JEE MAIN - Mathematics (2011 - No. 15)
$${{{d^2}x} \over {d{y^2}}}$$ equals:
$$ - {\left( {{{{d^2}y} \over {d{x^2}}}} \right)^{ - 1}}{\left( {{{dy} \over {dx}}} \right)^{ - 3}}$$
$${\left( {{{{d^2}y} \over {d{x^2}}}} \right)^{}}{\left( {{{dy} \over {dx}}} \right)^{ - 2}}$$
$$ - \left( {{{{d^2}y} \over {d{x^2}}}} \right){\left( {{{dy} \over {dx}}} \right)^{ - 3}}$$
$${\left( {{{{d^2}y} \over {d{x^2}}}} \right)^{ - 1}}$$
ব্যাখ্যা
$${{{d^2}x} \over {d{y^2}}} = {d \over {dy}}\left( {{{dx} \over {dy}}} \right)$$
$$ = {d \over {dx}}\left( {{{dx} \over {dy}}} \right){{dx} \over {dy}}$$
$$ = {d \over {dx}}\left( {{1 \over {dy/dx}}} \right){{dx} \over {dy}}$$
$$ = - {1 \over {{{\left( {{{dy} \over {dx}}} \right)}^2}}}.{{{d^2}y} \over {d{x^2}}}.{1 \over {{{dy} \over {dx}}}}$$
$$ = - {1 \over {{{\left( {{{dy} \over {dx}}} \right)}^3}}}{{{d^2}y} \over {d{x^2}}}$$
$$ = {d \over {dx}}\left( {{{dx} \over {dy}}} \right){{dx} \over {dy}}$$
$$ = {d \over {dx}}\left( {{1 \over {dy/dx}}} \right){{dx} \over {dy}}$$
$$ = - {1 \over {{{\left( {{{dy} \over {dx}}} \right)}^2}}}.{{{d^2}y} \over {d{x^2}}}.{1 \over {{{dy} \over {dx}}}}$$
$$ = - {1 \over {{{\left( {{{dy} \over {dx}}} \right)}^3}}}{{{d^2}y} \over {d{x^2}}}$$