Higher derivatives

1. Feb 26, 2006

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 $$n!$$

2. Feb 26, 2006

VietDao29

Then just carry on differentiating some more times:
f(x) := xn
f'(x) = nxn - 1.
f''(x) = (f'(x))' = (nxn - 1)' = n(n - 1)xn - 2.
f'''(x) = (f''(x))' = (n(n - 1)xn - 2)' = n(n - 1)(n - 2)xn - 3.
f''''(x) = n(n - 1)(n - 2)(n - 3)xn - 4.
f(v)(x) = n(n - 1)(n - 2)(n - 3)(n - 4)xn - 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))
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Another hint:
Consider g(x) := x2, so g''(x) = 2.
h(x) := x3, so h'''(x) = 6.
Can you go from here? :)

3. Feb 26, 2006

arildno

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

4. Feb 26, 2006

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 $$f^{(n)} (x) = n!(1) = n!$$