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

Answer: (D)

Given that, $y=e^{nx}$
Differeniating both sides with respect to x
$\frac{dy}{dx}=n{e}^{nx}$
Again differentiating w.r.t 'x',
$\frac{d^{2}y}{d{x}^2}=n.n{e}^{nx}=n^2{e}^{nx}$ .....(1)
$y=e^{nx}$
$nx={\log }_{e}y$
$x=\frac{1}{n}\log y$
Differentiating x with respect to y
$\frac{dx}{dy}=\frac{1}{n}.\frac{1}{y}$
Again differentiating with respect to y
$\frac{d^{2}x}{d{y}^2}=\frac{1}{n}.\left(\frac{-1}{y^2}\right)=\frac{-1}{n{y}^2}=-\frac{1}{n{e}^{2nx}}$ ......(2)
Multiplying equation (1) and (2)
$\left(\frac{d^{2}y}{d{x}^2}\right)\left(\frac{d^{2}x}{d{y}^2}\right)=\left(n^2{e}^{nx}\right)\left(\frac{-1}{n}{e}^{2nx}\right)=-n{e}^{-nx}$