Question

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Пожалуйста

Answer 1

$g'(x)=\frac{6x}{\pi } -2*\frac{1}{cos^2x}\\ \\g'(\frac{\pi }{3})=\frac{6}{\pi } *\frac{\pi }{3}-2*\frac{1}{0.5^2}=2-2*4=2-8=-6$

Answer 2

$g(x)=\frac{3x^{2}}{\pi }-2tgx-\pi\\\\g'(x)=\frac{3}{\pi}*(x^{2})'-2(tgx)'-\pi' =\frac{6}{\pi} x-\frac{2}{Cos^{2}x}\\\\g'(\frac{\pi }{3})=\frac{6}{\pi }*\frac{\pi }{3}-\frac{2}{Cos^{2}\frac{\pi }{3}}=2-\frac{2}{(\frac{1}{2})^{2}}=2-8=-6$