The chain rule applications
Derivative of parametric functions, parametric derivatives
Derivative of parametric functions example
Derivative of vector-valued functions
Derivative of parametric functions, 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  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
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 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.
Calculus contents C