Write the correct alternative in the following:
If $x=f(t)$ and $y=g(t)$, 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. $\frac{g^{\prime \prime}}{f^{\prime \prime}}$
D. $\frac{f^{\prime \prime} g^{\prime}-g^{\prime \prime} f^{\prime}}{\left(g^{\prime}\right)^{3}}$
Given:
$x=f(t)$ and $y=g(t)$
$\frac{\mathrm{dx}}{\mathrm{dt}}=\mathrm{f}^{\prime}(\mathrm{t}) ; \frac{\mathrm{dy}}{\mathrm{dt}}=\mathrm{g}^{\prime}(\mathrm{t})$
$\frac{d y}{d x}=\frac{\frac{d y}{d t}}{\frac{d x}{d t}}=\frac{g^{\prime}(t)}{f^{\prime}(t)}$
$\frac{\mathrm{d}^{2} \mathrm{y}}{\mathrm{dx}^{2}}=\frac{\frac{\mathrm{d}}{\mathrm{dt}}\left(\frac{\mathrm{dy}}{\mathrm{dx}}\right)}{\frac{\mathrm{dx}}{\mathrm{dt}}}$
$=\frac{1}{\mathrm{f}^{\prime}(\mathrm{t})}\left\{\frac{1}{\mathrm{f}^{\prime}(\mathrm{t})^{2}}\left(\mathrm{~g}^{\prime \prime}(\mathrm{t}) \mathrm{f}^{\prime}(\mathrm{t})-\mathrm{f}^{\prime \prime}(\mathrm{t}) g^{\prime}(\mathrm{t})\right)\right\}$
$=\frac{\left(\mathrm{g}^{\prime \prime}(\mathrm{t}) \mathrm{f}^{\prime}(\mathrm{t})-\mathrm{f}^{\prime \prime}(\mathrm{t}) \mathrm{g}^{\prime}(\mathrm{t})\right)}{\left(\mathrm{f}^{\prime}(\mathrm{t})\right)^{3}}$