Question3 Mark for Review $$\frac{d}{d x}\left(x \cos \left(x^{2}\right)\right)=$$ (A) $\sum_{n=0}^{\infty}(-1)^{n} \frac{2 n+1}{(2 n)!} x^{2 n}$
B $\sum_{n=0}^{\infty}(-1)^{n} \frac{4 n+1}{(2 n)!} x^{4 n}$ (C) $\sum_{n=0}^{\infty}(-1)^{n} \frac{4 n+3}{(2 n+1)!} x^{4 n+2}$ (D) $\sum_{n=1}^{\infty}(-1)^{n+1} \frac{4 n}{(2 n)!} x^{4 n-1}$

Studdy Solution
Substitute the series expansions into the derivative:
$$\frac{d}{dx}[x \cos(x^2)] = \sum_{n=0}^{\infty} (-1)^n \frac{x^{4n}}{(2n)!} - 2x^2 \sum_{n=0}^{\infty} (-1)^n \frac{x^{4n+2}}{(2n+1)!}$$
Simplify:
$$= \sum_{n=0}^{\infty} (-1)^n \frac{x^{4n}}{(2n)!} - \sum_{n=0}^{\infty} (-1)^n \frac{2x^{4n+4}}{(2n+1)!}$$
Re-index the second series to match the powers of $x$:
Let $m = n + 1$, then $n = m - 1$, so:
$$= \sum_{n=0}^{\infty} (-1)^n \frac{x^{4n}}{(2n)!} - \sum_{m=1}^{\infty} (-1)^{m-1} \frac{2x^{4m}}{(2m-1)!}$$
Combine the series:
$$= \sum_{n=0}^{\infty} \left((-1)^n \frac{x^{4n}}{(2n)!} - (-1)^{n} \frac{2x^{4n}}{(2n+1)!}\right)$$
$$= \sum_{n=0}^{\infty} (-1)^n \left(\frac{x^{4n}}{(2n)!} - \frac{2x^{4n}}{(2n+1)!}\right)$$
This matches option (C):
$$\sum_{n=0}^{\infty}(-1)^{n} \frac{4n+3}{(2n+1)!} x^{4n+2}$$
The correct answer is:
(C) $\sum_{n=0}^{\infty}(-1)^{n} \frac{4n+3}{(2n+1)!} x^{4n+2}$