is denoted by f^(k) (x).    Thus

(6)        f'(x)=na_0 x^n-1 + (n-l)a_1 x^n-2 + $$ +2a_n-2 x + a_n-1,

(7)       f''(x)=n(n-1)a_0 x^n-2 + (n-1) (n-2)a_1 x^n-3 +  ... +2a_n-2,

etc.    Hence we have

(8)        f(x+h) =f(x) +f'(x)h+f''(x) h^2 / 1·2 + f'''(x)h^3 / 1·2·3

         + ... + f^(r) (x)h^r / r! + ... + f^(n) (x)h^n / n!,

where r! is the symbol, read r factorial, for the product 1·2·3 ... (r-1)r.
Here r is a positive integer, but we include the case r = 0 by the definition,
0!=1.

This formula (8) is known as Taylor's theorem for the present case of
a polynomial f(x) of degree n. We call f'(x) the (first) derivative of f(x),
and f''(x) the second derivative of f(x), etc. Concerning the fact that
f''(x) is equal to the first derivative of f'(x) and that, in general, the kth
derivative f^(k) (x)  of f(x) is equal to the first derivative of f^(k-1) (x), see
Exs. 6-9 of the next set.

In view of (8), the limit of (4) as h approaches zero is f'(x). Hence
f'(x) is the slope of the tangent to the graph of y=f(x) at the point (x, y).

In (5) and (6), let every a be zero except a_0. Thus the derivative of
a_0 x^n is na_0 x^n-1, and hence is obtained by multiplying the given term by
its exponent n and then diminishing its exponent by unity. For example,
the derivative of 2x^3 is 6x^2.

Moreover, the derivative of f(x) is equal to the sum of the derivatives
of its separate terms.    Thus the derivative of x^3 + 4x^2 -11 is 3x^2 + 8x,
as found also in § 55.

EXERCISES

1.  Show that the slope of the tangent to y =8x^3 - 22x^2 +13x -2 at (x, y) is 24x^2 -
44x + 13, and that the bend points are (0.37, 0.203), (1.46, -5.03), approximately.
Draw the graph.

2.  Prove that the bend points of y=x^3 -2x -5 are (.82, -6.09), (-.82, -3.91),
approximately.    Draw the graph and locate the real roots.

3.  Find the bend points of y = x^3 -6x^2 +8x+8.    Locate the real roots.

4.  Locate the real roots of f(x) =x^4 +x^3 -x-2 = 0.

Hints: The abscissas of the bend points are the roots of f'(x) =4x^3 +3x^2 -1=0.
The bend points of y =f'(x) are (0, -1) and (-1/2, -3/4), so that f'(x)= 0 has a single real
root (it is just less than 1/2). The single bend point of y=f(x) is (1/2, -37/16), approximately. 
