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:

$$ frac {d ^ 2y} {dx ^ 2} =6t (1 )= 6t$$



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