найдите значение производной функции 1)y=12cos x в точке x0=-пи/6 2)y=19tgx в точке x0=-пи/6 3)y=tgx/16в точке x0=-пи/3 4)y=2tgx+3 в точке х0=пи/6
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1) y'=-12sinx y'(x₀)=-12sin(π/6)=-12*1/2=-62) y'=19/cos²x y'(x₀)=19/cos²(-π/6)=19/(√3/2)²=4*19/3=76/3=25 1/3
3)y'=1/16cos²x y'(x₀)=1/16cos²(π/3)=1/16(1/2)²=1/4
4) y'=2/cos²x y'(x₀)=2/cos²(π/6)=2/(√3/2)²=2*4/3=8/3= 2 2/3