How do you find f^6(0) where f(x)=xe^x?

Feb 6, 2017

${f}^{\left(6\right)} \left(0\right) = \left(0 + 6\right) {e}^{0} = 6.$

Explanation:

$f \left(x\right) = x {e}^{x}$

By the Product Rule, $f ' \left(x\right) = x \left({e}^{x}\right) ' + \left({e}^{x}\right) \left(x\right) '$

$\therefore f ' \left(x\right) = x {e}^{x} + {e}^{x} = \left(x + 1\right) {e}^{x} \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots . . \left(1\right)$

This means that, $\left\{x {e}^{x}\right\} ' = \left(x + 1\right) {e}^{x} \ldots \ldots \ldots \ldots . \left(\star\right)$

$\text{Now, } f ' ' \left(x\right) = \left\{f ' \left(x\right)\right\} ' = \left\{x {e}^{x} + {e}^{x}\right\} ' = \left\{x {e}^{x}\right\} ' + \left({e}^{x}\right) '$

$= \left(x + 1\right) {e}^{x} + {e}^{x} , \ldots \ldots \ldots \ldots \ldots \ldots . . \left[\because , \left(\star\right)\right]$

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

$\text{As, "f''(x)=xe^x+2e^x," we have, by } \left(\star\right) ,$

$f ' ' ' \left(x\right) = \left\{f ' ' \left(x\right)\right\} ' = \left\{x {e}^{x}\right\} ' + \left(2 {e}^{x}\right) ' = \left(x + 1\right) {e}^{x} + 2 {e}^{x} , i . e . ,$

$f ' ' ' \left(x\right) = \left(x + 3\right) {e}^{x}$.......................................(3)

$\text{Generalising, } {f}^{\left(n\right)} \left(x\right) = \left(x + n\right) {e}^{x} , n \in \mathbb{N} .$

$\text{In Particular, "f^((6))(x)=(x+6)e^x," giving,}$

${f}^{\left(6\right)} \left(0\right) = \left(0 + 6\right) {e}^{0} = 6.$

Mar 12, 2017

We can also use the Maclaurin series for ${e}^{x}$:

e^x=sum_(n=0)^oox^n/(n!)=1+x+x^2/(2!)+x^3/(3!)+x^4/(4!)+...

Multiplying this by $x$ we see that:

xe^x=x(1+x+x^2/(2!)+x^3/(3!)+x^4/(4!)+...)=x+x^2+x^3/(2!)+x^4/(3!)+x^5/(4!)+...

Or:

xe^x=sum_(n=0)^oox^(n+1)/(n!)

A general Maclaurin series is given by:

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

So when $n = 6$ the $6$th term of any general Maclaurin series is f^6(0)/(6!)x^6.

Also note that the $n = 5$ term of the $f \left(x\right) = x {e}^{x}$ series is x^6/(5!).

Equating their coefficients:

f^6(0)/(6!)x^6=x^6/(5!)

f^6(0)=(6!)/(5!)=6#