Question

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. $n{e}^{nx}$
• B. $n{e}^{-nx}$
• C. 1
• D. $-n{e}^{-nx}$

Answer 1

The correct option is D $-n{e}^{-nx}$

$y=e^{nx}$
$\Rightarrow \frac{dy}{dx}={ne}^{nx}$

$\Rightarrow \frac{d^{2}y}{d{x}^2}={n^{2}e}^{nx}=n^{2}y$

$x=\frac{\log y}{n}$

$\frac{d^{2}x}{d{y}^2}=-\frac{1}{n{y}^2}$ .

$\left(\frac{d^{2}y}{d{x}^2}\right)\left(\frac{d^{2}x}{d{y}^2}\right)=-{ne}^{nx}$