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QUESTION 1.
𝒅
(−𝟐√𝟓𝒔𝟐 + 𝟔)
𝒅𝒔
Take the constant out: (𝑎 ⋅ 𝑓)′ = 𝑎 ⋅ 𝑓 ′
𝑑
= −2 (√5𝑠 2 + 6)
𝑑𝑠
𝑑𝑓(𝑢) 𝑑𝑓 𝑑𝑢
Apply the chain rule:
=
⋅
𝑑𝑥
𝑑𝑢 𝑑𝑥
𝑓 = √𝑢, 𝑢 = 5𝑠 2 + 6
𝑑
𝑑
= −2
(√𝑢) (5𝑠 2 + 6)
𝑑𝑢
𝑑𝑠
1
= −2 ⋅
⋅ 10𝑠
2 √𝑢
Substitute back 𝑢 = 5𝑠 2 + 6
1
= −2 ⋅
⋅ 10𝑠
2√5𝑠 2 + 6
=−
1 ⋅ 2 ⋅ 10𝑠
2√5𝑠 2 + 6
=−
𝟏𝟎𝒔
√𝟓𝒔𝟐 + 𝟔
𝑑
1
(√𝑢) =
𝑑𝑢
2 √𝑢
𝑑
(5𝑠 2 + 6) = 10𝑠
𝑑𝑠
QUESTION 2.
𝒅
𝟐𝒙
(
)
𝒅𝒙 √𝒙𝟐 + 𝟗
Take the constant out: (𝑎 ⋅ 𝑓)′ = 𝑎 ⋅ 𝑓 ′
𝑑
𝑥
=2 (
)
𝑑𝑥 √𝑥 2 + 9
𝑓
𝑓 ′ ⋅ 𝑔 − 𝑔′
Apply the Quotient Rule: ( )′ =
𝑔
𝑔2
𝑑
𝑑
(𝑥)√𝑥 2 + 9 −
(√𝑥 2 + 9)𝑥
𝑑𝑥
𝑑𝑥
= 2⋅
(√𝑥 2 + 9)2
=
1
⋅ 2𝑥
2√𝑢
Substitute back 𝑢 = 𝑥 2 + 9
1
=
⋅ 2𝑥
2√𝑥 2 + 9
=
𝑥
√𝑥 2 + 9
1 ⋅ √𝑥 2 + 9 −
= 2⋅
=
𝑥
+9
√𝑥 2
(√𝑥 2 + 9)2
𝟏𝟖
(𝒙𝟐 + 𝟗)√𝒙𝟐 + 𝟗
𝑑
(𝑥) = 1
𝑑𝑥
𝑑
(√𝑥 2 + 9)
𝑑𝑥
𝑑𝑓(𝑢) 𝑑𝑓 𝑑𝑢
Apply the chain rule:
=
⋅
𝑑𝑥
𝑑𝑢 𝑑𝑥
𝑓 = √𝑢, 𝑢 = 𝑥 2 + 9
𝑑
𝑑
=
(√𝑢) (𝑥 2 + 9)
𝑑𝑢
𝑑𝑥
𝑑
1
(√𝑢) =
𝑑𝑢
2√𝑢
𝑑 2
(𝑥 + 9)
𝑑𝑥
Apply the Sum/Difference Rule: (𝑓 ± 𝑔)′ = 𝑓 ′ ± 𝑔
𝑑 2
𝑑
=
(𝑥 ) +
(9)
𝑑𝑥
𝑑𝑥
𝑑 2
(𝑥 ) = 2𝑥
𝑑𝑥
𝑑
(9) = 0
𝑑𝑥
𝑑 2
(𝑥 + 9) = 2𝑥
𝑑𝑥
QUESTION 3
𝒅
(𝟒 𝐭𝐚𝐧(𝟓𝒙))
𝒅𝒙
Take the constant out: (𝑎 ⋅ 𝑓)′ = 𝑎 ⋅ 𝑓 ′
𝑑
= 4 (tan(5𝑥))
𝑑𝑥
𝑑𝑓(𝑢) 𝑑𝑓 𝑑𝑢
Apply the chain rule:
=
⋅
𝑑𝑥
𝑑𝑢 𝑑𝑥
𝑓 = tan (𝑢), 𝑢 = 5𝑥
𝑑
𝑑
=4
(tan (𝑢)) (5𝑥)
𝑑𝑢
𝑑𝑥
= 4sec 2 (𝑢) ⋅ 5
Substitute back 𝑢 = 5
= 4sec 2 (5𝑥) ⋅ 5
Simplify
= 𝟐𝟎𝐬𝐞𝐜 𝟐 (𝟓𝒙)
𝑑
(tan (𝑢)) = sec 2 (𝑢)
𝑑𝑢
𝑑
(5𝑥) = 5
𝑑𝑥
QUESTION 4
𝑑
(4𝑥 2 𝑦 + 9𝑦 2 𝑥)
𝑑𝑥
𝒅
𝒅
(𝟒𝒙𝟐 𝒚 + 𝟗𝒚𝟐 𝒙) =
(−𝟏)
Apply the Sum/Difference Rule: (𝑓 ± 𝑔)′ = 𝑓 ′ ± 𝑔′
𝒅𝒙
𝒅𝒙
𝑑
𝑑
=
(4𝑥 2 𝑦) +
(9𝑦 2 𝑥)
𝑑
𝑑𝑥
𝑑𝑥
(−1) = 0
𝑑
𝑑
𝑑𝑥
(4𝑥 2 𝑦) = 4(2𝑥𝑦 +
(𝑦)𝑥 2 )
𝑑
𝑑
𝑑
𝑑𝑥
𝑑𝑥
(−1) = 0 = 4(2𝑥𝑦 +
(𝑦)𝑥 2 ) + 9(2𝑥𝑦 (𝑦) + 𝑦 2 )
𝑑
𝑑
𝑑𝑥
𝑑...