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Решите задачу:

$y=ln (tg \frac{x}{3}), x= \pi \\\\ \frac{dy}{dx} =\frac{1}{tg \frac{x}{3}}* \frac{1}{cos^2 \frac{x}{3} } * \frac{1}{3}= \frac{cos\frac{x}{3}}{sin\frac{x}{3}}* \frac{1}{cos^2 \frac{x}{3} }*\frac{1}{3}= \frac{1}{3*sin\frac{x}{3}* cos\frac{x}{3} }=\frac{1}{ \frac{3}{2} *2*sin\frac{x}{3}* cos\frac{x}{3} }=\frac{1}{ \frac{3}{2} *sin\frac{2x}{3}}=\frac{2}{ 3sin\frac{2x}{3}} \\ \frac{dy}{dx} ( \pi )=\frac{2}{ 3sin\frac{2 \pi }{3}} =\frac{2}{ 3*\frac{ \sqrt{3} }{2}} =\frac{4}{ 3*\sqrt{3} } =\frac{4\sqrt{3} } {9} \\ \frac{d^2y}{dx^2} =(\frac{2}{ 3sin\frac{2x}{3}})'= \frac{2}{3} * (sin^{-1} \frac{2x}{3} )'=- \frac{2}{3} *sin^{-2} \frac{2x}{3}*cos \frac{2x}{3} * \frac{2}{3} =- \frac{4*cos \frac{2x}{3}}{9*sin^{2} \frac{2x}{3}} \\ \frac{d^2y}{dx^2} ( \pi )=- \frac{4*cos \frac{2 \pi }{3}}{9*sin^{2} \frac{2 \pi }{3}} =- \frac{4*(-\frac{1 }{2})}{9*( \frac{ \sqrt{3} }{2} )^{2} } =- \frac{-2}{9* \frac{3}{4} } =\frac{2}{\frac{27}{4} } = \frac{8}{27}$