The chain rule applications
      Derivative of parametric functions, parametric derivatives
         Derivative of parametric functions example
      Derivative of vector-valued functions
Derivative of parametric functions or parametric derivatives
When Cartesian coordinates of a curve is represented as functions of the same variable (usually written t), they are called the parametric equations.
Thus, parametric equations in the xy-plane
 x = x (t and  y = y (t)   or   x = f (t and   y = g (t),
denote the x and y coordinate of the graph of a curve in the plane.
Assume that f and g 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
                                         xf (t         or             tf -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 derivative of the parametric functions,
Derivative of parametric functions example
Example:   Write equation of the line tangent to the curve x = t + 1 and  y = - t2 + 4 at the point  t = 1.
Solution:  The equation  x = t + 1 solve for t and plug into y = - t2 + 4, thus
t = x - 1  =>   y = -  t2 + 4,     y = - (x - 1)2 + 4  
i.e.,  y - 4 = - (x - 1)2  or  y = - x2 + 2x + 3  translated parabola with the vertex V(x0, y0), so V(1, 4).
When plotting points of a parametric curve by increasing t, the graph of the function is traced out in the direction of motion.
The derivative of the given parametric equations at  t = 1  is the slope of the tangent line,
since  t = x - 1  then   m = y' (x) =  - 2(x - 1) = - 2(2 - 1) =  - 2,   m = - 2.
Therefore, the equation of the line tangent to the given parametric curve at  t = 1 or the point P1(2, 3) is
y - y1 = m(x - x1),      y - 3 = - 2(x - 2)    =>    y  = - 2x + 7.
Derivative of vector-valued functions
If the radius vector r of a point in a plane depends on a parameter t, say t represents time, so that its magnitude and direction change continuously while t changes, then its arrow sweeping out a curve.
Let r (t) denotes its value at the moment t and  r (t + h) represents its value at  t + h, and P and P1 are the corresponding points of the curve, as is shown in the figure below.
The increment   Dr = r (t + h) -  r (t) is the vector that falls in the direction of the secant line PP1 and points from P to P1. The difference quotient
obtained by division with the scalar h, is the vector of the
same direction but of different length.
The limit of the difference quotient as h 0
is the derivative vector of the vector-valued function that falls 
in the direction of the line tangent to the curve at P.
If  x (t) and  y (t) are the scalar components of the vector r (t) then, according to rules of vector algebra,
Dr = [x (t + h) -  x (t)] i + [y (t + h) -  y (t)] j
and by use of the definition of the derivative
is the derivative vector, where  x' (t)  and  y' (t)  are its scalar components, and where
is its length or magnitude.
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