How do you find the second derivative of parametric function?

Hint: In order to determine to determine the second derivative of parametric function or equations , we have to use the chain rule twice and with the help of the result obtained for first derivative , you will see that the second derivative is equal to division of the derivative with respect to $ t $ of the first derivative with the derivative of $ x $ with respect to t.

Complete step-by-step answer:
To find out the second derivative of a parametric function or equation lets understand what are parametric equations.
Parametric equations are equations in which the dependent variable of derivative i.e. $ x\,and\,y $ are dependent on some other independent third variable $ (t) $ .
 $ x = x\left( t \right) $ and $ y = y\left( t \right) $
Now the first derivative of dependent variable $ y $ with respect to the another dependent variable $ x $ comes to be
 $
  \dfrac{{dy}}{{dx}} = \dfrac{{\dfrac{{dy}}{{dt}}}}{{\dfrac{{dx}}{{dt}}}} \\
  \dfrac{{dy}}{{dx}} = \dfrac{{y'(t)}}{{x'(t)}} \;
  $
Here, $ \dfrac{{dx}}{{dt}} = x'(t) $ denotes the derivative of parametric equation $ x $ with respect to $ t $ and similarly $ \dfrac{{dy}}{{dt}} = y'(t) $ denotes the derivative of parametric equation $ y $ with respect to $ t $ .
Now to calculate the second derivative of parametric equations, we have to use the chain rule twice.
 $
  \dfrac{{{d^2}y}}{{d{x^2}}} = \dfrac{d}{{dx}}\left( {\dfrac{{dy}}{{dx}}} \right) = \dfrac{d}{{dx}}\left( {\dfrac{{\dfrac{{dy}}{{dt}}}}{{\dfrac{{dx}}{{dt}}}}} \right) \\
  \dfrac{{{d^2}y}}{{d{x^2}}} = \dfrac{d}{{dt}}\left( {\dfrac{{\dfrac{{dy}}{{dt}}}}{{\dfrac{{dx}}{{dt}}}}} \right)\dfrac{{dt}}{{dx}} \\
  \dfrac{{{d^2}y}}{{d{x^2}}} = \dfrac{{\dfrac{d}{{dt}}\left( {\dfrac{{\dfrac{{dy}}{{dt}}}}{{\dfrac{{dx}}{{dt}}}}} \right)}}{{\dfrac{{dx}}{{dt}}}} \;
  $
Therefore, to find out the second derivative of the parametric function, find out the derivative with respect to $ t $ of the first derivative and after that divide it by the derivative of $ x $ with respect to t.

Note: 1. Calculus consists of two important concepts one is differentiation and other is integration.
2.What is Differentiation?
It is a method by which we can find the derivative of the function .It is a process through which we can find the instantaneous rate of change in a function based on one of its variables.
Let y = f(x) be a function of x. So the rate of change of $ y $ per unit change in $ x $ is given by:
 $ \dfrac{{dy}}{{dx}} $ .
1.Don’t forget to cross-check your answer at least once.
2.Differentiation is basically the inverse of integration.
3. $ x $ and $ y $ are the dependent variables of the derivative and both $ x $ and $ y $ are dependent on some independent variable $ t $ .