I suppose you mean
[tex]f(x)=\cos(x^2)[/tex]
Recall that
[tex]\cos x=\displaystyle\sum_{n=0}^\infty(-1)^n\frac{x^{2n}}{(2n)!}[/tex]
which converges everywhere. Then by substitution,
[tex]\cos(x^2)=\displaystyle\sum_{n=0}^\infty(-1)^n\frac{(x^2)^{2n}}{(2n)!}=\sum_{n=0}^\infty(-1)^n\frac{x^{4n}}{(2n)!}[/tex]
which also converges everywhere (and we can confirm this via the ratio test, for instance).
a. Differentiating the Taylor series gives
[tex]f'(x)=\displaystyle4\sum_{n=1}^\infty(-1)^n\frac{nx^{4n-1}}{(2n)!}[/tex]
(starting at [tex]n=1[/tex] because the summand is 0 when [tex]n=0[/tex])
b. Naturally, the differentiated series represents
[tex]f'(x)=-2x\sin(x^2)[/tex]
To see this, recalling the series for [tex]\sin x[/tex], we know
[tex]\sin(x^2)=\displaystyle\sum_{n=0}^\infty(-1)^{n-1}\frac{x^{4n+2}}{(2n+1)!}[/tex]
Multiplying by [tex]-2x[/tex] gives
[tex]-x\sin(x^2)=\displaystyle2x\sum_{n=0}^\infty(-1)^n\frac{x^{4n}}{(2n+1)!}[/tex]
and from here,
[tex]-2x\sin(x^2)=\displaystyle 2x\sum_{n=0}^\infty(-1)^n\frac{2nx^{4n}}{(2n)(2n+1)!}[/tex]
[tex]-2x\sin(x^2)=\displaystyle 4x\sum_{n=0}^\infty(-1)^n\frac{nx^{4n}}{(2n)!}=f'(x)[/tex]
c. This series also converges everywhere. By the ratio test, the series converges if
[tex]\displaystyle\lim_{n\to\infty}\left|\frac{(-1)^{n+1}\frac{(n+1)x^{4(n+1)}}{(2(n+1))!}}{(-1)^n\frac{nx^{4n}}{(2n)!}}\right|=|x|\lim_{n\to\infty}\frac{\frac{n+1}{(2n+2)!}}{\frac n{(2n)!}}=|x|\lim_{n\to\infty}\frac{n+1}{n(2n+2)(2n+1)}<1[/tex]
The limit is 0, so any choice of [tex]x[/tex] satisfies the convergence condition.
