Question

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

Answer 1

Let's have a look.

Explanation:

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

`y=f(x)`

`:.color(red)(y=x^(2n))`

Now, let `y_n` be the `n^(th)` derivative of the function `y`.

`:.y_1=2nx^(2n-1)`

`:.y_2=2n(2n-1)nx^(2n-2)`

`:.y_3=2n(2n-1)(2n-2)nx^(2n-3)`
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Following the sequences...
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`:.color(red)(y_n=[2n(2n-1)....(2n-n+1)]nx^(2n-n))`

`:.color(red)(y_n=nx^n[2n(2n-1)....(n+1)])`

Hope it Helps:)

Answer 2

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

Explanation:

`y = x^(2n)`

We can calculate the first few derivatives directly:

`y^((1)) = (2n)x^(2n-1)`
`y^((2)) = (2n)(2n-1)x^(2n-2)`
`y^((3)) = (2n)(2n-1)(2n-2)x^(2n-3)`

From which we may conclude:

`y^((n)) = (2n)(2n-1)(2n-2)...(2n-(n+1))x^(2n-n)`
`\ \ \ \ \ \ \ = (2n)(2n-1)(2n-2)...(2n-n-1)x^(2n-n)`
`\ \ \ \ \ \ \ = (2n)(2n-1)(2n-2)...(n-1)x^(2n-n)`
`\ \ \ \ \ \ \ = ((2n)(2n-1)(2n-2)...(n-1)n(n-1)...1x^(2n-n)) / (n(n-1)...1)`
`\ \ \ \ \ \ \ = ((2n)!) / (n!) \ x^n`