Question

y=x^2n n-th derivative should be in this form y=2^n*{1.3.5....(2n-1)}x^n

Answer 1

First of all you should know this,
$\frac{d}{dx}( x^{n} ) = n ( x^{(n - 1)} )$

Lets say $y^{k}$ is $k^{th}$ derivative, then

For y = x^(2n)

$y^{1}$ = (2n) * x^(2n-1)
$y^{2}$ = (2n) * (2n - 1) * x^(2n-2)
$y^{3}$ = (2n) * (2n - 1) * (2n - 2) * x^(2n - 3)

So, similarly
$y^{n}$ = (2n) * (2n - 1) * (2n - 2) * (2n - 3) * ........ * (2n - (n - 1)) * x^(2n - n)
It simplifies to,
$y^{n}$ = (2n) * (2n - 1) * (2n - 2) * (2n - 3) * ....... * (n + 1) * x^(n)

Now, lets re-form this,
$y^{n}$ = (2n) * (2n - 2) * (2n - 4) * .... * (2) * (2n - 1) * (2n - 3) * (2n - 5) * ..... * (n + 1) * x^(n)

So, you have n even factors that have 2 common in them,
So,
$y^{n}$ = 2^(n) * (n) * (n - 1) * (n - 2) * ... * 1 * (2n - 1) * (2n - 3) * (2n - 5) * ..... * (n + 1) * x^(n)

This is what you need to prove,
$y^{n}$ = 2^(n) * {1.2.3....(2n - 1)} * x^(n)