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desheva66
@desheva66
July 2022
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1.Вычислить производную в точке
2.Найти экстремумыфункции
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mikael2
F'=12x²+6 f'(1)=18
f=1/√x+1/x=x^(-1/2)+x^(-1) f'=-1/2x^(3/2)-x^(-2) f'(1) =-1/2-1=-3/2
f=xcosx u=x u'=1 v=cosx v'=-sinx'
f'=(uv)'=u'v+v'u=cosx-xsinx
f'(π/2)=cosπ/2-π/2*sinπ/2=0-π/2*1=-π/2
------------------
y=x⁵/5-x⁴+5 y'=x⁴-4x³=x³(x-4)=0 точки экстремума х=0 и х=4
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Answers & Comments
f=1/√x+1/x=x^(-1/2)+x^(-1) f'=-1/2x^(3/2)-x^(-2) f'(1) =-1/2-1=-3/2
f=xcosx u=x u'=1 v=cosx v'=-sinx'
f'=(uv)'=u'v+v'u=cosx-xsinx
f'(π/2)=cosπ/2-π/2*sinπ/2=0-π/2*1=-π/2
------------------
y=x⁵/5-x⁴+5 y'=x⁴-4x³=x³(x-4)=0 точки экстремума х=0 и х=4