2ndn + 2dn\\$
$\displaystyle ddt = 2d{n^2}\\$
$\displaystyle {d^3}t = 0$

i know how dt, $\displaystyle {d^3}t$ is calculated but have no idea regarding ddt,..... what does it even mean
• Jun 23rd 2013, 08:42 AM
Plato
Re: derivative problem
Quote:

Originally Posted by mpx86
$\displaystyle t = {n^2} + 2n\\$
$\displaystyle dt = 2ndn + 2dn\\$
$\displaystyle ddt = 2d{n^2}\\$
$\displaystyle {d^3}t = 0$

i know how dt, $\displaystyle {d^3}t$ is calculated but have no idea regarding ddt,..... what does it even mean

$\displaystyle \\t=n^2+2n\\ \frac{dt}{dn}=2n+2\text{ first derivative} \\\frac{d^2t}{dn^2}=2\text{ second derivative}\\\frac{d^3t}{dn^3}=0\text{ third derivative}$
• Jun 23rd 2013, 09:01 AM
mpx86
Re: derivative problem
Quote:

Originally Posted by Plato
$\displaystyle \\t=n^2+2n\\ \frac{dt}{dn}=2n+2\text{ first derivative} \\\frac{d^2t}{dn^2}=2\text{ second derivative}\\\frac{d^3t}{dn^3}=0\text{ third derivative}$
but i saw $\displaystyle ${\rm{ }}d(dt) = 2d{n^2}$$ in a book and m curious to know how the author solved it...