d^2 x d y^2 = - Vidyadip

MCQ

$\frac{{{d^2}x}}{{d{y^2}}} = $

A
$ - {\left( {\frac{{{d^2}y}}{{d{x^2}}}} \right)^{ - 1}}{\left( {\frac{{dy}}{{dx}}} \right)^{ - 3}}$
B
$\;\left( {\frac{{{d^2}y}}{{d{x^2}}}} \right){\left( {\frac{{dy}}{{dx}}} \right)^{ - 2}}$
✓
$ - \left( {\frac{{{d^2}y}}{{d{x^2}}}} \right){\left( {\frac{{dy}}{{dx}}} \right)^{ - 3}}$
D
$\;{\left( {\frac{{{d^2}y}}{{d{x^2}}}} \right)^{ - 1}}$

✓

Answer

Correct option: C.

$ - \left( {\frac{{{d^2}y}}{{d{x^2}}}} \right){\left( {\frac{{dy}}{{dx}}} \right)^{ - 3}}$

c
$\frac{{{d^2}x}}{{d{y^2}}} = \frac{d}{{dy}}\left( {\frac{{dx}}{{dy}}} \right) = \frac{d}{{dy}}\left( {\frac{1}{{\frac{{dy}}{{dx}}}}} \right)$

$ = \frac{d}{{dx}}\left( {\frac{1}{{\frac{{dy}}{{dx}}}}} \right) \cdot \frac{{dx}}{{dy}} = - \frac{1}{{{{\left( {\frac{{dy}}{{dx}}} \right)}^2}}} \cdot \frac{{{d^2}y}}{{d{x^2}}} \cdot \frac{1}{{\left( {\frac{{dy}}{{dx}}} \right)}}$

$=-\frac{1}{\left(\frac{d y}{d x}\right)^{3}} \cdot \frac{d^{2} y}{d x^{2}}=-\left(\frac{d y}{d x}\right)^{-3}\left(\frac{d^{2} y}{d x^{2}}\right) $

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