Applications of differentiation - the graph of a function and its derivatives
Taylor's theorem (Taylor's formula) - The extended mean value theorem
Maclaurin's formula or Maclaurin's theorem
Representing polynomial using Maclaurin's and Taylor's formula
Taylor's theorem (Taylor's formula) - The extended mean value theorem
Suppose  f  is continuous on the closed interval [x0, x0 + h] with continuous derivatives to (n - 1)th order on the interval and its nth derivative defined on (x0, x0 + h) then,
is called Taylor's theorem.
To prove Taylor's theorem we substitute  x0 = ax0 + h = b  and introduce the function
where Q is still undetermined and  p > 0.
Since j (b) = 0 we can define Q by setting j (a) = 0 too,
then we compute j ' (x) to apply Rolle's theorem.
Note that successive terms cancel (by the product rule) when differentiating j (x), so we get
Thus, by Rolle's theorem there is some intermediate point c inside the interval [a, b] such that  j' (c) = 0,
 which gives
By plugging the Q back into above equation j (a) = 0 while returning substitutions,
 a = x0,  b = x0 + h,  b - a = h  and where
we obtain Taylor's theorem to be proved.
The last term in Taylor's formula
is called the remainder and denoted Rn since it follows after n terms.
By plugging,  a)   p = n  into Rn  we get
the Lagrange form of the remainder,
while if     b)   p = 1    we get
the Cauchy form of the remainder.
Thus, by substituting x0 + h = x  obtained is
Taylor's formula with Lagrange form of the remainder.
Therefore, Taylor's formula gives values of a function f inside the interval [x0, x0 + h] using its value and the values of its derivatives to (n - 1)th order at the point x0 in the form
f (x) = Pn - 1(x - x0) + R n
where, Pn - 1(x - x0) is (n - 1)th order Taylor polynomial for f given by the first n terms in the above formula and Rn is one of the given remainders.
Maclaurin's formula or Maclaurin's theorem
The formula obtained from Taylor's formula by setting x0 = 0
that holds in an open neighborhood of the origin, is called Maclaurin's formula or Maclaurin's theorem.
Consider the polynomial   fn (x) = anxn + an - 1xn - - 1 + · · · a3x3 + a2x2 + a1x + a0,
let evaluate the polynomial and its successive derivatives at the origin,
f (0) = a0,     f '(0) = 1· a1,     f ''(0) = 1· 2a2,     f '''(0) = 1· 2· 3a3,  . . .  ,  f (n)(0) = n!an
 we get the coefficients,
Therefore, the Taylor polynomial of a function f centered at x0 is the polynomial of degree n which has the same derivatives as f at x0, up to order n.
If a function f is infinitely differentiable on an interval about a point x0  or the origin, as are for example ex and  sin x, then
P0 (x) = f (x0),
P1 (x) = f (x0) + (x - x0) f ' (x0),
P0, P1, P2, . . . is a sequence of increasingly approximating polynomials for  f.
Representing polynomial using Maclaurin's and Taylor's formula
If given is an n-th degree polynomial
Pn(x) = anxn + an - 1xn - 1 + · · ·a2x2 + a1x + a0    then,
Pn(0) = a0Pn' (0) = a1Pn''(0) = 2! a2Pn'''(0) = 3! a3, . . . ,  Pn(n)(0) = n! an  and  Pn(n + 1) = 0
so, the coefficients of the polynomial
therefore, applying Maclaurin's formula, every polynomial can be written as
since Pn(n + 1) = 0, the remainder vanishes.
Representing polynomial using Maclaurin's and Taylor's formula, examples
Example:  Represent the quintic  y = 2x5 + 3x4 - 5x3 + 8x2 - 9x + 1 using Maclaurin's formula.
Solution:  Let write all successive derivatives of the given quintic and evaluate them at the origin,
y' (x) = 10x4 + 12x3 - 15x2 + 16x - 9,           y' (0) = - 9
y'' (x) = 40x3 + 36x2 - 30x + 16,                    y'' (0) = 16
y''' (x) = 120x2 + 72x - 30                            y''' (0) = - 30
yIV (x) = 240x + 72                                       yIV (0) = 72
yV (x) = 240,                                                   yV (0) = 240
yVI = 0  and the last term of the polynomial  a0 y(0) = P5(0) = 1,
then substitute obtained values into Maclaurin's formula
Example:  Represent the quartic  yx4 - 4x3 + 4x2 + x - 4 at x0 = 2 using Taylor's formula.
Solution:  Let write all successive derivatives of the given quartic and evaluate them at x0 = 2,
y' (x) = 4x3 - 12x2 + 8x + 1,              y' (2) = 1
y'' (x) = 12x2 - 24x + 8,                    y'' (2) = 8
y''' (x) = 24x - 24                            y''' (2) = 24
yIV (x) = 24                                           yIV = 24
yV (x) = 0     and the last term,   a0 y(2) = P4(2) = - 2,
then substitute obtained values into Taylor's formula
Calculus contents D