β€² = sin( 2 + ) = βˆ’2 + 3 sin( 2 + ) = βˆ’2 + 3 cos( 2 + )

𝑑𝑦
𝑦 β€² = 𝑑π‘₯
sin(π‘₯ 2 + 𝑦) = 𝑦 βˆ’2 + π‘₯ 3
sin(π‘₯ 2 + 𝑦) = 𝑦 βˆ’2 + π‘₯ 3
𝑑𝑦
𝑑𝑦
cos(π‘₯ 2 + 𝑦) βˆ™ (2π‘₯ + ) = βˆ’2𝑦 βˆ’3
+ 3π‘₯ 2
𝑑π‘₯
𝑑π‘₯
𝑑𝑦
𝑑𝑦
2
2
2π‘₯ cos(π‘₯ + 𝑦) + cos(π‘₯ + 𝑦)
= βˆ’2𝑦 βˆ’3
+ 3π‘₯ 2
𝑑π‘₯
𝑑π‘₯
𝑑𝑦
𝑑𝑦
2
βˆ’3
2
cos(π‘₯ + 𝑦)
+ 2𝑦
= 3π‘₯ βˆ’ 2π‘₯ cos(π‘₯ 2 + 𝑦)
𝑑π‘₯
𝑑π‘₯
𝑑𝑦
(cos(π‘₯ 2 + 𝑦) + 2𝑦 βˆ’3 )
= 3π‘₯ 2 βˆ’ 2π‘₯ cos(π‘₯ 2 + 𝑦)
𝑑π‘₯
𝑑𝑦 3π‘₯ 2 βˆ’ 2π‘₯ cos(π‘₯ 2 + 𝑦)
=
𝑑π‘₯
cos(π‘₯ 2 + 𝑦) + 2𝑦 βˆ’3
π’…π’š πŸ‘π’™πŸ βˆ’ πŸπ’™ 𝐜𝐨𝐬(π’™πŸ + π’š)
=
𝒅𝒙
𝐜𝐨𝐬(π’™πŸ + π’š) + πŸπ’šβˆ’πŸ‘
1
4
25
𝑦
π‘₯
𝑑π‘₯
1
= βˆ’ 𝑓𝑑/𝑠𝑒𝑐
𝑑𝑑
4
𝑑𝑦
=?
𝑑𝑑
𝑑 = 12 𝑠𝑒𝑐
1
β†’ π‘₯ = 10 βˆ’ (12) = 10 βˆ’ 3 = 7
4
2
√
β†’ 𝑦 = 25 βˆ’ 72 = 24
π‘₯ 2 + 𝑦 2 = 252
𝑑π‘₯
𝑑𝑦
2π‘₯
+ 2𝑦
=0
𝑑𝑑
𝑑𝑑
1
𝑑𝑦
2(7) (βˆ’ ) + 2(24)
=0
4
𝑑𝑑
1
𝑑𝑦
2(7) (βˆ’ ) + 2(24)
=0
4
𝑑𝑑
𝑑𝑦
7
=
β‰ˆ 0.073 𝑓𝑑/𝑠𝑒𝑐
𝑑𝑑 96
πŸ•
πŸ—πŸ”
𝒇𝒕/𝒔𝒆𝒄