Question

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

Answer 1

f(x) = e^sin 2x => f(0) = e^0 = 1

f'(x) = 2(cos 2x) e^sin2x => f'(0) = 2(1)e^0 = 2(1)(1) = 2

f''(x) = 2 cos 2x [2 (cos 2x)e^sin 2x] + e^sin 2x [(-4sin 2x)]
= 4 (cos 2x)^2 e^sin 2x - 4 (sin 2x) e^sin 2x

f''(0) = 4(1)(1) - 4(0)(1) = 4

The first three terms of the Maclaurin Series expansion are as follows:

1 + 2x + (4/2!) x^2 = 1 + 2x + 2x^2