Question

Буду очень благодарна

Answer 1

Ответ:

Пошаговое объяснение:

(uv)' = u'v + uv'

$\displaystyle y'=(e^{x^2/2}*arcsin(x))' = xe^{x^2/2}*arcsin(x) +\frac{e^{x^2/2}}{\sqrt{1-x^2} }$

$\displaystyle y'_{(\sqrt{2} /2)}=\frac{\sqrt{2} }{2}*e^{1/4} *\frac{\pi}{4} +\frac{e^{1/4}}{\sqrt{1-\frac{2}{4} } } =\frac{\pi\sqrt[4]{e} }{4\sqrt{2} } +\sqrt{2} *\sqrt[4]{e} =\sqrt[4]{e} \bigg (\frac{\pi}{4\sqrt{2} } +\sqrt{2} \bigg )$