JEE Main 2021DifferentiationDifferentiationEasyQuestionIf e y + xy = e, the ordered pair (dydx,d2ydx2)\left( {{{dy} \over {dx}},{{{d^2}y} \over {d{x^2}}}} \right)(dxdy,dx2d2y) at x = 0 is equal to :OptionsA(1e,−1e2)\left( {{1 \over e}, - {1 \over {{e^2}}}} \right)(e1,−e21)B(−1e,1e2)\left( { - {1 \over e},{1 \over {{e^2}}}} \right)(−e1,e21)C(−1e,−1e2)\left( { - {1 \over e}, - {1 \over {{e^2}}}} \right)(−e1,−e21)D(1e,1e2)\left( {{1 \over e},{1 \over {{e^2}}}} \right)(e1,e21)Check AnswerHide SolutionSolutiony = 1 ⇒\Rightarrow⇒ x = 0 eydydx+xdydx+y=0{e^y}{{dy} \over {dx}} + x{{dy} \over {dx}} + y = 0eydxdy+xdxdy+y=0 ⇒edydx+1=0⇒dydx=−1e \Rightarrow e{{dy} \over {dx}} + 1 = 0 \Rightarrow {{dy} \over {dx}} = - {1 \over e}⇒edxdy+1=0⇒dxdy=−e1 ⇒eyd2ydx2+ey(dydx)2+xd2ydx2+2dydx=0 \Rightarrow {e^y}{{{d^2}y} \over {d{x^2}}} + {e^y}{\left( {{{dy} \over {dx}}} \right)^2} + x{{{d^2}y} \over {d{x^2}}} + 2{{dy} \over {dx}} = 0⇒eydx2d2y+ey(dxdy)2+xdx2d2y+2dxdy=0 x = 0, y = 1 ⇒ed2ydx2+e(−1e)2+0+2(−1e)=0 \Rightarrow e{{{d^2}y} \over {d{x^2}}} + e{\left( { - {1 \over e}} \right)^2} + 0 + 2\left( { - {1 \over e}} \right) = 0⇒edx2d2y+e(−e1)2+0+2(−e1)=0 ⇒d2ydx2=1e2 \Rightarrow {{{d^2}y} \over {d{x^2}}} = {1 \over {{e^2}}}⇒dx2d2y=e21