1. ## nth derivative formula

find an explicit formula for the $nth$ derivative of $1/x$

so, i know $f^{0}(x)=1/x$
$f^{1}(x)=-1/x^2$
$f^{2}(x)= 2/x^3$
$f^{3}(x)=-6/x^4$

i'm not sure i see the pattern..

2. $f(x) = \frac{1}{x} = x^{-1}$

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

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

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

I suggest looking now into that handy little ! operator.

3. 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 :/

Easier way to get that negative, though:

When n is -4, we have a negative coefficient. -1^(n - 1)

$(-1)^{n-1}(n - 1)!x^{-n}$

Edit: Whoops. Technically, this isn't nth derivative. But you don't have to change much. =p

$(-1)^{n}(n!)x^{-(n + 1)}$

So, first derivative, we have -1x^-2, right? Formula says, first derivative means n = 1.

$(-1)^1(1!)x^{-2} = -x^{-2}$

And n = 2?

$(-1)^2(2!)x^{-3} = 2x^{-3}$

Seems to work

5. okay, so.. i show work by just showing the calculation of the derivatives that you showed earlier?

6. That should be good enough. But I don't know your instructor.

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

Show a few derivatives. Recognize patterns and relationships between numbers and the order of each derivative.

$(1)(2)(3) = 3!$ occurs @ 3rd-order derivative. So that's how you get $n!$

$x^{-4}$ @ 3rd order derivative. So that's $-(n + 1)$ or $(1 - n)$

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 $(-1)^n$

Edit: Another way to show the $(-1)^n$:

$(-1)(-2)(-3)$ is the same as $(-1)(-1)(-1)(1)(2)(3)$, right? So that's $(-1)^3(1)(2)(3) = (-1)^33!$ right there.

7. this was a very big help, thanks so much

8. 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.