Question

Решить 2.1 ,2.2 ......

Answer 1

Только производные:
2.1
$y=sin \frac{x}{2}+cos \frac{x}{2} \\ y'= \frac{1}{2} cos \frac{x}{2}- \frac{1}{2}sin \frac{x}{2}$

2.2
$y= \sqrt[3]{(4+3x)^2}=(4+3x)^{ \frac{2}{3} } \\ y'= \frac{2}{3}(4+3x)^{ \frac{2}{3}- \frac{3}{3} }*3=2(4+3x)^{- \frac{1}{3} }= \frac{2}{ \sqrt[3]{4+3x} }$

2.3
$y= \sqrt{2x-sin2x} \\ y'= \frac{1}{2 \sqrt{2x-sin2x} }*(2-2cos2x)= \frac{2(1-cos2x)}{2 \sqrt{2x-sin2x} }= \frac{1-cos2x}{ \sqrt{2x-sin2x} }$

2.4
$y= \sqrt[4]{1+cos^{2}x} =(1+cos^{2}x)^{ \frac{1}{4} } \\ y'= \frac{1}{4}(1+cos^2x)^{ \frac{1}{4}- \frac{4}{4} }*(2cosx*(-sinx))= \\ = \frac{1}{4}(1+cos^2x)^{- \frac{3}{4} }*(-2sinxcosx)=- \frac{sin2x}{4 \sqrt[4]{(1+cos^2x)^3} }$

2.5
$y=sin \sqrt{x} \\ y'=cos \sqrt{x} * \frac{1}{2 \sqrt{x} } = \frac{cos \sqrt{x} }{2 \sqrt{x} }$

2.6
$y= \frac{1}{(1+cos4x)^5}=(1+cos4x)^{- 5 } \\ y'=-5(1+cos4x)^{-5-1}*(-4sin4x)= \frac{20sin4x}{(1+cos4x)^6}$

2.7
не понятно условие

2.8
$y=x \sqrt{x^2-1} \\ y'=x'*( \sqrt{x^2-1} )+x*( \sqrt{x^2-1} )'= \sqrt{x^2-1}+ \frac{x}{2 \sqrt{x^2-1} }*2x= \\ = \sqrt{x^2-1}+ \frac{x^2}{ \sqrt{x^2-1} }= \frac{x^2-1+x^2}{ \sqrt{x^2-1} }= \frac{2x^2-1}{ \sqrt{x^2-1} }$

2.9
$y= \sqrt{4x+sin4x} \\ y'= \frac{1}{2 \sqrt{4x+sin4x} }*(4+4cos4x)= \frac{4(1+cos4x)}{2 \sqrt{4x+sin4x} }= \\ = \frac{2(1+cos4x)}{ \sqrt{4x+sin4x} }= \frac{2+2cos4x}{ \sqrt{4x+sin4x} }$

2.10
$y= \frac{1+sin2x}{1-sin2x} \\ y'= \frac{(1+sin2x)'*(1-sin2x)-(1+sin2x)*(1-sin2x)'}{(1-sin2x)^2}= \\ =\frac{2cos2x(1-sin2x)-(1+sin2x)*(-2cos2x)}{(1-sin2x)^2}= \\ = \frac{2cos2x-2sin2xcos2x+2cos2x+2sin2xcos2x}{(1-sin2x)^2}= \\ = \frac{4cos2x}{(1-sin2x)^2}$

2.11
$y= \sqrt{ \frac{x}{2}-sin \frac{x}{2} } \\ y'= \frac{1}{2 \sqrt{ \frac{x}{2}-sin \frac{x}{2} } }*( \frac{1}{2}- \frac{1}{2}cos \frac{x}{2} )= \\ = \frac{ \frac{1}{2}(1-cos \frac{x}{2} ) }{2 \sqrt{ \frac{x}{2}-sin \frac{x}{2} } } = \frac{1-cos \frac{x}{2} }{4 \sqrt{ \frac{x}{2}-sin \frac{x}{2} } }$

2.12
$y= \sqrt{1+cos^{2}x^2} \\ y'= \frac{1}{2 \sqrt{1+cos^2x^2} }*(1+2cosx^2*(-sinx^2)*2x)= \\ = \frac{1-2xsin2x^2}{2 \sqrt{1+cos^2x^2} }$

2.13
$y= \frac{ \sqrt{4x+1} }{x^2} \\ y'= \frac{( \sqrt{4x+1} )'*x^2- \sqrt{4x+1}*(x^2)' }{(x^2)^2}= \frac{ (\frac{1}{2 \sqrt{4x+1} }*4)*x^2-2x \sqrt{4x+1} }{x^4}= \\ \\ = \frac{ \frac{2x^2}{ \sqrt{4x+1} }-2x \sqrt{4x+1} }{x^4}= \frac{ \frac{2x^2-2x(4x+1)}{ \sqrt{4x+1} } }{x^4}= \frac{2x^2-8x^2-2x}{x^4 \sqrt{4x+1} }= \\ = \frac{-6x^2-2x}{x^4 \sqrt{4x+1} }= \frac{-x(6x+2)}{x^4 \sqrt{4x+1} }=- \frac{6x+2}{x^3 \sqrt{4x+1} }$

2.14
$y=sin^2x^3 \\ y'=2sinx^3*cosx^3*3x^2=3x^2sin2x^3$

2.15
$y=tgx+ \frac{2}{3}tg^3x + \frac{1}{5}tg^5x \\ y'= \frac{1}{cos^2x} + \frac{2}{3}*3tg^2x* \frac{1}{cos^2x}+ \frac{1}{5}*5tg^4x* \frac{1}{cos^2x}= \\ \\ = \frac{1}{cos^2x}+ \frac{2tg^2x}{cos^2x}+ \frac{5tg^4x}{cos^2x}= \frac{1+2tg^2x+5tg^4x}{cos^2x}$

2.16
$y=sinx^3+cos^3x \\ y'=cosx^3*3x^2+3cos^2x*(-sinx)=3x^2cosx^3-3cos^2xsinx$

2.17
$y=(1+ \frac{1}{ \sqrt[3]{x} } )^3=(1+x^{- \frac{1}{3} })^3 \\ y'=3(1+ \frac{1}{ \sqrt[3]{x} } )^2*(- \frac{1}{3}x^{- \frac{4}{3} } )=- \frac{(1+ \frac{1}{ \sqrt[3]{x} })^2 }{ \sqrt[3]{x^4} } =- \frac{(1+ \frac{1}{ \sqrt[3]{x} } )^2}{x \sqrt[3]{x} }$

2.18
$y= \frac{cosx}{1+2sinx} \\ y'= \frac{(cosx)'(1+2sinx)-cosx*(1+2sinx)'}{(1+2sinx)^2}= \frac{-sinx(1+2sinx)-cosx*(2cosx)}{(1+2sinx)^2}= \\ \\ = \frac{-sinx-2sin^2x-2cos^2x}{(1+2sinx)^2}= \frac{-sinx-2(sin^2x+cos^2x)}{(1+2sinx)^2}= \\ \\ = \frac{-sinx-2}{(1+2sinx)^2}=- \frac{sinx+2}{(1+2sinx)^2}$

2.19
$y=x- \frac{2}{x}- \frac{1}{3x^3}=x-2x^{-1}- \frac{1}{3} x^{-3} \\ y'=1-2*(- \frac{1}{x^2} )- \frac{1}{3}*(-3x^{-4})= \\ =1+ \frac{2}{x^2}+ \frac{1}{x^4}= (1+ \frac{1}{x^2} )^2$

2.20
$y= \frac{x^2-1}{x^2+1} \\ y'= \frac{(x^2-1)'*(x^2+1)-(x^2-1)*(x^2+1)'}{(x^2+1)^2}= \frac{2x(x^2+1)-2x(x^2-1)}{(x^2+1)^2}= \\ \\ = \frac{2x(x^2+1-x^2+1)}{(x^2+1)^2} = \frac{4x}{(x^2+1)^2}$