Section 2.3 Rules for Finding Derivatives
• EXAMPLE S
Find DrV
+
113
.
SOLUTION
Dry
=
)
2
4
D(
+ Dr()
4 + 1)Dr(2)
(x
—
4 + 1)
2D(x
4 + 1)
(x
-
—
(x + i)(O)
(2)(4x) + 0)
(3)(1)
4 + 1)2
(x
2
x
-
—Sx
3
2
+
4
(x
1)
2
x
—
EXAMPLE 6
(x)
1
xDr(3) —31)
+
I
Show that the Power Rule holds for negative integral expo
nents; that is,
ic”
=
D(x)
Dr(i)
_
We saw as part of Example 5 that D(3/x)
way to see the same thing.
n-I
=
x”O—1,ix’
=
=
“1
—3/x. Now we have another
( loncepts Review
I. The derivative of a product of two functions is the first
plus the
lies
times the derivative of the first. In
q ihuls.
D [f(x)g(x)]
=
3. The second term (the term involving Ii) in the expansion
of (x + h)’ is
It is this fact that leads to the formula
D,
.
2. The derivative of a quotient is the
times the derivaP’, of the numerator minus the numerator times the derivative
1
he
all divided by the
In symbols,
1’, f(v)/g(x)j =
[i”] =
4. L is called a linear operator if L(kf)
derivative operator denoted by
LU + g) =
such an operator.
[
Problem Set 2.3
Iii
Ii ,h1cnis 1—44, tind D,v its j,,t
I. y
=
3. 3’
S. y
=
7. y
=
the nih’s of this section.
2
2x
2.
lix
4. y
2
2x
y =
= TX
6. y
=
8. y
=
17.
=
19. y
=
2
no
10. y
a
x
2 +
13.
v
14. v
IS. y
2x
12. v
x + x + 2 +
=
3x 4
=
irx 7
l
—
—
2r7
2x
16. yx’+5x
7
—
5
—2
—
=
5x + irx +
5x
—7
—ITX
-
—I))
-,
2x
=
23. v
=
2 + 1)
x(x
=
(2x + 1)2
4x
25.
4 + x
3x
27. v=
y
—
2x
+
v
(v +
30. y
=
4
(x
31. y
=
2
(5x
32. y
=
4
(3x + 2x)(x
+
1)
—
—
3
2x)(x
18. v
=
20. y
=
22. v
=
24. v
=
26. v
t
+
2
(x
+
2)(x
=
29.
+
x
21. v
—
3a
=
1
—
I
I
9.
+
3
3x
+
2
7)(3x
28. v
3x + 1)
—
—
2
2x
+ 1)
2x + 1)
3x + 1)
2x6
1
3
=
x.
2
3x
+
—
—
3x(x
(—3x
x
-
2
3
—
1)
+ 2)2
+ 1)
2
— I)(x
4
(x
and
is