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ironwoman9
@ironwoman9
July 2022
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к графику функции y=sinx+cosx в точке с абсциссой x=п/2 проведена касательная. найдите тангенс угла наклона касательной к оси ох
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kirichekov
Verified answer
Y'(x₀)=tgα.
y'(x)=(sinx+cox)'=cosx-sinx
y'(π/2)=cos(π/2)-sin(π/2)=0-1=-1
ответ: tgα=-1
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Answers & Comments
Verified answer
Y'(x₀)=tgα.y'(x)=(sinx+cox)'=cosx-sinx
y'(π/2)=cos(π/2)-sin(π/2)=0-1=-1
ответ: tgα=-1