Ответ:
Urinmaning burchak koeffitsiyenti uning musbat \(Ox\) yo‘nalishi bilan hosil qilgan burchagining tangensiga teng. Shuning uchun
\(f'(-5)=\tan 120^\circ=-\sqrt3\).
\(d_1\) va \(d_2\) parallel bo‘lgani sababli ularning burchak koeffitsiyentlari teng:
\(g'(-5)=f'(-5)=-\sqrt3\).
Ko‘paytmaning hosilasi qoidasiga ko‘ra, \(h'(-5)=f'(-5)g(-5)+f(-5)g'(-5)\). Berilgan nuqtalardan \(f(-5)=\sqrt3\) va \(g(-5)=3\sqrt3\).
\(h'(-5)=(-\sqrt3)(3\sqrt3)+(\sqrt3)(-\sqrt3)=-9-3=-12\).
Javob: \(h'(-5)=-12\).
