Q. $\frac{d^{2}x}{dy^{2}}\$equals
A) $\left( \frac{d^{2}y}{dx^{2}} \right)^{- 1}$
B) $- \left( \frac{d^{2}y}{dx^{2}} \right)^{- 1}\left( \frac{dy}{dx} \right)^{- 3}$
C) $\left( \frac{d^{2}y}{dx^{2}} \right)\left( \frac{dy}{dx} \right)^{- 2}$
D) $- \left( \frac{d^{2}y}{dx^{2}} \right)\left( \frac{dy}{dx} \right)^{- 3}\$
Explanation
(d)
Since, $\frac{dx}{dy} = \frac{1}{dy/dx} = \left( \frac{dy}{dx} \right)^{- 1}$
$\Rightarrow \ \ \frac{d}{dy}\ \left( \frac{dx}{dy} \right)\ = \ \frac{d}{dx}\ \left( \frac{dy}{dx} \right)^{- 1}\frac{dx}{dy}$
$\Rightarrow \ \ \frac{d^{2}x}{dy^{2}} = - \left( \frac{d^{2}y}{dx^{2}} \right)\left( \frac{dy}{dx} \right)^{- 2}\left( \frac{dx}{dy} \right) = - \left( \frac{d^{2}y}{dx^{2}} \right)\left( \frac{dy}{dx} \right)^{- 3}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.