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. $n^2{e}^{nx}$
• C. $-n{e}^{nx}$
• D. $-n{e}^{-nx}$

Answer 1

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

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

$\left(\frac{d^{2}y}{d{x}^2}\right)\left(\frac{d^{2}y}{d{y}^2}\right)=?$ $\frac{d^{2}y}{dx}=n^2{e}^{nx}$

$lny=nx$

$x=\frac{lny}{n}\Rightarrow \frac{dx}{dy}=\frac{1}{yn}$

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

$\left(\frac{d^{2}y}{d{x}^2}\right)\left(\frac{d^{2}y}{d{x}^2}\right)=n^2{e}^{nx}x\left(\frac{-1}{e^{nx}x}\right)$

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