**Maclaurin series** is a special case of Taylor series that is centered at **a=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)x)/(1!)+(f^2(0))/(2!)x^2+(f^3(0))/(3!)x^3+(f^4(0))/(4!)x^4 +...`

To determine the `4th` Maclaurin polynomial from the given function `f(x)=e^(4x)` ,

we may apply derivative formula for exponential function:` d/(dx) e^u = e^u * (du)/(dx)`

Let `u =4x` then `(du)/(dx)= 4 `

Applying the values on the derivative formula for exponential function, we get:

`d/(dx) e^(4x) = e^(4x) *4`

Applying `d/(dx) e^(4x)= 4e^(4x)` for each `f^n(x)` , we get:

`f'(x) = d/(dx) e^(4x)`

`=e^(4x) * 4`

`= 4e^(4x)`

`f^2(x) = 4 *d/(dx) e^(4x)`

`= 4*4e^(4x)`

`=16e^(4x)`

`f^3(x) = 16*d/(dx) e^(4x)`

`= 16*4e^(4x)`

`=64e^(4x)`

`f^4(x) = 64*d/(dx) e^(4x)`

`= 64*4e^(4x)`

`=256e^(4x)`

Plug-in `x=0` , we get:

`f(0) =e^(4*0) =1`

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

`f^2(0) =16e^(4*0)=16`

`f^3(0) =64e^(4*0)=64`

`f^4(0) =2564e^(4*0)=256`

Note: `e^(4*0)=e^0 =1` .

Plug-in the values on the formula for Maclaurin series.

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

`= 1+4/(1!)x+16x^2+64x^3+256/(4!)x^4`

`=1+ 4/1x +16/(1*2)x^2 + 64/(1*2*3)x^3 +256/(1*2*3*4)x^4`

`=1+ 4/1x +16/2x^2 + 64/6x^3 +256/24x^4`

`= 1+4x+ 8x^2 + 32/3x^3 + 32/3x^4`

The **4th Maclaurin polynomial** for the given function `f(x)= e^(4x)` will be:

`e^(4x) =1+4x+ 8x^2 + 32/3x^3 + 32/3x^4`

or `P_4(x) =1+4x+ 8x^2 + 32/3x^3 + 32/3x^4`

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