### Formula for nth derivative

Started doing this in calculus. It's actually really fun. It feels almost like programming. And that's my random statement of the day.
There's a formula for it? I thought you just kept taking the derivative. I'm in AP Calculus BC and we're on Infinite Series right now - we did nth derivative last semester. Am I forgetting something?
Not a general formula that applies to any f, but given an f, find a formula that works for any nth prime.

You're way ahead of me (like 8 chapters in the book we use) so this is probably old stuff for you. But it's kind of fun for me.
Last edited on
By 'prime' I'm assuming you mean the prime symbol, like f -> f' -> f'' -> f(3) -> etc.

Don't get too used to that notation, the dy/dx notation is preferred because it works better in so many cases. It also makes nth derivative much easier:
y -> dy/dx -> d²y/d -> d³y/d -> etc...

It also makes separation of parts (something you'll see later) much easier to comprehend - it 'looks' like you multiply both sides by dx, but the "d/dx" is one symbol and *technically* you can't split it, but it makes so much sense to think of it that way (it was actually designed that way). You also have to use the notation for integrals (the opposite of derivatives).

Edit: prettied up formatting a bit
Last edited on
Well, you could build the definition of the nth derivative (by piling the definition of derivative on top of itself n times) and you can find a pattern in derivatives of most simple functions, but in general, there is no way to get the nth derivative, other than taking the derivative n times. If you don't believe me, try finding a pattern in derivatives of tan(x).
I actually just talked to the professor about this, and tan(x) and sec(x) came up. He claims there is a pattern, but it's not a nice one and it's just easier to take the derivative n times.

@L B, yea we're familiar with Leibnez (spelling?) notation. It was just easier to say prime then figure out how to make that look nice on a forum.
People would know what you were talking about if you said "dy/dx" ;)
 Leibnez (spelling?)

Leibniz.

Basically the coolest person ever.
L B wrote:
...it 'looks' like you multiply both sides by dx, but the "d/dx" is one symbol and *technically* you can't split it, but it makes so much sense to think of it that way (it was actually designed that way)....

d/dx is the differential operator and is considered one "thing," it's true. When the operator is apllied to a function, however, it represents the ratio of the (infinitesimal) changes of the dependent variable to the independent variable. It is this "duality" that makes many amazing things possible. For example, the derivative of the arcsin function.

Let y = arcsin x
then x = sin y

d/dx[arcsin x] = dy/dx = 1/ (dx/dy) = 1/ (d/dy[sin y]) = 1/ cos y

1/ cos y = 1/ sqrt(1 - sin2 y) = 1/ sqrt(1 - x2)

In conclusion: Math :)
@Resident Biscuit, it depends on what you call a pattern. See http://en.wikipedia.org/wiki/Bernoulli_number#Taylor_series_of_tan_and_tanh - Bernoulli numbers are not that easy to compute.
Speaking of derivatives, if someone is really bored and wants to work out the third derivative of e^x / x and tell me what you get, that would be cool.

I've done it three times and keep getting the wrong answer according to wolframalpha. My work seems right, but I guess it's not. It gets messy pretty quickly.
@ResidentBiscuit :

d3/dx3 [ex/x] =

ex/x ( -6/x3 + 4/x2 - 2/x + 1)
Last edited on
Hmm I keep getting:

(ex(-3x2 + 5x - 6))/ x4
Last edited on
 third derivative of e^x / x

The n^{th} derivative of (fg) equals f^(n)g + ... + C_{n,k} f^(n-k) g^(k)+...+f g^(n),
where C_{n,k}= n!/((n-k)! k!) is the Newtwon binomial "n choose k".

Now (e^x/x) is simply e^x *(1/x). You should already know the derivatives of e^x and (1/x).

Therefore

 `(e^x/x)'''= e^x/x - 3e^x/x^2+ 2*3 e^x/x^3 - 2*3 e^x/x^4= e^x/x - 3e^x/x^2+ 6 e^x/x^3 - 6 e^x/x^4`

[Edit:] Wolfram alpha agrees.
Last edited on
Yea I've been using wolfram to check my answers. It also appears to use the product ex(1/x), instead of just applying the quotient rule directly.

That shouldn't get a different answer though.
I never use the quotient rule. I always use the product rule with a negative exponent for the denominator.

y = exx-1

dy/dx = (d/dxex)(x-1) + (ex)(d/dxx-1) = exx-1 - exx-2 = y - yx-1

d²y/d = dy/dx - ((dy/dx)(x-1) + (y)(-x-2)) = dy/dx - dy/dxx-1 + yx-2

d³y/d = d²y/d - ((d²y/d)(x-1) + (dy/dx)(-x-2)) + ((dy/dx)(x-2) + (y)(-2x-3))
= d²y/d - d²y/dx-1 + 2dy/dxx-2 - 2yx-3
(not verified)

Then you just plug in and simplify. It's more writing than math - that's the thing about math, when you understand it, your only challenge is tedium with some problems. Don't even get me started on Rotation of Axes...
Last edited on
Topic archived. No new replies allowed.