If f(x)=∑n=1∞(−1)nx2nnf(x) = \sum_{n=1}^{\infty} \frac{(-1)^n x^{2n}}{n}f(x)=∑n=1∞n(−1)nx2n, what is f′(x)f'(x)f′(x)?
−2x1+x2\frac{-2x}{1+x^2}1+x2−2x
∑n=1∞(−1)nx2n−1\sum_{n=1}^{\infty} (-1)^n x^{2n-1}∑n=1∞(−1)nx2n−1
−11+x2\frac{-1}{1+x^2}1+x2−1
2x1+x2\frac{2x}{1+x^2}1+x22x