# Finding the nth derivative

• February 6th 2011, 10:23 AM
joatmon
Finding the nth derivative
Find the nth derivative of $f(x) = x^n$ by calculating the first few derivatives and observing the pattern that occurs.

So, here's what I've done so far:

$f'(x) = nx^{x-1})$

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

$f'''(x) = (n^3-2n^2+2n)x^{n-3}$

$f^{(4)}(x) = (n^4 - 5n^3 + 8n^2 - 6n)x^{n-4}$

The pattern of the exponents is clear to me, but the pattern of the coefficients is not. Can anybody help me? Thanks.
• February 6th 2011, 10:31 AM
Unknown008

$f'(x) = nx^{n-1}$

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

$f'''(x) = n(n-1)(n-2)x^{n-3}$

$f''''(x) = n(n-1)(n-2)(n-3)x^{n-4}$
• February 6th 2011, 10:39 AM
VonNemo19
Quote:

Originally Posted by joatmon
Find the nth derivative of $f(x) = x^n$ by calculating the first few derivatives and observing the pattern that occurs.

So, here's what I've done so far:

$f'''(x) = (n^3-2n^2+2n)x^{n-3}$

$f^{(4)}(x) = (n^4 - 5n^3 + 8n^2 - 6n)x^{n-4}$

The pattern of the exponents is clear to me, but the pattern of the coefficients is not. Can anybody help me? Thanks.

$f'=nx^{n-1}$

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

$f'''=n(n-1)(n-2)x^{n-3}$

$f^{(4)}=n(n-1)(n-2)(n-3)x^{n-3}$

So, looks like maybe the coefficients are following some kind of reverse factorial...

Something like $f^{(k)}=\frac{n!}{(n-k)!}x^{n-k}$

I could be wrong because I have to go and I haven't checked my work here, but maybe...
• February 6th 2011, 10:56 AM
TheCoffeeMachine
First, consider the nth derivative of $x^m$ (for clarity).
Finding the first few derivatives indicate that we've:

$\displaystyle (x^m)^{(n)} = x^{m-n}\prod_{k=0}^{n-1}(n-k)$.

This can be proven by induction. Put $m = n$ to get:

\begin{aligned}\displaystyle (x^n)^{(n)} = x^{n-n}\prod_{k=0}^{n-1}(n-k) = \prod_{k=0}^{n-1}(n-k) = n!\end{aligned}
• February 6th 2011, 11:22 AM
joatmon
All of these replies are helpful. Unknown008, that helps a lot to not multiply them out. I can see the sequence now clearly. What I don't know is how to write this in mathematical notation. The other two posts are a little beyond my abilities. I don't understand the notation used by CoffeeMachine, and I'm not sure that I agree with VonNemo's (although it might be right). Can anybody help me with how to write this out? Thanks.
• February 6th 2011, 11:37 AM
Unknown008
You might not be familar with the factorial notation, which is:

$n! = n(n-1)(n-2)(n-3)...(n-(n-1))$

For example,

$5! = 5\times4\times 3 \times 2 \times 1$

In the derivative, you put n! in the numerator, but you get everything till 1, but that's not in the derivative! So you divide by another factorial, such that the factorial in the denominator will cancel out anything that should not be in the numerator.

Example, you want 9 x 8 x 7 x 6

But you have 9!

What you do, is divide by 5!, so that:

$\dfrac{9!}{5!} = \dfrac{9\times8\times7\times6\times5\times4\times3 \times2\times1}{5\times4\times 3 \times 2 \times 1}$

What is unneeded after 6 is cancelled out. (Smile)
• February 6th 2011, 11:46 AM
joatmon
I'm somewhat familiar with factorial notation, but not to the extent that I feel comfortable writing this like you have done. Here is what I am going to do:

$\frac{d^ky}{dx^k} = n(n-1)(n-2)...(n-k+1) x^{n-k}$

Is the Leibnitz notation correct? They are throwing this notation at us in the book and in lecture, but never really telling us how to write it for ourselves.

Thanks.
• February 6th 2011, 06:00 PM
TheCoffeeMachine
Quote:

Originally Posted by joatmon
Is the Leibnitz notation correct?

If $y = x^n$, then what you've found is the kth derivative $y$. So if that's what you wanted, then it's absolutely fine. But wasn't your question, as stated in the OP, to find the nth derivative of $x^n$? If so, then you need to replace $k$ by $n$ in the equation you have found for the kth derivative of $y$, and then note that $x^{n-n} = 1$. The product $n(n-1)(n-2)...(n-k+1)$ evaluates to $n!$