
nth derivative formula
find an explicit formula for the $\displaystyle nth$ derivative of $\displaystyle 1/x$
so, i know $\displaystyle f^{0}(x)=1/x$
$\displaystyle f^{1}(x)=1/x^2$
$\displaystyle f^{2}(x)= 2/x^3$
$\displaystyle f^{3}(x)=6/x^4$
i'm not sure i see the pattern..(Wondering)

$\displaystyle f(x) = \frac{1}{x} = x^{1}$
$\displaystyle f(x)' = 1x^{2}$
$\displaystyle f(x)'' = (1)(2)x^{3}$
$\displaystyle f(x)''' = (1)(2)(3)x^{4}$
I suggest looking now into that handy little ! operator.

sorry, i'm a little confused.. is there such a thing as (n)! ? is that even what you're suggesting? you have to forgive me, i'm not very good at this :/

Google the Gamme function.
Easier way to get that negative, though:
When n is 4, we have a negative coefficient. 1^(n  1)
$\displaystyle (1)^{n1}(n  1)!x^{n}$
Edit: Whoops. Technically, this isn't nth derivative. But you don't have to change much. =p
$\displaystyle (1)^{n}(n!)x^{(n + 1)}$
So, first derivative, we have 1x^2, right? Formula says, first derivative means n = 1.
$\displaystyle (1)^1(1!)x^{2} = x^{2}$
And n = 2?
$\displaystyle (1)^2(2!)x^{3} = 2x^{3}$
Seems to work

okay, so.. i show work by just showing the calculation of the derivatives that you showed earlier?
thanks for your help!

That should be good enough. But I don't know your instructor.
$\displaystyle f^1(x) = (1)x^{2}$
$\displaystyle f^2(x) = (1)(2)x^{3}$
$\displaystyle f^3(x) = (1)(2)(3)x^{4}$
Show a few derivatives. Recognize patterns and relationships between numbers and the order of each derivative.
$\displaystyle (1)(2)(3) = 3!$ occurs @ 3rdorder derivative. So that's how you get $\displaystyle n!$
$\displaystyle x^{4}$ @ 3rd order derivative. So that's $\displaystyle (n + 1)$ or $\displaystyle (1  n)$
First order derivative has a negative coefficient, second has positive, third negative, fourth positive, etc. You know that every even gives a positive and every odd gives a negative. Hence $\displaystyle (1)^n$
Edit: Another way to show the $\displaystyle (1)^n$:
$\displaystyle (1)(2)(3)$ is the same as $\displaystyle (1)(1)(1)(1)(2)(3)$, right? So that's $\displaystyle (1)^3(1)(2)(3) = (1)^33!$ right there.

this was a very big help, thanks so much(Rofl)

Once you have "guessed" a formula for the nth derivative of 1/x, it should not be difficult to prove it by induction on n.