If f(x)=∑n=0∞(−1)nx2n(2n)!f(x) = \sum_{n=0}^{\infty} \frac{(-1)^n x^{2n}}{(2n)!}f(x)=∑n=0∞(2n)!(−1)nx2n, what is f′(x)f'(x)f′(x)?
∑n=0∞(−1)n+1x2n+1(2n+1)!\sum_{n=0}^{\infty} \frac{(-1)^{n+1} x^{2n+1}}{(2n+1)!}∑n=0∞(2n+1)!(−1)n+1x2n+1
−sin(x)-\sin(x)−sin(x)
∑n=1∞(−1)n⋅2n⋅x2n−1(2n)!\sum_{n=1}^{\infty} \frac{(-1)^n \cdot 2n \cdot x^{2n-1}}{(2n)!}∑n=1∞(2n)!(−1)n⋅2n⋅x2n−1