Verified Solution • 2025–26 Session moderate

QJEEM11Ma420b: JEE Main 2011 Mathematics - Class 12 JEEM11Ma420b Solution: equals:

d2xdy2{{{d^2}x} \over {d{y^2}}} equals:

Multiple Choice Options

  • A −(d2ydx2)−1(dydx)−3- {\left( {{{{d^2}y} \over {d{x^2}}}} \right)^{ - 1}}{\left( {{{dy} \over {dx}}} \right)^{ - 3}}
  • B (d2ydx2)(dydx)−2{\left( {{{{d^2}y} \over {d{x^2}}}} \right)^{}}{\left( {{{dy} \over {dx}}} \right)^{ - 2}}
  • C −(d2ydx2)(dydx)−3- \left( {{{{d^2}y} \over {d{x^2}}}} \right){\left( {{{dy} \over {dx}}} \right)^{ - 3}}
  • D (d2ydx2)−1{\left( {{{{d^2}y} \over {d{x^2}}}} \right)^{ - 1}}

Step-by-Step Verified Solution

d2xdy2=ddy(dxdy){{{d^2}x} \over {d{y^2}}} = {d \over {dy}}\left( {{{dx} \over {dy}}} \right)

=ddx(dxdy)dxdy= {d \over {dx}}\left( {{{dx} \over {dy}}} \right){{dx} \over {dy}}

=ddx(1dy/dx)dxdy= {d \over {dx}}\left( {{1 \over {dy/dx}}} \right){{dx} \over {dy}}

=−1(dydx)2.d2ydx2.1dydx= - {1 \over {{{\left( {{{dy} \over {dx}}} \right)}^2}}}.{{{d^2}y} \over {d{x^2}}}.{1 \over {{{dy} \over {dx}}}}

=−1(dydx)3d2ydx2= - {1 \over {{{\left( {{{dy} \over {dx}}} \right)}^3}}}{{{d^2}y} \over {d{x^2}}}

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