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

##### 1 Answer

Mar 14, 2017

#### Explanation:

We can first construct the Maclaurin series for the function:

#1/(1-x)=sum_(n=0)^oox^n#

Replacing

#1/(1+x^2)=sum_(n=0)^oo(-x^2)^n=sum_(n=0)^oo(-1)^nx^(2n)#

Multiplying by

#x/(1+x^2)=xsum_(n=0)^oo(-1)^nx^(2n)=sum_(n=0)^oo(-1)^nx^(2n+1)#

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

#x/(1+x^2)=x-x^3+x^5-x^7+x^9+...#

A general Maclaurin series is given by:

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

When

In the Maclaurin series for

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

And:

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