Question

Which of the following represents the Maclaurin series expansion of the functions y=e^(sin 2x) truncated to the third term?

Answer 1

have you taken the derivatives?

for the third term you will need to take until the third second derviative

Answer 2

whoops, i meant for the third terms take until the second derivative

Answer 3

first term, just plug in 0 into the function e^(sin 2*0), which is e^0 or just 1, so that is the first term

Answer 4

I meant "at the third term"

Answer 5

ok, then take the y''

it is nasty, but can be done

Answer 6

y'=2cos(2x)*e^(sin 2x)

Answer 7

now use product rule to find y''

y''=-4sin(2x)*e^(sin 2x)+4cos^2(2x)*e^(sin 2x)

plug in 0 and you should get 4

Answer 8

now since the third term will have x^2, you also have to divide by 2!, which means the coefficient will become 4/2=2

so the third term should be 2x^2 if I haven't made an algebra mistake

Answer 9

Would this answer be correct? 1+2x+2x^(2)

Answer 10

if you wanted all 3 terms, then yes that is correct

Answer 11

Thank you !!!!

Answer 12

hi alma,do you have the selections of answers?

Answer 13

f(x)=f(0)+f ' (0)x +f ''(0)x^2/(1*2) the 3rd term is
=f ''(0)x^2/(1*2) therefore after doing the 2nd derivative set x=0