Question

How do you find the nth derivative of the function `f(x)=x^n`?

Answer 1

`n!`

Explanation:

`f(x)=x^n`
`f'(x)=nx^(n-1)`'
`f''(x)=n(n-1)x^(n-2)`
...
`f^((n))(x)=n(n-1)...3.2.1x^0`
`=n!`

Or by induction on `n` if you want a formal proof.

Answer 2

Each derivative gives us a pattern.

`f'(x) = nx^(n-1)`

`f''(x) = n(n-1)x^(n-2)`

`f'''(x) = n(n-1)(n-2)x^(n-3)`

and so on until `n - k = 0` where `k` is the order of the derivative. When we finish, we get:

`f^((k))(x) = n(n-1)(n-2)cdots(n-k+1)x^(n-k)`

When we go all the way to `n = k`, then:

`color(blue)(f^((n))(x) = n(n-1)(n-2)cdots(1)cancel(x^0)^(1))`

which is a constant equaling `color(blue)(n!)`, as `n! = n(n-1)(n-2)cdots(2)(1)`, and `x^0 = 1` which doesn't affect `n!`.