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## Answers

Lets say is derivative, then

For

**y = x^(2n)**

**= (2n) * x^(2n-1)**

**= (2n) * (2n - 1) * x^(2n-2)**

**= (2n) * (2n - 1) * (2n - 2) * x^(2n - 3)**

So, similarly

**= (2n) * (2n - 1) * (2n - 2) * (2n - 3) * ........ * (2n - (n - 1)) * x^(2n - n)**

It simplifies to,

**= (2n) * (2n - 1) * (2n - 2) * (2n - 3) * ....... * (n + 1) * x^(n)**

Now, lets re-form this,

**= (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,

**= 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,

**= 2^(n) * {1.2.3....(2n - 1)} * x^(n)**