STD 12 - 5. continuity and differentiation — JEE STD 11 Science — Question

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}}$