JEE Advance - Mathematics (2007 - No. 35)
$${{{d^2}x} \over {d{y^2}}}$$ equals
$${\left( {{{{d^2}y} \over {d{x^2}}}} \right)^{ - 1}}$$
$$ - {\left( {{{{d^2}y} \over {d{x^2}}}} \right)^{ - 1}}{\left( {{{dy} \over {dx}}} \right)^{ - 3}}$$
$$\left( {{{{d^2}y} \over {d{x^2}}}} \right){\left( {{{dy} \over {dx}}} \right)^{ - 2}}$$
$$ - \left( {{{{d^2}y} \over {d{x^2}}}} \right){\left( {{{dy} \over {dx}}} \right)^{ - 3}}$$
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