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
 A: $$
\begin{array}{lcl}
\frac{1}{1-\sin(x)} &=& 1+\sin(x)+\sin^2(x)+\sin^3(x)+\sin^4(x)+O(x^5) \\
&=& 1+\bigg(x-\frac{x^3}{6}\bigg)+\bigg(x-\frac{x^3}{6}\bigg)^2+
\bigg(x-\frac{x^3}{6}\bigg)^3+\bigg(x-\frac{x^3}{6}\bigg)^4+O(x^5) \\
&=& 1+\bigg(x-\frac{x^3}{6}\bigg)+\bigg(x^2-\frac{x^4}{3}\bigg)+
\bigg(x^3\bigg)+\bigg(x^4\bigg)+O(x^5) \\
&=& 1+x+x^2+\frac{5}{6}x^3+\frac{2}{3}x^4+O(x^5)
\end{array}
$$
We deduce
$$
\begin{array}{lcl}
f(x) &=& \bigg(1+x+x^2+\frac{5}{6}x^3+\frac{2}{3}x^4\bigg)
\bigg(1+x+\frac{x^2}{2}+\frac{x^3}{6}+\frac{x^4}{24}\bigg) +O(x^5) \\
&=& 1+2x+\frac{5}{2}x^2+\frac{5}{2}x^3+\frac{53}{24}x^4+O(x^5) \\
\end{array}
$$
So we see that $f^{(4)}(0)=53$.
A: 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 $4!\cdot a_4$.
A: 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 $g(x)=\dfrac1{1-\sin x}$
