Given this function : $s (x)=\sum_{k=0}^\infty (-1)^k\frac {x^{2k+1}}{(2k+1)!} $ , its derivative $c (x)=s' (x) $. How to show that $c'(x)=-s (x) $ ?
This question is related to Relationship Between Sine as a Series and Sine in Triangles
Is $c'(x)$ the double of derivative $ \sum_{k=0}^\infty (-1)^k\frac {x^{2k+1}}{(2k+1)!} $ ? So I take the derivative of $ \sum_{k=0}^\infty (-1)^k\frac {x^{2k+1}}{(2k+1)!} $ to arrive at $c(x)$ and then take derivative of $c(x)$ to arrive at c'(x) ?
Yes this is exactly the idea. \begin{align*} s(x) = \sum_{k=0}^\infty (-1)^k \frac{x^{2k+1}}{(2k+1)!}. \end{align*} Hence we have that: \begin{align*} c(x) = s'(x) = \sum_{k=0}^\infty (-1)^k \frac{(2k+1)x^{2k}}{(2k+1)!} = \sum_{k=0}^\infty(-1)^k \frac{x^{2k}}{(2k)!}. \end{align*} Now taking yet another derivative we get: \begin{align*} c'(x) &= \sum_{k=1}^\infty (-1)^k \frac{2kx^{2k-1}}{(2k)!} = \sum_{k=0}^\infty (-1)^{k+1}\frac{2(k+1)x^{2(k+1)-1}}{(2(k+1))!} \\ &= - \sum_{k=0}^\infty(-1)^k\frac{x^{2k+1}}{(2k+1)!} = -s(x). \end{align*}
