Question

If $x=f\left(t\right)$ and $y=g\left(t\right)$, then $\frac{d^{2}y}{d{x}^2}$ is equal to

• A. $\frac{f^{\prime }{g}^{\prime \prime }-g^{\prime }{f}^{\prime \prime }}{{\left(f^{\prime }\right)}^3}$
• B. $\frac{f^{\prime }{g}^{\prime \prime }-g^{\prime }{f}^{\prime \prime }}{{\left(f^{\prime }\right)}^2}$
• C. $\frac{g^{\prime }}{f^{\prime \prime }}$
• D. $\frac{f^{\prime }{g}^{\prime }-g^{\prime \prime }{f}^{\prime \prime }}{{\left(f^{\prime }\right)}^2}$

Answer 1

The correct option is A $\frac{f^{\prime }{g}^{\prime \prime }-g^{\prime }{f}^{\prime \prime }}{{\left(f^{\prime }\right)}^3}$

$x=f\left(t\right),y=g\left(t\right)\Rightarrow \frac{dy}{dt}=g^{\prime },\frac{dx}{dt}=f^{\prime }\Rightarrow \frac{dy}{dx}=\frac{\frac{dy}{dt}}{\frac{dx}{dt}}=\frac{g^{\prime }}{f^{\prime }}\Rightarrow \frac{d^{2}y}{d{x}^2}=\frac{d}{dx}\left(\frac{dy}{dx}\right)\Rightarrow \frac{d^{2}y}{d{x}^2}=\frac{d}{dt}\left(\frac{dy}{dx}\right)\frac{dt}{dx}\Rightarrow \frac{d^{2}y}{d{x}^2}=\frac{f^{\prime }{g}^{\prime \prime }-g^{\prime }{f}^{\prime \prime }}{{\left(f^{\prime }\right)}^2}\times \frac{1}{f^{\prime }}\therefore \frac{d^{2}y}{d{x}^2}=\frac{f^{\prime }{g}^{\prime \prime }-g^{\prime }{f}^{\prime \prime }}{{\left(f^{\prime }\right)}^3}$