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ail4
@ail4
July 2022
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Найти производную f(x)=sin^2 4x и вычислить ее значение в точке x1 П\ 16
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Беня2018
F'(x)=2(sin4x)*(sin4x)'=2(sin4x)*(cos4x)*(4x)'=4*2sin4xcos4x=
= применяем формулу sin2a=2sina*cosa= 4sin8x
f(π/16)=4sin(8*π/16)=4sinπ/2=4*1=4
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= применяем формулу sin2a=2sina*cosa= 4sin8x
f(π/16)=4sin(8*π/16)=4sinπ/2=4*1=4