# What is the n^(th) derivative of  y = x^(2n) ?

Jan 16, 2018

Let's have a look.

#### Explanation:

Let $y$ be a function of $x$, such that,

$y = f \left(x\right)$

$\therefore \textcolor{red}{y = {x}^{2 n}}$

Now, let ${y}_{n}$ be the ${n}^{t h}$ derivative of the function $y$.

$\therefore {y}_{1} = 2 n {x}^{2 n - 1}$

$\therefore {y}_{2} = 2 n \left(2 n - 1\right) n {x}^{2 n - 2}$

$\therefore {y}_{3} = 2 n \left(2 n - 1\right) \left(2 n - 2\right) n {x}^{2 n - 3}$
.
.
Following the sequences...
.
.
.
.
$\therefore \textcolor{red}{{y}_{n} = \left[2 n \left(2 n - 1\right) \ldots . \left(2 n - n + 1\right)\right] n {x}^{2 n - n}}$

$\therefore \textcolor{red}{{y}_{n} = n {x}^{n} \left[2 n \left(2 n - 1\right) \ldots . \left(n + 1\right)\right]}$

Hope it Helps:)

Jan 17, 2018

 y^((n)) = ((2n)!) / (n!) \ x^n

#### Explanation:

$y = {x}^{2 n}$

We can calculate the first few derivatives directly:

${y}^{\left(1\right)} = \left(2 n\right) {x}^{2 n - 1}$
${y}^{\left(2\right)} = \left(2 n\right) \left(2 n - 1\right) {x}^{2 n - 2}$
${y}^{\left(3\right)} = \left(2 n\right) \left(2 n - 1\right) \left(2 n - 2\right) {x}^{2 n - 3}$

From which we may conclude:

${y}^{\left(n\right)} = \left(2 n\right) \left(2 n - 1\right) \left(2 n - 2\right) \ldots \left(2 n - \left(n + 1\right)\right) {x}^{2 n - n}$
$\setminus \setminus \setminus \setminus \setminus \setminus \setminus = \left(2 n\right) \left(2 n - 1\right) \left(2 n - 2\right) \ldots \left(2 n - n - 1\right) {x}^{2 n - n}$
$\setminus \setminus \setminus \setminus \setminus \setminus \setminus = \left(2 n\right) \left(2 n - 1\right) \left(2 n - 2\right) \ldots \left(n - 1\right) {x}^{2 n - n}$
$\setminus \setminus \setminus \setminus \setminus \setminus \setminus = \frac{\left(2 n\right) \left(2 n - 1\right) \left(2 n - 2\right) \ldots \left(n - 1\right) n \left(n - 1\right) \ldots 1 {x}^{2 n - n}}{n \left(n - 1\right) \ldots 1}$
 \ \ \ \ \ \ \ = ((2n)!) / (n!) \ x^n