Question

Let $y=e^{2x}$ . Then $\left(\frac{d^{2}y}{d{x}^2}\right)\left(\frac{d^{2}x}{d{y}^2}\right)$ is

• A. 1
• B. $e^{-2x}$
• C. $2e^{-2x}$
• D. $-2e^{-2x}$

Answer 1

Answer: (D)

$y=e^{2x}\therefore \frac{dy}{dx}=2e^{2x}$ and $\frac{d^{2}y}{d{x}^2}=4e^{2x}$
$\frac{dx}{dy}=\frac{1}{2e^{2x}}=\frac{1}{2y}$
$\therefore \frac{d^{2}x}{d{y}^2}=-\frac{1}{2y^2}=-\frac{1}{2}{e}^{-4x}$
$\therefore \frac{d^{2}y}{d{x}^2}.\frac{d^{2}x}{d{y}^2}=4e^{2x}\left(\frac{-e^{-2x}}{2e^{2x}}\right)=-2e^{-2x}$