JEE Main 2021
Differentiation
Differentiation
Easy
Question
For x > 1, if (2x) 2y = 4e 2x2y , then (1 + log e 2x) 2 is equal to :
Options
Solution
(2x) 2y = 4e 2x-2y 2yn2x = n4 + 2x 2y y = y ' = y '
For x > 1, if (2x) 2y = 4e 2x2y , then (1 + log e 2x) 2 is equal to :
(2x) 2y = 4e 2x-2y 2yn2x = n4 + 2x 2y y = y ' = y '