Generalizing derivatives and integrals

Start with a simple expression, $x^k$ , and take a few derivatives:

$\frac{\partial}{\partial x} x^k = k x^{k-1}$
$\frac{\partial^2}{\partial x^2} x^k = (k-1) k x^{k-2}$
$\frac{\partial^3}{\partial x^3} x^k = (k-2) (k-1) k x^{k-3}$
$\frac{\partial^4}{\partial x^4} x^k = (k-3) (k-2) (k-1) k x^{k-4}$

A pattern is emerging:

$\frac{\partial^n}{\partial x^n} x^k = (k-n+1)...(k-1)k x^{k-n} = c x^{k-n}$ , where c is the coefficient.

Now the hard part is finding the pattern in the coefficient. This needs to be taken out of the ‘…’ form. Focus on that:

$c = (k-n+1)...(k-1)k$

This is a series of numbers, each one larger than the next. This looks like a factorial, so divide k! by that:

$\frac{k!}{c} = \frac{k!}{(k-n+1)...(k-1)k} = \frac{1(2)(3)...k}{(k-n+1)...(k-1)k}$

Note how the top goes from 1 to k and the bottom goes from k-n+1 to k. That means that k-n+1 to k is a subset of 1 to k, so just divide that part out:

$\frac{k!}{c} = \frac{1(2)(3)...k}{(k-n+1)...(k-1)k} = 1(2)(3)...(k-n) = (k-n)!$

Now solve for c:

$\frac{1}{c} = \frac{(k-n)!}{k!}$ , so $c = \frac{k!}{(k-n)!}$

Puts this back into the original equation to get:

$\frac{\partial^n f(x)}{\partial x^n} = c x^{k-n} = \frac{k!}{(k-n)!} x^{k-n}$

So, the n^th derivative of $x^k$ is equal to:

$\frac{k!}{(k-n)!} x^{k-n}$

Verify this with a couple of derivatives:

$\frac{\partial}{\partial x} x^2 = \frac{k!}{(k-n)!} x^{k-n} = \frac{2!}{(2-1)!} x^{2-1} = \frac{2}{1} x^1 = 2x$

$\frac{\partial^2}{\partial x^2} 4 x^4 = 4 (\frac{k!}{(k-n)!}) x^{k-n} = 4 (\frac{4!}{(4-2)!}) x^{4-2} = 4 (\frac{24}{2}) x^2 = 4(12) x^2 = 48 x^2$

And a couple of integrals:

$\frac{\partial^{-1}}{\partial x^{-1}} x^2 = \frac{k!}{(k-n)!} x^{k-n} = \frac{2!}{(2+1)!} x^{2+1} = \frac{2}{6} x^3 = \frac{1}{3} x^3$

$\frac{\partial^{-7}}{\partial x^{-7}} 3 x^4 = 3 (\frac{k!}{(k-n)!}) x^{k-n} = 3 (\frac{4!}{(4+7)!}) x^{4+7} = 3 (\frac{24}{39916800}) x^{11} = \frac{1}{554400} x^{11}$

3 Responses to Generalizing derivatives and integrals

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Josh Franz says:

2009-01-06 at 10:19 am

Wow… I must admit, when I first saw this, I doubted I would comprehend any of this, but I actually understand it now that I went through it. Where did you find this? I’m definitely going to use this somehow sometime.
David says:

2009-01-11 at 4:08 pm

@Josh Franz:
I was just playing around with calculus for no reason whatsoever.

Maybe you can use it if one of your teachers is being a jerk and decides to ask for the 50th antiderivative of a function.

Generalizing derivatives and integrals

3 Responses to Generalizing derivatives and integrals

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