Question

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

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

Answer 1

Answer: (B)

We have, $x=\varphi (t),y=\psi (t)$
$\therefore \frac{dy}{dx}=\frac{dy/dt}{dx/dt}=\frac{{\psi }^{\prime }}{{\varphi }^{\prime }}$
$\Rightarrow \frac{d^{2}y}{d{x}^2}=\frac{d}{dx}\left(\frac{{\psi }^{\prime }}{{\varphi }^{\prime }}\right)=\frac{d}{dt}\left(\frac{{\Psi }^{\prime }}{{\varphi }^{\prime }}\right)\frac{dt}{dx}$
$=\frac{{\varphi }^{\prime }{\Psi }^{\prime \prime }-{\psi }^{\prime }{\varphi }^{\prime \prime }}{{\left({\varphi }^{\prime }\right)}^2}\cdot \frac{1}{{\varphi }^{\prime }}=\frac{{\varphi }^{\prime }{\Psi }^{\prime \prime }-{\psi }^{\prime }{\varphi }^{\prime \prime }}{{\left({\varphi }^{\prime }\right)}^3}$