Question

$\frac{d^{2}x}{d{y}^2}$ equals

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

Answer 1

Answer: (D)

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

Recalling that $\frac{dx}{dy}=\frac{1}{\frac{dy}{dx}}$, we see that

$\frac{d^{2}x}{a{y}^2}=\frac{d}{dy}\left(\frac{dx}{dy}\right)=\frac{d}{dy}\left(\frac{1}{\frac{dy}{dx}}\right)$

Now by quotient rule, we have that

$\frac{d}{dy}\left(\frac{1}{\frac{dy}{dx}}\right)=\frac{-\frac{d}{dy}\left(\frac{dy}{dx}\right)}{{\left(\frac{dy}{dx}\right)}^2}$

$=-\frac{\frac{dx}{dy}.\frac{d}{dx}\left(\frac{dy}{dx}\right)}{{\left(\frac{dy}{dx}\right)}^2}$

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

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