i cant believe i got this wrong but I did and I dont know why.

find the 2nd order Taylor Polynomial of$\displaystyle f(x)= cos(x^2) $ centered at zero

f(x) = cos x , then f(0) = 1

f'(x) = -sinx, f'(0) =0

f"(x) = -cosx f"(0) = -1

second order TP of cosx is:

$\displaystyle 1 + 0x - \frac{x^2}{2!}$

Replace x with x^2, second order TP of $\displaystyle cos(x^2)$ :

$\displaystyle 1 - \frac{x^4}{2!}$

But my book says that P"(x) = 1

What gives?