[Math] How to prove derivative of $\cos x$ is $-\sin x$ using power series

Question

How to prove derivative of $\cos x$ is $-\sin x$ using power series?

So $\sin x=\sum \limits_{n=0}^\infty\dfrac{(-1)^nx^{2n+1}}{(2n+1)!}$ and $\cos x=\sum \limits_{n=0}^\infty\dfrac{(-1)^nx^{2n}}{(2n)!}$

Then $\cos'(x)=\sum \limits_{n=0}^\infty\left(\dfrac{(-1)^nx^{2n}}{(2n)!}\right)'=\sum \limits_{n=0}^\infty\dfrac{(-1)^nx^{2n-1}}{(2n-1)!}=…$

Then I don't to how to go to $\sin x$, could someone help?

Answer 1

$$\sum_{i=1}^{+\infty}a_i=\sum_{i=0}^{+\infty}a_{i+1} $$

$$\cos (x)=\sum_{n=0}^{+\infty}\frac {(-1)^nx^{2n}}{(2n)!} $$ $$=1+\sum_{n=0}^{+\infty}\frac {(-1)^{n+1}x^{2n+2}}{(2n+2)!} $$

$$\cos'(x)=0+\sum_{n=0}^{+\infty}\frac {(-1)^{n+1}x^{2n+1}}{(2n+1)!} $$ $$=-\sin (x ) $$

since $(-1)^n=-(-1)^{n+1} $.