Question:
If $x=t^{2}$ and $y=t^{3}$, where $a$ is a constant, then find $\frac{d^{2} y}{d x^{2}}$ at $x=\frac{1}{2}$
Solution:
Given:
$x=t^{2} ; y=t^{3}$
$\frac{\mathrm{dy}}{\mathrm{dt}}=3 \mathrm{t}^{2} ; \frac{\mathrm{dx}}{\mathrm{dt}}=2 \mathrm{t}$
$\frac{\mathrm{dy}}{\mathrm{dx}}=\frac{\frac{\mathrm{dy}}{\mathrm{dt}}}{\frac{\mathrm{dx}}{\mathrm{dt}}}=\frac{3 \mathrm{t}}{2}$
$\frac{\mathrm{d}^{2} \mathrm{y}}{\mathrm{dx}^{2}}=\frac{\frac{\mathrm{d}}{\mathrm{dt}}\left(\frac{\mathrm{dy}}{\mathrm{dx}}\right)}{\frac{\mathrm{dx}}{\mathrm{dt}}}=\frac{\frac{3 \mathrm{t}}{2}}{2 \mathrm{t}}$
$=\frac{3}{4}$