Question

$\frac{d^{2}x}{d{y}^2}$is equal to

• A. ${\left(\frac{d^{2}y}{d{x}^2}\right)}^{-1}$
• B. $-\left(\frac{d^{2}y}{d{x}^2}\right)-1{\left(\frac{dy}{dx}\right)}^{-3}$
• C. $-\left(\frac{d^{2}y}{d{x}^2}\right){\left(\frac{dy}{dx}\right)}^{-3}$
• D. $noneofthese$

Answer 1

The correct option is C

$-\left(\frac{d^{2}y}{d{x}^2}\right){\left(\frac{dy}{dx}\right)}^{-3}$

Explanation for the correct option:

Differentiation of derivative operator:

Let, $$\begin{gathered} \frac{dx}{dy}=\frac{1}{\left({\displaystyle \frac{dy}{dx}}\right)} \\ \frac{d^{2}x}{d^{2}y}=\frac{d}{dy}\left(\frac{dx}{dy}\right) \\ =\frac{d}{dy}\frac{1}{\left({\displaystyle \frac{dy}{dx}}\right)} \end{gathered}$$

Use quotient rule

$$\begin{gathered} \frac{d}{dy}\left(\frac{1}{{\displaystyle \frac{dy}{dx}}}\right)=\frac{d}{dy}\left(\frac{1}{\left({\displaystyle \frac{dy}{dx}}\right)}\right) \\ =-\left(\frac{dx}{dy}\right)\times \frac{1}{{\left({\displaystyle \frac{dy}{dx}}\right)}^2}\times \frac{d^{2}y}{d{x}^2} \\ =-\left(\frac{dx}{dy}\right)\times \frac{\left({\displaystyle \frac{d^{2}y}{d{x}^2}}\right)}{{\left({\displaystyle \frac{dy}{dx}}\right)}^2} \\ =-\frac{1}{\left({\displaystyle \frac{dy}{dx}}\right)}\times \frac{\left({\displaystyle \frac{d^{2}y}{d{x}^2}}\right)}{{\left({\displaystyle \frac{dy}{dx}}\right)}^2} \\ =-\frac{\left({\displaystyle \frac{d^{2}y}{d{x}^2}}\right)}{{\left({\displaystyle \frac{dy}{dx}}\right)}^3} \\ \mathbf{}\mathbf{=}\mathbf{-}\mathbf{\left(}\frac{{\mathbf{d}}^{\mathbf{2}}\mathbf{y}}{\mathbf{d}{\mathbf{x}}^{\mathbf{2}}}\mathbf{\right)}{\mathbf{\left(}\frac{\mathbf{d}\mathbf{y}}{\mathbf{d}\mathbf{x}}\mathbf{\right)}}^{\mathbf{-}\mathbf{3}} \end{gathered}$$

Hence option(C) is correct.