MCQ
If $y = {e^{nx}}$, then $\left( {\frac{{{d^2}y}}{{d{x^2}}}} \right)\left( {\frac{{{d^2}x}}{{d{y^2}}}} \right)$ is equal to
- A$ne^{nx}$
- B$ne^{-nx}$
- C$1$
- ✓$-ne^{-nx}$
Differeniating both sides with respect to $x$
$\frac{{dy}}{{dx}} = n{e^{nx}}$
Again differentiating w.r.t $'x'$,
$\frac{{{d^2}y}}{{d{x^2}}} = n.n{e^{nx}} = {n^2}{e^{nx}}\,\,\,\,\,\,\,\,....\left( 1 \right)$
$y = {e^{nx}}$
$nx = {\log _e}y$
$x = \frac{1}{n}\log y$
differentiating $x$ with respect to $y$
$\frac{{dx}}{{dy}} = \frac{1}{n}.\frac{1}{y}$
Again differentiating with respect to $y$
$\frac{{{d^2}x}}{{d{y^2}}} = \frac{1}{n}.\left( {\frac{{ - 1}}{{{y^2}}}} \right) = \frac{{ - 1}}{{n{y^2}}} = - \frac{1}{{n{e^{2nx}}}}\,.....\left( 2 \right)$
Multiplying equation $(1)$ and $(2)$
$\left( {\frac{{{d^2}y}}{{d{x^2}}}} \right)\left( {\frac{{{d^2}x}}{{d{y^2}}}} \right)$
$ = \left( {{n^2}{e^{nx}}} \right)\left( {\frac{{ - 1}}{n}{e^{2nx}}} \right) = - n{e^{ - nx}}$
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