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Differential calculus - derivatives
 Higher derivatives of parametric functions
  Assume that  f (t) and g (t) are differentiable and f '(t) is not 0 then, given parametric curve can be expressed as y = y (x) and this function is differentiable at x, that is
                                         x = f (t)          or             t = f -1(x),
        by plugging into          y = g (t)     obtained is      y = g [f -1(x)].

Therefore, we use the chain rule and the derivative of the inverse function to find the first derivative of parametric functions,
or
The second derivative of parametric functions,
The third derivative of parametric functions,
 Higher derivatives of parametric functions example
 Example:  Find the second derivative of the parametric functions  x = ln t and  y = t3 + 1.
 Solution:   Since
                then
 Higher order differentials
The first order differential of a function y = f (x) at a point x
dy = f '(x)dx = y'dx
is defined as the linear part of the increment of the function expressed as the product of the derivative of the function and the corresponding increment of the independent variable.
The second order differential is defined as the differential of the first order differential with respect to
                                      the same increment dx, written      d 2 y = d(dy) = d(y'dx) = y'' (dx)2.
Similarly defined are the third and higher order differentials,     d 3 y =  y''' (dx)3,
                                                                                        ·  ·  ·  ·  ·  ·  ·  ·  ·  ·
                                                                      therefore,      d n y =  y (n) (dx)n.
Furthermore, if  y = f (u), where  u = g (x), then
d 2 y =  y'' (du)2 + y' d 2 u,
                              d 3 y =  y''' (du)3 + 3 y'' du d 2 u +  y' d 3 u, etc.
Note that, here primes denote derivatives with respect to u.
 Higher order differentials examples
 Example:  Find  d 2 y  of the function  y = cos 3x.
 Solution:   Therefore, given  y = f(u) = cos u,  where  u = g(x) = 3x.
The second differential we calculate using the formula
d 2 y =  y'' (du)2 + y' d 2 u,   where  d 2 u = g'' (x) (dx)2.
Since,      y' (u) =  - 3sin 3x  then,   y'' (u) =  - 9cos 3x
 and         g' (x) = (3x)' = 3   then,   g'' (x) = 0,  so that   d 2 u = 0.    Thus,  d 2 y = - 9cos 3x (du)2.
 Example:  Find  d 2 y  of the function  y = sin x · ln x.
 Solution:   The second differential   d 2 y =  y'' (dx)2,  since
 
 
 
 
 
 
 
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