Use implicit differentiation in finding y’.
1. 𝒙𝟐 𝒚 + 𝒙𝒚𝟐 = 𝟔
STEP 1:
𝑑 𝑑 𝑑
(𝑥 2 𝑦) + (𝑥𝑦 2 ) = (6)
𝑑𝑥 𝑑𝑥 𝑑𝑥
𝑑 𝑑 𝑑 𝑑
( (𝑥 2 𝑦) + (𝑦)𝑥 2 ) + (𝑥𝑦 2 ) = (6)
𝑑𝑥 𝑑𝑥 𝑑𝑥 𝑑𝑥
2𝑥𝑦 + 𝑦 ′ 𝑥 2 + 𝑦 2 + 2𝑥𝑦𝑦′ = 0
STEP 2:
2𝑥𝑦 + 𝑦 ′ 𝑥 2 + 𝑦 2 + 2𝑥𝑦𝑦′ = 0
𝑦 ′ 𝑥 2 + 2𝑥𝑦𝑦 ′ = −𝑦 2 − 2𝑥𝑦
𝑥𝑦 ′ 𝑥 + 𝑥𝑦′(2𝑦) = −𝑦 2 − 2𝑥𝑦
𝑥𝑦 ′ (𝑥 + 2𝑦) = −𝑦 2 − 2𝑥𝑦
𝑥𝑦 ′ (𝑥 + 2𝑦) −𝑦 2 −2𝑥𝑦
= +
𝑥(𝑥 + 2𝑦) 𝑥(𝑥 + 2𝑦) 𝑥(𝑥 + 2𝑦)
−𝒚(𝟐𝒙 + 𝒚)
𝒚′ =
𝒙(𝒙 + 𝟐𝒚)
Thus, the derivative of the given equation 𝒙𝟐 𝒚 + 𝒙𝒚𝟐 = 𝟔
is
−𝒚(𝟐𝒙 + 𝒚)
𝒚′ =
𝒙(𝒙 + 𝟐𝒚)
𝒙−𝟏
2. 𝒚𝟐 = 𝒙+𝟏
1. 𝒙𝟐 𝒚 + 𝒙𝒚𝟐 = 𝟔
STEP 1:
𝑑 𝑑 𝑑
(𝑥 2 𝑦) + (𝑥𝑦 2 ) = (6)
𝑑𝑥 𝑑𝑥 𝑑𝑥
𝑑 𝑑 𝑑 𝑑
( (𝑥 2 𝑦) + (𝑦)𝑥 2 ) + (𝑥𝑦 2 ) = (6)
𝑑𝑥 𝑑𝑥 𝑑𝑥 𝑑𝑥
2𝑥𝑦 + 𝑦 ′ 𝑥 2 + 𝑦 2 + 2𝑥𝑦𝑦′ = 0
STEP 2:
2𝑥𝑦 + 𝑦 ′ 𝑥 2 + 𝑦 2 + 2𝑥𝑦𝑦′ = 0
𝑦 ′ 𝑥 2 + 2𝑥𝑦𝑦 ′ = −𝑦 2 − 2𝑥𝑦
𝑥𝑦 ′ 𝑥 + 𝑥𝑦′(2𝑦) = −𝑦 2 − 2𝑥𝑦
𝑥𝑦 ′ (𝑥 + 2𝑦) = −𝑦 2 − 2𝑥𝑦
𝑥𝑦 ′ (𝑥 + 2𝑦) −𝑦 2 −2𝑥𝑦
= +
𝑥(𝑥 + 2𝑦) 𝑥(𝑥 + 2𝑦) 𝑥(𝑥 + 2𝑦)
−𝒚(𝟐𝒙 + 𝒚)
𝒚′ =
𝒙(𝒙 + 𝟐𝒚)
Thus, the derivative of the given equation 𝒙𝟐 𝒚 + 𝒙𝒚𝟐 = 𝟔
is
−𝒚(𝟐𝒙 + 𝒚)
𝒚′ =
𝒙(𝒙 + 𝟐𝒚)
𝒙−𝟏
2. 𝒚𝟐 = 𝒙+𝟏
