Respuesta :

imtiazkalwar

Answer:

A. [tex]-2\sqrt{2\pi }[/tex]

Step-by-step explanation:

[tex]f(x)=sin(x^{2}+\pi )[/tex]

[tex]f^{'}(x)=cos(x^{2} +\pi ) 2x[/tex]

[tex]f^{'}(\sqrt{2\pi } )=cos(2\pi +\pi )2\sqrt{2\pi }[/tex]

[tex]f^{'} (\sqrt{2\pi } )=-1 (2\sqrt{2\pi } )[/tex]