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

Question

94917 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-01-28T19:47:39

$n!$ $n!$

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

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

Answer 2 · 2017-01-28T21:24:06

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 $n!$ .

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