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Question 1

Question 2

A)
$y= \frac{3x+2}{ \sqrt{x^2+3x+1} } \\ \\ y'= \frac{(3x+2)' \sqrt{x^2+3x+1}-(3x+2)( \sqrt{x^2+3x+1} )' }{( \sqrt{x^2+3x+1} )^2}= \\ \\ = \frac{3 \sqrt{x^2+3x+1} - \frac{(3x+2)(2x+3)}{2 \sqrt{x^2+3x+1} } }{x^2+3x+1}= \\ \\ = \frac{6(x^2+3x+1)-(3x+2)(2x+3)}{2(x^2+3x+1) \sqrt{x^2+3x+1} }=$
$= \frac{6x^2+18x-6-6x^2-4x-9x-6}{(2x^2+6x+2) \sqrt{x^2+3x+1} } = \\ \\ = \frac{5x-12}{(2x^2+6x+2) \sqrt{x^2+3x+1} }$

б)
$y=(2^{tg3x}-sec3x)^5 \\ \\ y'=5(2^{tg3x}-sec3x)^4*(2^{tg3x}*ln2* \frac{3}{cos^23x}- \frac{3sin3x}{cos^23x} )= \\ \\ =5(2^{tg3x}-sec3x)^4*( \frac{3*2^{tg3x}*ln2-3sin3x}{cos^23x} )$

в)
$y=ln(tg \frac{1}{ \sqrt{x} }) \\ \\ y'= \frac{1}{tg \frac{1}{ \sqrt{x} } } * \frac{1}{cos^2 \frac{1}{ \sqrt{x} } }*(- \frac{1}{2x \sqrt{x} } ) = \\ \\ = \frac{cos \frac{1}{ \sqrt{x} } }{sin \frac{1}{ \sqrt{x} } } * \frac{1}{cos^2 \frac{1}{ \sqrt{x} } } *(- \frac{1}{2x \sqrt{x} } )=$
$=- \frac{1}{(2sin \frac{1}{ \sqrt{x} }*cos \frac{1}{ \sqrt{x} } )(x \sqrt{x} )} = \\ \\ =- \frac{1}{x \sqrt{x} *sin \frac{2}{ \sqrt{x} } }$

г)
$y=ln \sqrt[3]{ \frac{10-3x^2}{x^3-10x} } \\ \\ y'= \frac{1}{ \sqrt[3]{ \frac{10-3x^2}{x^3-10x} } }* \frac{1}{3}*( \frac{10-3x^2}{x^3-10x} )^{- \frac{2}{3} }*( \frac{-6x(x^3-10x)-(10-3x^2)(3x^2-10)}{(x^3-10x)^2} )=$
$= \frac{1}{3}*( \frac{10-3x^2}{x^3-10x} )^{- \frac{1}{3} }*( \frac{10-3x^2}{x^3-10x} )^{- \frac{2}{3} }*( \frac{-6x^4+60x^2-30x^2-9x^4+100-30x^2}{(x^3-10x)^2} )= \\ \\ = \frac{1}{3}*( \frac{10-3x^2}{x^3-10x} )^{-1}*( \frac{-15x^4+100}{(x^3-10x)^2} )= \\ \\ = \frac{1}{3}* \frac{x^3-10x}{10-3x^2}* \frac{100-15x^4}{(x^3-10x)^2}=$
$= \frac{100-15x^4}{3(10-3x^2)(x^3-10x)}= \frac{100-15x^4}{3(10x^3-3x^5-100x+30x^3)}= \\ \\ = \frac{100-15x^4}{3(-3x^5+40x^3-100x)}= \frac{100-15x^4}{120x^3-9x^5-300x}$