# Thread: derivative problem

1. ## 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

2. ## Re: derivative problem

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

Your notation is awful.

$\displaystyle \\t=n^2+2n\\ \frac{dt}{dn}=2n+2\text{ first derivative} \\\frac{d^2t}{dn^2}=2\text{ second derivative}\\\frac{d^3t}{dn^3}=0\text{ third derivative}$

3. ## Re: derivative problem

Originally Posted by Plato
Your notation is awful.

$\displaystyle \\t=n^2+2n\\ \frac{dt}{dn}=2n+2\text{ first derivative} \\\frac{d^2t}{dn^2}=2\text{ second derivative}\\\frac{d^3t}{dn^3}=0\text{ third derivative}$

i know this ......
but i saw $\displaystyle ${\rm{ }}d(dt) = 2d{n^2}$$ in a book and m curious to know how the author solved it...

4. ## Re: derivative problem

IMPLICIT FUNCTION. find dy /dx if x²y-5x=3