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5. continuity and differentiation — Maths STD 12 Science — Question

Gujarat BoardEnglish MediumSTD 12 ScienceMaths5. continuity and differentiation1 Mark
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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