d) Find the 5th degree Taylor polynomial centered at x = 0 for the function y = \tiny \frac{x}{1+x}

Answer:[tex]P(x) = x - x^2 + x^3 - x^4+x^5[/tex]Step-by-step explanation:Let us first remember how a Taylor polynomial looks like:Given a differentiable function [tex]f[/tex] then we can find its Taylor series to the [tex]nth[/tex] degree as follows:[tex]P(x) = f(x_{0}) + f'(x_{0}).(x-x_{0}) + \frac{f''(x_{0})}{2!}.(x-x_{0})^2+.....+\frac{f^n(x_{0})}{n!}.(x-x_{0})^n + R_{n}(x).(x-x_{0})^n[/tex]Where [tex]R_{n}(x)[/tex] represents the Remainder and [tex]f^n(x)[/tex] is the [tex]nth[/tex] derivative of [tex]f[/tex].So let us find those derivatives.[tex]f(x) = \frac{x}{1+x}\\f'(x) = \frac{1}{(1+x)^2}\\f''(x) = \frac{-2}{(1+x)^3}\\f'''(x) = \frac{6}{(1+x)^4}\\f''''(x) = \frac{-24}{(1+x)^5}\\f'''''(x) = \frac{120}{(1+x)^6}[/tex]The only trick for this derivatives is for the very first one:[tex]f'(x) = \frac{1}{1+x} - \frac{x}{(1+x)^2}\\f'(x) = \frac{(1+x) - x}{(1+x)^2} = \frac{1}{(1+x)^2}\\[/tex]Then it's only matter of replacing on the Taylor Series and replacing [tex]x_{0}=0[/tex]