The general process I would use is to take all the derivatives you need to do, first.

That way you can simplify them first **out of context**, **when things aren't yet complicated** (you'll see). Then, plug them into the Taylor series formula, and use a calculator to numerically simplify the result.

**The result would be this.**

The formula for the Taylor series is:

#sum_(n=0)^(oo)f^n(a)/(n!)(x-a)^n#

Some things to remember are:

- #x->a# for #f(x) -> f(a)# but
*not* #(x-a)#.
- #n# increases but #a# stays the same.
- Plug in #a#
**last**.
- Remember to do the
**Chain Rule** and **Product Rule**!

So first, I would take the #n#th derivative of this. I want to stop at #n = 3#, though (you'll see why). Here goes:

#f^0(x) = color(blue)(f(x) = 1/(1+x^3))#

#color(blue)(f'(x)) = -1(1+x^3)^(-2)(3x^2) = (-3x^2)(1+x^3)^-2#

#color(blue)( = -(3x^2)/(1+x^3)^2)#

#color(blue)(f''(x)) = -[3x^2(-2(1+x^3)^-3(3x^2)) + (1+x^3)^-2(6x)]#

# = 18x^4(1+x^3)^-3 - 6x(1+x^3)^-2#

#color(blue)( = (18x^4)/(1+x^3)^3 - (6x)/(1+x^3)^2)#

#color(blue)(f'''(x)) = [18x^4(-3(1+x^3)^-4(3x^2)) + ((1+x^3)^-3)72x^3] - [6x(-2(1+x^3)^-3(3x^2)) + ((1+x^3)^-2)6]#

# = [-162x^6(1+x^3)^-4 + 72x^3(1+x^3)^-3] - [-36x^3(1+x^3)^-3 + 6(1+x^3)^-2]#

# = -162x^6(1+x^3)^-4 + 72x^3(1+x^3)^-3 + 36x^3(1+x^3)^-3 - 6(1+x^3)^-2#

# = -162x^6(1+x^3)^-4 + 108x^3(1+x^3)^-3 - 6(1+x^3)^-2#

#color(blue)( = -(162x^6)/(1+x^3)^4 + (108x^3)/(1+x^3)^3 - 6/(1+x^3)^2)#

I'm not going to simplify these anymore though :-P Alright, let's use them. Plug in #a# and whip out your calculator!

#sum_(n=0)^(oo)f^n(a)/(n!)(x-a)^n = sum_(n=0)^(oo)f^n(a)*(1/(n!))(x-a)^n#

#= (1/(1+a^3))(1/(0!))(x-a)^0 + #

#(-(3a^2)/(1+a^3)^2)(1/(1!))(x-a)^1 + #

#((18a^4)/(1+a^3)^3 - (6a)/(1+a^3)^2)(1/(2!))(x-a)^2 + #

#(-(162a^6)/(1+a^3)^4 + (108a^3)/(1+a^3)^3 - 6/(1+a^3)^2)(1/(3!))(x-a)^3#

#+ ...#

At #a = 1# for simplicity (you see why!):

#= 1/2 - 3/4(x-1) + (18/8 - 12/8)/2(x-1)^2 + (-162/16 + 216/16 - 24/16)/6(x-1)^3#

#= color(blue)(1/2 - 3/4(x-1) + 3/8(x-1)^2 + 5/16(x-1)^3 + ...)#

If you look here, Wolfram Alpha gives the same result! YEAH!

A fuller answer is:

#= color(blue)(1/2 + (-3/4)(x-1) + 3/8(x-1)^2 + 5/16(x-1)^3 + (-21/32)(x-1)^4 + 21/64(x-1)^5 + ...)#