Continuity and DifferentiabilityHard
Question
If x2 + y2 = 1, then
Options
A.yy′′ -2(y′)2 + 1 = 0
B.yy′′ -(y′)2 + 1 = 0
C.yy′′ +(y′)2 - 1 = 0
D.yy′′ + 2(y′)2 - 1 = 0
Solution
Given x2 + y2 = 1
On differentiating w.r.t.x, we get
2x + 2yy′ = 0
⇒ x + yy′ = 0
Again, differentiating w.r.t.x, we get
1 + y′y′ + yy′ = 0
⇒ 1 + (y′)2 + yy′ = 0
On differentiating w.r.t.x, we get
2x + 2yy′ = 0
⇒ x + yy′ = 0
Again, differentiating w.r.t.x, we get
1 + y′y′ + yy′ = 0
⇒ 1 + (y′)2 + yy′ = 0
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