Page 84 - Schaum's Outline of Theory and Problems of Advanced Calculus
P. 84
CHAP. 4] DERIVATIVES 75
where the two representations of the slope of the tangent line have been equated. The solution of this
relation for x 1 is
f ðx 0 Þ
x 1 ¼ x 0
0
f ðx 0 Þ
Starting with the tangent line to the graph at P 1 ½x 1 ; f ðx 1 Þ and repeating the process, we get
f ðx 1 Þ f ðx 0 Þ f ðx 1 Þ
x 2 ¼ x 1 ¼ x 0
0 0 0
f ðx 1 Þ f ðx 0 Þ f ðx 1 Þ
and in general
n
X
f ðx k Þ
x n ¼ x 0
0
k¼0 f ðx k Þ
Under appropriate circumstances, the approximation x n to the root r can be made as good as
desired.
Note: Success with Newton’s method depends on the shape of the function’s graph in the neighbor-
hood of the root. There are various cases which have not been explored here.
Solved Problems
DERIVATIVES
3 þ x
, x 6¼ 3. Evaluate f ð2Þ from the definition.
0
4.1. (a) Let f ðxÞ¼
3 x
1 5 þ h 1 6h 6
f ð2Þ¼ lim f ð2 þ hÞ f ð2Þ ¼ lim 5 ¼ lim ¼ lim ¼ 6
0
h!0 h h!0 h 1 h h!0 h 1 h h!0 1 h
Note:Byusing rules of differentiation we find
d d
dx dx ð3 xÞð1Þ ð3 þ xÞð 1Þ
ð3 xÞ ð3 þ xÞ ð3 þ xÞ ð3 xÞ 6
0
2 2 2
f ðxÞ¼ ¼ ¼
ð3 xÞ ð3 xÞ ð3 xÞ
at all points x where the derivative exists. Putting x ¼ 2, we find f ð2Þ¼ 6. Although such rules are
0
often useful, one must be careful not to apply them indiscriminately (see Problem 4.5).
ffiffiffiffiffiffiffiffiffiffiffiffiffiffi
p
2x 1. Evaluate f ð5Þ from the definition.
0
(b) Let f ðxÞ¼
p ffiffiffiffiffiffiffiffiffiffiffiffiffiffi
9 þ 2h 3
f ð5Þ¼ lim f ð5 þ hÞ f ð5Þ ¼ lim
0
h!0 h h!0 h
p ffiffiffiffiffiffiffiffiffiffiffiffiffiffi p ffiffiffiffiffiffiffiffiffiffiffiffiffiffi
9 þ 2h 3 9 þ 2h þ 3 9 þ 2h 9 2 1
¼ lim p ffiffiffiffiffiffiffiffiffiffiffiffiffiffi ¼ lim p ffiffiffiffiffiffiffiffiffiffiffiffiffiffi ¼ lim p ffiffiffiffiffiffiffiffiffiffiffiffiffiffi ¼
h!0 h 9 þ 2h þ 3 h!0 hð 9 þ 2h þ 3Þ h!0 9 þ 2h þ 3 3
d 1=2 d
0 1=2 1
2
By using rules of differentiation we find f ðxÞ¼ dx ð2x 1Þ ¼ ð2x 1Þ dx ð2x 1Þ¼
1=2 0 1=2 1
3
ð2x 1Þ . Then f ð5Þ¼ 9 ¼ .
3
2
4.2. (a) Show directly from definition that the derivative of f ðxÞ¼ x is 3x .
1
d p ffiffiffi
(b) Show from definition that xÞ¼ p ffiffiffi.
dx 2 x
