Differentiating each formula relative to $latex t$, we have:
$ latex dfrac {dy} {dt} =1 ~ ~$ $latex ~ ~ dfrac {dy} {dt} =3t ^ 2$
Using the chain policy, we have
$$ frac {dy} {dx} =frac {dy} {dt} frac {dt} {dx} =3t ^ 2$$
Now, we obtain each regard to this formula relative to $latex x$ to locate the 2nd acquired $latex dfrac {d ^ 2y} {dx ^ 2} $:
$$ frac {d} {dx} left( frac {dy} {dx} right)= frac {d} {dx} (3t ^ 2)$$
Since we can not obtain $latex 3t ^ 2$ relative to $latex x$, we utilize the chain policy:
$$ frac {d} {dx} (3t ^ 2)= frac {d} {dt} (3t ^ 2) frac {dt} {dx} $$
$$= 6t frac {dt} {dx} $$
Substituting this, we have:
$$ frac {d ^ 2y} {dx ^ 2} =6t frac {dt} {dx} $$
Since $latex frac {dx} {dt} =1$, we can compose:
$$ frac {dx} {dt} =frac {1} {frac {dx} {dt}} =1$$
Then, the 2nd by-product is:
$$ frac {d ^ 2y} {dx ^ 2} =6t (1 )= 6t$$