Объяснение:
1) Решениеy=(4·x-9)^5
((4·x-9)^5)' = 20(4·x-9^)4
Поскольку:
((4·x-9)5)' = 5·(4·x-9)^5-^1((4·x-9))' = 20(4·x-9)^4
(4·x-9)' = 4
20(4·x-9)^4
y=(x2-3x+1)7
2) Решение:((x2-3x+1)7)' = (-7·3x·ln(3)+14·x)(x2-3x+1)6
Поскольку:
((x2-3x+1)7)' = 7·(x2-3x+1)7-1((x2-3x+1))' = (-7·3x·ln(3)+14·x)(x2-3x+1)6
(x2-3x+1)' = (x2)' + (-3x)' + (1)' = 2·x + (-3x·ln(3)) = -3x·ln(3)+2·x
(x2)' = 2·x2-1(x)' = 2·x
(x)' = 1
Здесь:
Решение ищем по формуле:
(af(x))' = af(x)*ln(a)*f(x)'
(-3x)' = -3x·ln(3)(x)' = -3x·ln(3)
(x)' = 1
(-7·3x·ln(3)+14·x)(x2-3x+1)6
3) Решение:y=(sin(x))^3
(sin(x)^3)' = 3·sin(x)^2·cos(x)
Поскольку:
(sin(x)^3)' = 3·(sin(x))^3-1((sin(x)))' = 3·sin(x)^2·cos(x)
(sin(x))' = cos(x)
3·sin(x)2·cos(x)
1)arcsin 0 =0
2)arccos 1= 0 ;
3)arcsin√2/2 =π/4 ;
4)arccos 3 не существует угол косинус которой =3 ;
5)arcsin (-1) = -π/2 ;
6)arccos(-√3/2) = π -π/6 = 5π/6 ;
7)arctg 0 = 0 ;
8)arctg 1 =π/4 ;
9)arctg(-√3) = - π/3 ;
10)arcctg(-√3/3) = π -π/3= 2π/3 ;
11)arcsin(-1/2)+arccos 1 = -π/6 +0 = -π/6 ;
12) (arcsin -1)/2+ arccos 1 = -π/4+0= -π/4;
13)cos ( arccos 1) =1;
14)sin(arcsin√2/2) =√2/2 ;
15)arcsin (sin π/4) =arcsin(√2/2) =π/4 ;
16)arccos ( cos(-π/4))=arccos ( cos(π/4))=arccos (√2/2))=π/4 ;
17)cos (arcsin(-1/3))=cos(arccos(√8/3)= √8/3 =2√2/3 ;
18)tg(arccos(-1/4)) =tq(arctq(-√15) = - √15; 1+tq²α= 1/cos²α
19)sin(arcctg(-2)) =sin(arcsin(1/√5)=1/√5 ;
20) arcsin(cos π/9) =arcsin(sin(π/2 - π/9))=arcsin(sin7π/18) =7π/18 .
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Объяснение:
