Question

If $y=\frac{f\left(x\right)}{\varphi \left(x\right)}$ and $z=\frac{f^{\prime }\left(x\right)}{{\varphi }^{\prime }\left(x\right)}$, then $\frac{f^{\prime \prime }}{f}-\frac{{\varphi }^{\prime \prime }}{\varphi }+\frac{2(y-z)}{f\varphi }{\left({\varphi }^{\prime }\right)}^2=$

• A. $\frac{d^{2}y}{d{x}^2}$
• B. $\frac{1}{y}.\frac{d^{2}y}{d{x}^2}$
• C. $y.\frac{d^{2}y}{d{x}^2}$
• D. None of these

Answer 1

The correct option is B $\frac{1}{y}.\frac{d^{2}y}{d{x}^2}$

$\frac{dy}{dx}=\frac{f^{\prime }\varphi -f{\varphi }^{\prime }}{{\varphi }^2}$

$\frac{d^{2}y}{{dx}^2}=\frac{(f^{\prime \prime }\varphi -f{\varphi }^{\prime \prime }){\varphi }^2-(f^{\prime }\varphi -f{\varphi }^{\prime })2\varphi {\varphi }^{\prime }}{{\varphi }^4}$

$\therefore \frac{1}{y}\frac{d^{2}y}{{dx}^2}=\frac{f^{\prime \prime }\varphi -f{\varphi }^{\prime \prime }}{f\varphi }-\frac{2(f^{\prime }\varphi -f{\varphi }^{\prime })}{f{\varphi }^2.{\varphi }^{\prime }}{\left({\varphi }^{\prime }\right)}^2$

$=\frac{f^{\prime \prime }}{f}-\frac{{\varphi }^{\prime \prime }}{\varphi }+\frac{2(y-z)}{f\varphi }.{\left({\varphi }^{\prime }\right)}^2$