Find the Maclaurin series for the given function f(x) = e-6x ...

The maclaurin series is:

$$\begin{gathered} \mathrm{f}\left(\mathrm{x}\right)={\mathrm{e}}^{-6\mathrm{x}} , \mathrm{f}\left(0\right)=1 \\ \mathrm{f}\text{'}\left(\mathrm{x}\right)=-6{\mathrm{e}}^{-6\mathrm{x}} , \mathrm{f}\text{'}\left(0\right)=-6\left(1\right)=-6 \\ \mathrm{f}\text{'}\text{'}\left(\mathrm{x}\right)=36{\mathrm{e}}^{-6\mathrm{x}} , \mathrm{f}\text{'}\text{'}\left(0\right)=36\left(1\right)=36 \\ \mathrm{f}\text{'}\text{'}\text{'}\left(\mathrm{x}\right)=-216{\mathrm{e}}^{-6\mathrm{x}} , \mathrm{f}\text{'}\text{'}\text{'}\left(0\right)=-6\left(1\right)=-216 \\ \mathrm{f}\text{'}\text{'}\text{'}\text{'}\left(\mathrm{x}\right)=1296{\mathrm{e}}^{-6\mathrm{x}} , \mathrm{f}\text{'}\text{'}\text{'}\text{'}\left(0\right)=1296\left(1\right)=1296 \end{gathered}$$

so,$f\left(x\right)=1-6x+\frac{36}{2!}{x}^2-\frac{216}{3!}{x}^3+\frac{1296}{4!}{x}^4+....$

The maclaurin series of $\mathrm{f}\left(\mathrm{x}\right)={\mathrm{e}}^{-6\mathrm{x}}$ is:

$f\left(x\right)=1-6x+\frac{36}{2!}{x}^2-\frac{216}{3!}{x}^3+\frac{1296}{4!}{x}^4+....$