Answers

2014-12-14T03:34:25+05:30
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)
1 5 1