How do you find f^5(0) where f(x)=x/(1+x^2)?

Mar 14, 2017

${f}^{\left(5\right)} \left(0\right) = 120$

Explanation:

We can first construct the Maclaurin series for the function:

$\frac{1}{1 - x} = {\sum}_{n = 0}^{\infty} {x}^{n}$

Replacing $x$ with $- {x}^{2}$:

$\frac{1}{1 + {x}^{2}} = {\sum}_{n = 0}^{\infty} {\left(- {x}^{2}\right)}^{n} = {\sum}_{n = 0}^{\infty} {\left(- 1\right)}^{n} {x}^{2 n}$

Multiplying by $x$:

$\frac{x}{1 + {x}^{2}} = x {\sum}_{n = 0}^{\infty} {\left(- 1\right)}^{n} {x}^{2 n} = {\sum}_{n = 0}^{\infty} {\left(- 1\right)}^{n} {x}^{2 n + 1}$

We can write out the first terms of the Maclaurin series:

$\frac{x}{1 + {x}^{2}} = x - {x}^{3} + {x}^{5} - {x}^{7} + {x}^{9} + \ldots$

A general Maclaurin series is given by:

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

When $n = 5$, we see that f^((5))(0)/(5!)x^5 is a term of a general Maclaurin series.

In the Maclaurin series for $\frac{x}{1 + {x}^{2}}$, we see that the term ${x}^{5}$ is included. Thus:

x^5=f^((5))(0)/(5!)x^5

And:

f^((5))(0)=5! =120