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ykovlevnickap
@ykovlevnickap
July 2022
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Найдите значение производной функции в точке x0:
а) f(x)=2tgx, x0= -3Pi/4
б) f(x)=(4x+1)/(x+3), x0= -2
в) f(x)=корень из 4x-7, x0= 2
г) f(x)=sin(3x-Pi/4), x0= Pi/4
д) f(x)=tg6x, x0= Pi/24
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ahta
A)f'(x)=(2tgx)'=2*(1/cos²x), f'(-3π/4)=2*(1/cos²(-3π/4)=2*(-2/√2)²=4
б)
, f'(-2)=11/1=11
в)
,
г)f'(x)=(sin(3x-π/4))'=3*cos(3x-π/4), f'(π/4)=3*cos(3*(π/4)-π/4)=3*cos(π/2)=0
д)f'(x)=(tg6x)'=6*(1/cos²(6x)), f'(π/24)=6*(1/cos²(6*π/24)=6*(1/cos²π/4)=6*(2/√2)²=6*4/2=12
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Answers & Comments
б), f'(-2)=11/1=11
в),
г)f'(x)=(sin(3x-π/4))'=3*cos(3x-π/4), f'(π/4)=3*cos(3*(π/4)-π/4)=3*cos(π/2)=0
д)f'(x)=(tg6x)'=6*(1/cos²(6x)), f'(π/24)=6*(1/cos²(6*π/24)=6*(1/cos²π/4)=6*(2/√2)²=6*4/2=12