# Question 90ee2

Mar 2, 2016

The first nonzero term of the Maclaurin series for the given function is $x$

#### Explanation:

The general form of the Maclaurin series for a function $f \left(x\right)$ is

sum_(n=0)^oof^((n))(0)/(n!)x^n

For the given function, at $n = 0$ we have

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

As this is not nonzero, we move on to $n = 1$

(f'(0))/(1!)x^1 = f'(0)*x#

The first derivative of $f$ is

$f ' \left(x\right) = - 2 {x}^{2} {e}^{- {x}^{2}} + {e}^{- {x}^{2}}$

Thus we have our first nonzero term as

$f ' \left(0\right) \cdot x = \left(- 2 \cdot {0}^{2} {e}^{- {0}^{2}} + {e}^{- {0}^{2}}\right) x$

$= \left(0 + 1\right) x$

$= x$