Question

I need help I been stuck on this assignment for 2 weeks because of that question.

Answer 1

Answer:

A. $-2\sqrt{2\pi }$

Step-by-step explanation:

$f(x)=sin(x^{2}+\pi )$

$f^{'}(x)=cos(x^{2} +\pi ) 2x$

$f^{'}(\sqrt{2\pi } )=cos(2\pi +\pi )2\sqrt{2\pi }$

$f^{'} (\sqrt{2\pi } )=-1 (2\sqrt{2\pi } )$