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if $y=e^{nx},then\; \left ( \frac{d^{2}y}{dx^{2}} \right )\left ( \frac{d^{2}x}{dy^{2}} \right )$ is equal to :

Option 1)
$n\, e^{nx}\; \;$

Option 2)
$n\, e^{-nx}\; \;$

Option 3)
1

Option 4)
$-n\, e^{-nx}$

Answers (1)

best_answer

As we learnt in

Differential Equations -

An equation involving independent variable (x), dependent variable (y) and derivative of dependent variable with respect to independent variable
$\left (\frac{\mathrm{d} y}{\mathrm{d} x} \right )$

- wherein

eg:

$\frac{d^{2}y}{dx^{2}}- 3\frac{dy}{dx}+5x=0$

$y=e^{nx}$

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

$\\ \frac{dy}{dx}=n.e^{nx}$ Now, $\log y=nx$

$\frac{d^{2}y}{dx^{2}}=n^{2}.e^{nx}$ .............(i) $\frac{1}{y}=n.\frac{dx}{dy}$

$-\frac{1}{y^{2}}=n.\frac{d^{2}x}{dy^{2}}$ ...........(ii)

So that, $\frac{d^{2}y}{dx^{2}}\times \frac{d^{2}x}{dy^{2}}= n^{2}.e^{nx}\times (-\frac{1}{ny^{2}})$

$= \frac{-n.e^{nx}}{(e^{nx})^{2}}$

$= -n.e^{-nx}$

Option 1)

$n\, e^{nx}\; \;$

This option is incorrect.

Option 2)

$n\, e^{-nx}\; \;$

This option is incorrect.

Option 3)

1

This option is incorrect.

Option 4)

$-n\, e^{-nx}$

This option is correct.

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