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JEE Main 2021
Differentiation
Differentiation
Easy

Question

If e y + xy = e, the ordered pair (dydx,d2ydx2)\left( {{{dy} \over {dx}},{{{d^2}y} \over {d{x^2}}}} \right) at x = 0 is equal to :

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Solution

y = 1 \Rightarrow x = 0 eydydx+xdydx+y=0{e^y}{{dy} \over {dx}} + x{{dy} \over {dx}} + y = 0 edydx+1=0dydx=1e \Rightarrow e{{dy} \over {dx}} + 1 = 0 \Rightarrow {{dy} \over {dx}} = - {1 \over e} 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 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 d2ydx2=1e2 \Rightarrow {{{d^2}y} \over {d{x^2}}} = {1 \over {{e^2}}}

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