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schutovalesh
@schutovalesh
December 2022
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Найти уровнение касательной к функции
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fenix6810
y=y(x0)+y'(x0)(x-x0)
ctg'x=-1/sin^2x
ctg'(x0)=-1/sin^2(-П/4)=-2
y(-П/4)=-1
y=-1-2(x+п/4)=-1-2x-п/2
y=-2x-(1+п/2)
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Answers & Comments
ctg'x=-1/sin^2x
ctg'(x0)=-1/sin^2(-П/4)=-2
y(-П/4)=-1
y=-1-2(x+п/4)=-1-2x-п/2
y=-2x-(1+п/2)