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010998k
@010998k
August 2022
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oganesbagoyan
Verified answer
Уравнение касаеульной функции в точке (x₁ ; y
₁) имеет вид :
y - y
₁ = f '(x₁)*(x -x₁)
y
₁ =sin((π/2)/2) =sin(π/4) =√2 /2
f '(x) = (sinx/2)' =cosx/2*(x/2)'=1/2cosx/2 ;
f '(x₁)=1/2cos((π/2)/2) =1/2cosπ/4=1/2*√2/2 =√2/4
y - √2 /2
= √2/4 *(x -π/2)
2√2x - 8y +√2*(4 -π ) =0
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Answers & Comments
Verified answer
Уравнение касаеульной функции в точке (x₁ ; y₁) имеет вид :y - y₁ = f '(x₁)*(x -x₁)
y₁ =sin((π/2)/2) =sin(π/4) =√2 /2
f '(x) = (sinx/2)' =cosx/2*(x/2)'=1/2cosx/2 ;
f '(x₁)=1/2cos((π/2)/2) =1/2cosπ/4=1/2*√2/2 =√2/4
y - √2 /2 = √2/4 *(x -π/2)
2√2x - 8y +√2*(4 -π ) =0