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Question : 64 of 160
Marks:
+1 ,
-0
Solution:
x 3 − 2 x 2 y 2 + 5 x + y − 5 = 0 On differentiating w.r.t.
x , we get
3 x 2 − 4 x y 2 − 4 x 2 y ⋅ y ′ + 5 + y ′ − 0 = 0 ........(i)
On putting
x = 1 , y = 1 ,
‌ 3 − 4 − 4 y ′ + 5 + y ′ = 0 ‌ ⇒ ‌ ‌ y ′ = ‌ ‌ y ′ ( 1 ) = ‌ On differentiating Eq. (i) w.r.t.
x ,
6 x − 4 y 2 − 8 x y y ′ − 8 x y y ′ − 4 x 2 y ′ ⋅ y ′ − 4 x 2 y y ′ ′ + y ′ ′ = 0
On putting
x = 1 , y = 1 and
y ′ = ‌ to obtain
y ′ ′ ( 1 ) ,
‌ ⇒ 2 − ‌ − ‌ − 3 y ′ ′ = 0 ⇒ ‌ = 3 y ′ ′ ⇒ y ′ ′ = − ‌ ‌ or ‌ ‌ y ′ ′ ( 1 ) = − ‌
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