Maclaurin expansion of e^(sin x)

Let f(x) = e^(sin x). Maclaurin's series: f(x) = f(0) + x f′(0) + (x²/2!) f″(0) + (x³/3!) f‴(0) + (x⁴/4!) f⁗(0) + ...

f   = e^(sin x)
f′  = cos x · f
f″  = −sin x · f + cos²x · f            = f (cos²x − sin x)
f‴  = f′(cos²x − sin x) + f(−2 sin x cos x − cos x)
f⁗  = f″(cos²x − sin x) + 2f′(−2 sin x cos x − cos x)
        + f(−2cos 2x + sin x)

At x = 0 (sin 0 = 0, cos 0 = 1, f = 1):

f(0)  = 1
f′(0) = 1
f″(0) = 1·(1 − 0) = 1
f‴(0) = 1·1 + 1·(0 − 1) = 0
f⁗(0) = 1·1 + 2·1·(−1) + 1·(−2) = −3

So

f(x) = 1 + x + x²/2 + 0·x³ − 3x⁴/24 + ...

Answer: e^(sin x) = 1 + x + x²/2 − x⁴/8 + ...