Question

Помогите найти производную

Answer 1

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