Mathematics

The key is recognizing that you don't need to compute five derivatives by hand—instead, use the power series expansion of $p(x)$. Once you write $p(x)$ as a Taylor series around $x=0$, the coefficient of $x^5$ directly encodes the fifth derivative through the formula $p^{(5)}(0) = 5! \cdot (\text{coefficient of } x^5)$. Start with the geometric series $\frac{1}{1-u} = 1 + u + u^2 + \cdots$ and substitute $u = -x^2$.

The key is recognizing that you don't need to compute five derivatives by hand—instead, use the power series expansion of $p(x)$. Once you write $p(x)$ as a Taylor series around $x=0$, the coefficient of $x^5$ directly encodes the fifth derivative through the formula $p^{(5)}(0) = 5! \cdot (\text{coefficient of } x^5)$. Start with the geometric series $\frac{1}{1-u} = 1 + u + u^2 + \cdots$ and substitute $u = -x^2$.