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JEE Main202126 Aug 2021Morning ShiftMathematicsDifferentiationActual

If y = y ( x ) is an implicit function of x such that log e ( x + y ) = 4 x y , then d 2 y d x 2 at x = 0 is equal to

Correct answer

0

Step-by-step solution

Given: log e ( x + y ) = 4 x y When x = 0 , then y = 1 log e x + y = 4 x y ⇒ x + y = e 4 x y Now differentiate w.r.t. x 1 + y ' = e 4 x y 4 y + 4 x y '       … i At ( 0 , 1 ) ⇒ y ' ( 0 ) + 1 = 4 ⇒ y ' ( 0 ) = 3 Now, again differentiate equation ( i ) , we get y " = e 4 x y 4 y + 4 x y ′ 2 + e 4 x y 4 y ' + 4 y ' + 4 x y " At 0 , 1 y " ( 0 ) = 1 ( 4 × 1 + 0 ) 2 + 1 ( 4 × 3 + 4 × 3 + 0 ) ⇒ y " ( 0 ) = 16 + 24 = 4

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