напишите уравнение касательной к графику функции f(x)=cos^2 6x в точке X0=pi/24. Created by volchenok1993. matematika-ru

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volchenok1993 @volchenok1993

July 2022 1 5 Report

напишите уравнение касательной к графику функции f(x)=cos^2 6x в точке X0=pi/24

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$y=f(x_0)+f'(x_0)(x-x_0)$ - уравнение касательной

$y=cos^2 6x$ , $x_0= \frac{ \pi }{24}$

$y'=(cos^2 6x)'=(cos6x)'*(cos6x)'=$ $=-sin6x*cos6x*(6x)'+cos6x*(-sin6x)*(6x)'=$ $=-6sin6xcos6x-6sin6xcos6x=-12sin6xcos6x=-6sin12x$

$f( \frac{ \pi }{24})= cos^2(6* \frac{ \pi }{24})=cos^2 \frac{ \pi }{4} =( \frac{ \sqrt{2} }{2} )^2= \frac{1}{2}$

$f'( \frac{ \pi }{24})=-6sin(12* \frac{ \pi }{24})=-6*sin \frac{ \pi }{2} =-6*1=-6$

$y=0.5-6(x- \frac{ \pi }{24} )$

$y=0.5-6x+ \frac{ \pi }{4}$

$y=-6x+0.5+ \frac{ \pi }{4}$

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