How find this $f^{(4)}(0)$ let
$$f(x)=\dfrac{e^x}{1-\sin{x}}$$
Find the value of
$$f^{(4)}(0)=?$$
My try: let
$$\dfrac{e^x}{1-\sin{x}}=a_{0}+a_{1}x+a_{2}x^2+a_{3}x^3+a_{4}x^4+\cdots$$
so 
$$e^x=(1-\sin{x})(a_{0}+a_{1}x+a_{2}x^2+a_{3}x^3+a_{4}x^4+\cdots)$$
since
$$e^x=1+x+\dfrac{x^2}{2}+\dfrac{x^3}{3}+\dfrac{x^4}{4}+\cdots$$
$$1-\sin{x}=1-x+\dfrac{1}{3!}x^3-\dfrac{1}{5!}x^5+\cdots$$
Follow I fell very ugly,maybe this problem  have nice methods?
Thank you

First, note that your $e^x$'s expansion is wrong as Shuchang pointed out.
Why don't you compare the coefficients? You'll get
$$1=a_0$$
$$1=a_1-a_0$$
$$\frac{1}{2!}=a_2-a_1$$
$$\frac{1}{3!}=a_3-a_2+\frac{a_0}{3!}$$
$$\frac{1}{4!}=a_4-a_3+\frac{a_1}{3!}$$
Then, the answer is $4!\cdot a_4$.
