Решить 2.1 ,2.2 ......

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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
<img src="https://tex.z-dn.net/?f=y%3Dtgx%2B+%5Cfrac%7B2%7D%7B3%7Dtg%5E3x+%2B+%5Cfrac%7B1%7D%7B5%7Dtg%5E5x++%5C%5C+%0Ay%27%3D+%5Cfrac%7B1%7D%7Bcos%5E2x%7D+%2B+%5Cfrac%7B2%7D%7B3%7D%2A3tg%5E2x%2A+%5Cfrac%7B1%7D%7Bcos%5E2x%7D%2B+%5Cfrac%7B1%7D%7B5%7D%2A5tg%5E4x%2A+%5Cfrac%7B1%7D%7Bcos%5E2x%7D%3D+%5C%5C+%0A+%5C%5C+%0A%3D+%5Cfrac%7B1%7D%7Bcos%5E2x%7D%2B+%5Cfrac%7B2tg%5E2x%7D%7Bcos%5E2x%7D%2B+%5Cfrac%7B5tg%5E4x%7D%7Bcos%5E2x%7D%3D+%5Cfrac%7B1%2B2tg%5E2x%2B5tg%5E4x%7D%7Bcos%5E2x%7D++++++++" id="TexFormula14" title="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} " alt="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} " align="absmiddle" cla

m11m_zn (233k баллов) 14 Апр, 18