Derivatives of high order at 0 without differentiating

1. $\dfrac1{1+x}=\sum_{k\ge0}(-1)^kx^k$, so $\dfrac{x^2}{1+x}=\sum_{k\ge0}(-1)^kx^{k+2}$. The coefficient of $x^{10}$ is obtained for $k=8$: it is $(-1)^8=1$. Hence
$$f^{(10)}(0)=10!\times1=3\,628\,800.$$

2. From $\arctan x=\sum_{k\ge0}\frac{(-1)^k}{2k+1}x^{2k+1}$, the coefficient of $x^{2015}$ corresponds to $k=1007$: it is $-\frac1{2015}$. Hence
$$g^{(2015)}(0)=-\frac{2015!}{2015}=-2014!.$$

Remark. The general formula $g^{(2k+1)}(0)=(-1)^k(2k)!$ shows how fast the derivatives of $\arctan$ at $0$ grow, even though $\arctan$ is bounded.