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ksuuusha
@ksuuusha
August 2021
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алгебра 11 класс сложные функции помогите пожалуйста решить!!**
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Гоша68
Verified answer
1. f'=[(2x-1)(x+2)-(x^2-x)]/(x+2)^2=(x^2+4x-2)/(x+2)^2
f'(2)=(4+8-2)/4^2=10/16=5/8
2.
f'=[(2x-8)(x-3)-x^2+8x-15]/(x-3)^2=[2x^2+24-14x-x^2-8x+15]/(x-3)^2=
=(x^2-22x+39)/(x-3)^2
f'(2)=(4-44+39)/1=-1
3.
f'=2cos2x-2sin2x/cos2x=2cos2x-2tg2x
f'(П/6)=2*cosП/3-2tgП/3=1-2sqrt(3)
4.
(1+cosx)/(1-cosx)*[sinx(1+cosx)+sinx(1-cosx)]/(1+cosx)^2=
=2sinx/(1-cosx)(1+cosx)=2sinx/(1-cos^2x)=2/sinx
5. f'=-sinx/cosx=-tgx
f'(П/4)=-tgП/4=-1
6.
f'=5cos(2x+2/x)*(2-2/x^2)
f'(1)=5cos(4)*(2-2)=0
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Answers & Comments
Verified answer
1. f'=[(2x-1)(x+2)-(x^2-x)]/(x+2)^2=(x^2+4x-2)/(x+2)^2f'(2)=(4+8-2)/4^2=10/16=5/8
2.
f'=[(2x-8)(x-3)-x^2+8x-15]/(x-3)^2=[2x^2+24-14x-x^2-8x+15]/(x-3)^2=
=(x^2-22x+39)/(x-3)^2
f'(2)=(4-44+39)/1=-1
3.
f'=2cos2x-2sin2x/cos2x=2cos2x-2tg2x
f'(П/6)=2*cosП/3-2tgП/3=1-2sqrt(3)
4.
(1+cosx)/(1-cosx)*[sinx(1+cosx)+sinx(1-cosx)]/(1+cosx)^2=
=2sinx/(1-cosx)(1+cosx)=2sinx/(1-cos^2x)=2/sinx
5. f'=-sinx/cosx=-tgx
f'(П/4)=-tgП/4=-1
6.
f'=5cos(2x+2/x)*(2-2/x^2)
f'(1)=5cos(4)*(2-2)=0