Question

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

Answer 1

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

Explanation:

We can first construct the Maclaurin series for the function:

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

Replacing `x` with `-x^2`:

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

Multiplying by `x`:

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

In the Maclaurin series for `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`