# Generalizing derivatives and integrals :: 03 Jan 2009

Disclaimer: This is sloppy math. Don’t take it too seriously.

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 $$n^{\text{th}}$$ 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:

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

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