JEE Main201911 Jan 2019Morning ShiftMathematicsDifferentiationActual

If x _ e ( _ e x )-x²+y²=4(y>0), then d y d x at x=e is equal to :

Options

A(1+2 e) 2 4+e²
B(2 e-1) 2 4+e²
C(1+2 e) 4+e²
De 4+e²

Correct answer

B. (2 e-1) 2 4+e²

Step-by-step solution

Consider the equation, x _ e ( _ e x )-x²+y²=4 Differentiate both sides w.r.t. x _ e ( _ e x )+x 1 x _ e x -2 x+2 y d y d x =0 _ e ( _ e x )+ 1 _ e x -2 x+2 y d y d x =0 When x=e, y= 4+e² . Put these values in (1), 0+1-2 e+2 4+e² d y d x =0 d y d x = 2 e-1 2 4+e²

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