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

According to the general Leibniz rule, $\displaystyle(f\cdot g)^{(n)}=\sum_{k=0}^n{n\choose k}f^{(k)}g^{(n-k)}$ . In this case, $n=4$ , $f(x)$ is $e^x$, and $g(x)=\dfrac1{1-\sin x}$
