
Calculus  Error
The time of oscillation, T seconds, of a simple pendulum is very nearly equal to 2/L , where L is the length of the pendulum in metres. A pendulum is made to have a time of oscillation of 1 second. Find the effect on the time, if there is an error of 2mm in the length.

This is my solution, somehow answer is different when I calculate without using calculus.
$\displaystyle {\delta L} = 2mm = 0.002m$
$\displaystyle \dfrac{dT}{dL}\approx\dfrac{\delta T}{\delta L}$
$\displaystyle {\delta T} = \dfrac{\delta T}{\delta L} * {\delta L}$
$\displaystyle {\delta T} = \dfrac{2}{L^2} * 0.002$
$\displaystyle {\delta T} = \dfrac{0.004}{L^2}$
Given that oscillation time, T = 1 second, so L = 2m
$\displaystyle {\delta T} = \dfrac{0.004}{2^2}$
$\displaystyle {\delta T} = 0.001s$
But, the answer given is 0.004s longer.
Can someone check if my working is right?