The fifth derivative of arctan(x)

[KiwiKid]

Homework Statement

Show that the fifth derivative of arctan(x) equals (120x^4 - 240x^2 + 24)/((x^2 + 1)^5). Then, calculate the fifth order Taylor series of arctan(x) around x = 0.

Homework Equations

Differentiation rules; knowing how a Taylor series works.

The Attempt at a Solution

I tried the brute-force way of differrentiating those fractions over and over and over again, resulting in one big mess. My answer ultimately turned out to be something different than what was asked, which isn't particularly surprising, considering the number of calculations I had to perform. I didn't even get to the Taylor series part.

My question is: is there a non-obvious solution to this problem that I'm missing? If so, could anyone give me a hint? I've been stuck on this problem for over an hour now and I'm not exactly fond of the idea of redoing all my previous calculations.

 

[Ray Vickson]

Homework Statement

Show that the fifth derivative of arctan(x) equals (120x^4 - 240x^2 + 24)/((x^2 + 1)^5). Then, calculate the fifth order Taylor series of arctan(x) around x = 0.

Homework Equations

Differentiation rules; knowing how a Taylor series works.

The Attempt at a Solution

I tried the brute-force way of differrentiating those fractions over and over and over again, resulting in one big mess. My answer ultimately turned out to be something different than what was asked, which isn't particularly surprising, considering the number of calculations I had to perform. I didn't even get to the Taylor series part.

My question is: is there a non-obvious solution to this problem that I'm missing? If so, could anyone give me a hint? I've been stuck on this problem for over an hour now and I'm not exactly fond of the idea of redoing all my previous calculations.

You need to show at least some of your work.

 

[KiwiKid]

You need to show at least some of your work.

f(x) = arctan(x); The first derivative: f'(x) = 1/(x^2 + 1); Second derivative: f''(x) = -2x/((x^2+1)^2). ...And so on. Except that it's around the third derivative that it starts to get more complicated, and it becomes virtually impossible to keep differentiating without making mistakes. At least, that's what I think, because 'my' fifth derivative of arctan(x) didn't match what I was supposed to get at all.

 

[SammyS]

f(x) = arctan(x); The first derivative: f'(x) = 1/(x^2 + 1); Second derivative: f''(x) = -2x/((x^2+1)^2). ...And so on. Except that it's around the third derivative that it starts to get more complicated, and it becomes virtually impossible to keep differentiating without making mistakes. At least, that's what I think, because 'my' fifth derivative of arctan(x) didn't match what I was supposed to get at all.

Writing those as:

f'(x) = (x^2 + 1)^-1

and

f''(x) = -2x(x^2+1)^-2

May help. You can then use the product rule rather than the quotient rule.

 

[KiwiKid]

Writing those as:

f'(x) = (x^2 + 1)^-1

and

f''(x) = -2x(x^2+1)^-2

May help. You can then use the product rule rather than the quotient rule.

Wow, thanks. That's probably it. *starts writing it all down once more*

 

[Ray Vickson]

Wow, thanks. That's probably it. *starts writing it all down once more*

What will also help is to put the whole thing over a common denominator before differentiating again, so before taking the (k+1)th derivative, make sure you write the kth derivative in the form
$$\frac{P_k(x)}{(x^2+1)^k},$$ where P_k(x) is a polynomial.