Question

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