`f(x) = xe^x , n=4` Find the n'th Maclaurin polynomial for the function.

Maclaurin series is a special case of Taylor series that is centered at `c=0` . The expansion of the function about `0` follows the formula:

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

or

`f(x)= f(0)+(f'(0))/(1!)x+(f^2(0))/(2!)x^2+(f^3(0))/(3!)x^3+(f^4(0))/(4!)x^4 +...`

To determine the Maclaurin polynomial of degree `n=4` for the given function `f(x)=xe^x` , we may apply the formula for Maclaurin series.

To list `f^n(x)` up to `n=4` , we may apply the Product rule for differentiation: `d/(dx) (u*v) = u' *v +u*v'` and derivative property: `d/(dx) (f+g) = d/(dx) f +d/(dx) g` .

`f(x)=xe^x`

Let: `u =x` then `u' = 1`

`v = e^x`  then `v' = e^x`

`d/(dx) (xe^x) =(1*e^x) + (x*e^x)`

`=e^x +xe^x`

`f'(x)=d/(dx) (xe^x)`

`= e^x +xe^x`

`f^2(x) = d/(dx) (e^x +xe^x)`

`=d/(dx) e^x + d/(dx) xe^x`

`= e^x + (e^x+xe^x)`

`= 2e^x+xe^x`

`f^3(x) = d/(dx) ( 2e^x +xe^x)`

`=d/(dx) 2e^x + d/(dx) xe^x`

`= 2e^x + (e^x+xe^x)`

`= 3e^x+xe^x`

`f^4(x) = d/(dx) ( 3e^x +xe^x)`

`=d/(dx) 3e^x + d/(dx) xe^x`

`= 3e^x + (e^x+xe^x)`

`= 4e^x+xe^x`

Plug-in `x=0` for each `f^n(x)` , we get:

`f(0)=0*e^0 `

`=0*1`

`=0`

`f'(0)=e^0+0*e^0`

` =1 +0*1`

`=1`

`f^2(0)=2e^0+0*e^0`

`=2*1 +0*1`

`=2`

`f^3(0)=3e^0+0*e^0`

`=3*1 +0*1`

`=3`

`f^4(0)=4e^0+0*e^0`

`=4*1 +0*1`

`=4`

Plug-in the values on the formula for Maclaurin series, we get:

`sum_(n=0)^4 (f^n(0))/(n!) x^n`

`= f(0)+(f'(0))/(1!)x+(f^2(0))/(2!)x^2+(f^3(0))/(3!)x^3+(f^4(0))/(4!)x^4`

`= 0+1/(1!)x+2/(2!)x^2+3/(3!)x^3+4/(4!)x^4 `

`= 0+1/1x+2/2x^2+3/6x^3+4/24x^4`

`= 0+x+x^2+1/2x^3+1/6x^4`

`= x+x^2+1/2x^3+1/6x^4`

The Maclaurin polynomial of degree `n=4` for the given function `f(x)=xe^x ` will be:

`P(x)=x+x^2+1/2x^3+1/6x^4`

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