Continuity and DifferentiabilityMedium
Question
If x3 - y3 + 3xy2 - 3x2y + 1 = 0 , then at (0, 1) 
equals -
equals -Options
A.1
B.-1
C.2
D.0
Solution
x3 - y3 + 3xy2 - 3x2y + 1 = 0
Put f = x3 - y3 + 3xy2 - 3x2y + 1
= 3x2 - 0 + 3y2 - 6xy
= 0 + 3 - 0
again f = x3 - y3 + 3xy2 - 3x2y + 1
= 0 - 3y2 + 6xy - 3x2
= -3
so
= 1
Put f = x3 - y3 + 3xy2 - 3x2y + 1
= 3x2 - 0 + 3y2 - 6xy= 0 + 3 - 0again f = x3 - y3 + 3xy2 - 3x2y + 1
= 0 - 3y2 + 6xy - 3x2 = -3so
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