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)

$-2e^{-2x}$

$Differentiatingwithrespectto$x$,weget$

$y^{\prime }=2e^{2x}$

$Againdifferentiatingwithrespectto$x$,$

$y^{\prime \prime }=4e^{2x}$

$Nowwriting{}^{\prime }{x}^{\prime }intermsof{}^{\prime }{y}^{\prime }bytakinglogweget,$

$x=\frac{1}{2}\log y$

$Differentiatingwithrespectto$y$,$
$\frac{dx}{dy}=\frac{1}{2y}$

$Againdifferentiatingwithrespectto$y$,$
$\frac{d^{2}x}{d{y}^2}=\frac{-1}{2y^2}$
$(\frac{d^{2}y}{d{x}^2}\left)\right(\frac{d^{2}x}{d{y}^2})=4e^{2x}(-\frac{1}{2e^{4x}})$

$=-2e^{-2x}$