The coefficient of x^4 in the Maclaurin series is...

The question says:
"the coefficient of x^4 in the Maclaurin series for f(x) = e^(-x/2) is:"
How do I do this?

Use the Maclaurin series for e^x, then replace x by -x/2:

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

e^(-x/2) = 1 - x/2 + ((-x/2)^2)/2! + ((-x/2)^3)/3! + ((-x/2)^4)/4! + ...
= 1 - x/2 + (x^2)/8 - (x^3)/48 + (x^4)/384.

So the coefficient of x^4 in the Maclaurin series for f(x) = e^(-x/2) is 1/384.

Alternatively, the coefficient of x^4 in the Maclaurin series for f(x) = e^(-x/2) is
f^(4) (0) / 4!
= (-1/2)^4 e^(-x/2) / 4! @ x = 0, from repeated applications of the chain rule
= (-1/2)^4 e^(0) / 4!
= (1/16)/24
= 1/384, again.

Lord bless you today!