To understand this derivation you need to understand the concept of work done and have a good ability to rearrange and substitute into equations.

To derive most energy equations, you begin with the definition for work done:

$W = F d$    (1)

Newton’s second law of motion tells us that $F = ma$

Therefore equation (1) becomes:

$W = mad$    (2)

The concept of energy is based entirely around this equation and as such we must look at the situation regarding the kinetic energy of a object as “the amount of energy to get the object travelling from a stationary position to it moving with a velocity v”. So we must now apply SUVAT (or the equations of motion):

We assume that the object started at rest and accelerated with a uniform rate of $a$ to a speed of $v$. In doing so the object travelled through a distance of $d$. We do not care about how long it took the object to get to this speed because the amount of energy required t get it to that speed would be the same no matter how long it took.

distance: $s = d$
initial speed: $u = 0$
final speed: $v = v$
acceleration: $a = a$
time: $t = n/a$

Algebraically we know all of the quantities except time, so using the equation $v^{2} - u^{2} = 2as$ and substituting in we get:

$v^{2} = 2ad$

rearranging for d gives:

$d = \frac{v^{2}}{2a}$     (3)

Equation (3) can now be substituted into equation (2) to give:

$W = mad = ma \frac{v^{2}}{2a}$

The acceleration cancels (which is a good thing because the equation for kinetic energy does not have acceleration as it is not the energy required to make an object speed up or slow down but rather just the energy for it to be moving at some speed) to give:

$W = m \frac{v^{2}}{2}$

Which can then be rearranged to give:

$W = \frac{1}{2} m v^{2}$

Or in a similar way to how you are given it in your examinations:

$W = 0.5 m v^{2}$

Since this type of energy is kinetic energy, the $W$ can just be changed into $E_{k}$, really you can use whatever symbol you want but to help an examiner mark your examination, use $E_{k}$ !

$E_{k} = 0.5 m v^{2}$