Let $f \left({x}\right) = e^{\sin x}$. Then: {{begin-eqn}} {{eqn | l = \frac \d {\d x} f \left({x}\right)       | r = \cos x \, e^{\sin x}       | c = Chain Rule for Derivatives }} {{eqn | ll= \leadsto       | l = \frac {\d^2} {\d x^2} f \left({x}\right)       | r = \cos x \frac \d {\d x} e^{\sin x} + e^{\sin x} \frac \d {\d x} \cos x       | c = Product Rule }} {{eqn | r = \cos^2 x \, e^{\sin x} - \sin x \, e^{\sin x}       | c =  }} {{eqn | r = \left({\cos^2 x - \sin x}\right) e^{\sin x}       | c =  }} {{eqn | ll= \leadsto       | l = \frac {\d^3} {\d x^3} f \left({x}\right)       | r = \left({\cos^2 x - \sin x}\right) \frac \d {\d x} e^{\sin x} + e^{\sin x} \frac \d {\d x} \left({\cos^2 x - \sin x}\right)       | c = Product Rule }} {{eqn | r = \left({\cos^2 x - \sin x}\right) \cos x \, e^{\sin x} + e^{\sin x} \left({- 2 \cos x \sin x - \cos x}\right)       | c =  }} {{eqn | r = \left({\cos^3 x - 3 \sin x \cos x - \cos x}\right) e^{\sin x}       | c =  }} {{eqn | ll= \leadsto       | l = \frac {\d^4} {\d x^4} f \left({x}\right)       | r = \left({\cos^3 x - 3 \sin x \cos x - \cos x}\right) \frac \d {\d x} e^{\sin x} + e^{\sin x} \frac \d {\d x} \left({\cos^3 x - 3 \sin x \cos x - \cos x}\right)       | c = Product Rule }} {{eqn | r = \left({\cos^3 x - 3 \sin x \cos x - \cos x}\right) \cos x \, e^{\sin x} + e^{\sin x} \left({-3 \cos^2 x \sin x - 3 \left({-\sin^2 x + \cos^2 x}\right) + \sin x}\right)       | c =  }} {{eqn | r = \left({\cos^4 x - 3 \sin x \cos^2 x - \cos^2 x - 3 \cos^2 x \sin x + 3 \sin^2 x - 3 \cos^2 x + \sin x}\right) e^{\sin x}       | c =  }} {{eqn | r = \left({\cos^4 x - 6 \cos^2 x \sin x - 4 \cos^2 x + 3 \sin^2 x + \sin x}\right) e^{\sin x}       | c =  }} {{eqn | ll= \leadsto       | l = \frac {\d^5} {\d x^5} f \left({x}\right)       | r = \left({\cos^4 x - 6 \cos^2 x \sin x - 4 \cos^2 x + 3 \sin^2 x + \sin x}\right) \frac \d {\d x} e^{\sin x}       | c = Product Rule }} {{eqn | o =        | ro= +       | r = e^{\sin x} \frac \d {\d x} \left({\cos^4 x - 6 \cos^2 x \sin x - 4 \cos^2 x + 3 \sin^2 x + \sin x}\right)       | c =  }} {{eqn | r = \left({\cos^4 x - 6 \cos^2 x \sin x - 4 \cos^2 x + 3 \sin^2 x + \sin x}\right) \cos x \, e^{\sin x}       | c =  }} {{eqn | o =        | ro= +       | r = e^{\sin x} \left({-4 \cos^3 x \sin x - 6 \left({-2 \cos x \sin^2 x + \cos^3 x}\right) + 8 \cos x \sin x + 6 \cos x \sin x + \cos x}\right)       | c =  }} {{eqn | r = \left({\cos^5 x - 10 \cos^3 x \sin x - 10 \cos^3 x + 15 \cos x \sin^2 x + 15 \cos x \sin x + \cos x}\right) e^{\sin x}       | c =  }} {{end-eqn}} By definition of Taylor series: :$f \left({x}\right) \sim \displaystyle \sum_{n \mathop = 0}^\infty \frac {\left({x - \xi}\right)^n} {n!} f^{\left({n}\right)} \left({\xi}\right)$ and so expanding about $\xi = 0$: {{begin-eqn}} {{eqn | l = e^{\sin x}       | r = \frac {x^0} {0!} e^{\sin 0} + \frac {x^1} {1!} \cos 0 \, e^{\sin 0} + \frac {x^2} {2!} \left({\cos^2 0 - \sin 0}\right) e^{\sin 0}       | c =  }} {{eqn | o =        | ro= +       | r = \frac {x^3} {3!} \left({\cos^3 0 - 3 \sin 0 \cos 0 - \cos 0}\right) e^{\sin 0} + \frac {x^4} {4!} \left({\cos^4 0 - 6 \cos^2 0 \sin 0 - 4 \cos^2 0 + 3 \sin^2 0 + \sin 0}\right) e^{\sin 0}       | c =  }} {{eqn | o =        | ro= +       | r = \frac {x^5} {5!} \left({\cos^5 0 - 10 \cos^3 0 \sin 0 - 10 \cos^3 0 + 15 \cos 0 \sin^2 0 + 15 \cos 0 \sin 0 + \cos 0}\right) e^{\sin 0} + \cdots       | c =  }} {{eqn | r = 1 + x + \frac {x^2} 2 + \frac {x^3} 6 \left({1 - 0 - 1}\right) + \frac {x^4} {24} \left({1 - 0 - 4}\right) + \frac {x^5} {120} \left({1 - 0 - 10 + 1}\right) + \cdots       | c = Sine of Zero is Zero, Exponential of Zero, Cosine of Zero is One }} {{eqn | r = 1 + x + \frac {x^2} 2 - \frac {x^4} 8 - \frac {x^5} {15} + \cdots       | c =  }} {{end-eqn}} No pattern is immediately apparent. {{qed}}