Higher Derivatives: What is the Pattern for Finding the nth Derivative of x^n?

[Jeff Ford]

I'm a little stuck on this one

Find f^{(n)}(x) = x^n

I know f'(x) = n(x^{n-1})
and f''(x) = n(n-1)(x^{n-2}) and so on

But I can't seem to see the pattern that leads to the answer n!

 

[VietDao29]

I'm a little stuck on this one

Find f^{(n)}(x) = x^n

I know f'(x) = n(x^{n-1})
and f''(x) = n(n-1)(x^{n-2}) and so on

But I can't seem to see the pattern that leads to the answer n!

Then just carry on differentiating some more times:
f(x) := xⁿ
f'(x) = nx^{n - 1}.
f''(x) = (f'(x))' = (nx^{n - 1})' = n(n - 1)x^{n - 2}.
f'''(x) = (f''(x))' = (n(n - 1)x^{n - 2})' = n(n - 1)(n - 2)x^{n - 3}.
f''''(x) = n(n - 1)(n - 2)(n - 3)x^{n - 4}.
f^(v)(x) = n(n - 1)(n - 2)(n - 3)(n - 4)x^{n - 5}.
...
Now if k < n, what can you say about f^(k)(x)?
If k = n then f^(k)(x) = ? (i.e, what's f⁽ⁿ⁾(x))
--------------
Another hint:
Consider g(x) := x², so g''(x) = 2.
h(x) := x³, so h'''(x) = 6.
Can you go from here? :)

 

[arildno]

Well, you could prove by induction that we have:
f^{(l)}(x)=\frac{n!}{(n-l)!}x^{n-l}, l\leq{n}

 

[Jeff Ford]

I think I've got it. If I expanded it out far enough

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

Since n-(n-1) = 1 the coefficient becomes 1(2)(3)...(n-1)(n) = n! and x^{n-n} = x^0 = 1

So the whole things boils down to f^{(n)} (x) = n!(1) = n!