CALCULATING THE DERIVATIVE
Chapter 4
CALCULATING THE DERIVATIVE
4.1 Techniques for Finding Derivatives
1. y = 12x3 ? 8x2 + 7x + 5
dy = 12(3x3?1) ? 8(2x2?1) + 7x1?1 + 0 dx
= 36x2 ? 16x + 7
2.
y
=
8x3
?
5x2
?
x 12
dy dx
= =
8(3x3?1) ? 5(2x2?1) ?
24x2
?
10x
?
1 12
1 12
x1?1
3.
y
=
3x4
?
6x3
+
x2 8
+
5
dy dx
=
3(4x4?1)
?
6(3x3?1) +
1 8
(2x2?1)
+
0
=
12x3
?
18x2
+
1 4x
4. y = 5x4 + 9x3 + 12x2 ? 7x
dy = 5(4x4?1) + 9(3x3?1) dx
+ 12(2x2?1) ? 7(x1?1) = 20x3 + 27x2 + 24x ? 7
5. y = 6x3:5 ? 10x0:5
dy = 6(3:5x3:5?1) ? 10(0:5x0:5?1)
dx
=
21x2:5
? 5x?0:5
or
21x2:5
?
5 x0:5
6. f (x) = ?2x1:5 + 12x0:5 f 0(x) = ?2(1:5x1:5?1) + 12(0:5x0:5?1)
= ?3x0:5 + 6x?0:5 or
?
3x0:5
+
6 x0:5
7.
y
=
p 8x
+
6x3=4
= 8x1=2 + 6x3=4
?
?
dy = 8 dx
1 2
x1=2?1
+6
3 4
x3=4?1
=
4x?1=2
+
9 2
x?1=4
4
9
or x1=2 + 2x1=4
8. y = ?100px ? 11x2=3
= ?100x1=2 ? 11x2=3
?
?
dy = ?100 dx
1 2
x1=2?1
? 11
2 3
x2=3?1
=
?50x?1=2
?
22x?1=3 3
or
?50 x1=2
?
22 3x1=3
9. g(x) = 6x?5 ? x?1 g0(x) = 6(?5)x?5?1 ? (?1)x?1?1 = ?30x?6 + x?2
or
?30 x6
+
1 x2
10. y = 10x?3 + 5x?4 ? 8x
dy = 10(?3x?3?1) + 5(?4x?4?1) ? 8x1?1 dx
=
?30x?4
?
20x?5
?
8
or
?30 x4
?
20 x5
?
8
11. y = 5x?5 ? 6x?2 + 13x?1
dy = 5(?5x?5?1) ? 6(?2x?2?1) + 13(?1x?1?1) dx
= ?25x?6 + 12x?3 ? 13x?2
or
?25 x6
+
12 x3
?
13 x2
75 12. f(t) = t ? t3
= 7t?1 ? 5t?3 f 0(t) = 7(?1t?1?1) ? 5(?3t?3?1)
=
?7t?2
+
15t?4
or
?7 t2
+
15 t4
14 12 p
13. f(t) = t + t4 + 2
=
14t?1
+
12t?4
+
p 2
f 0(t) = 14(?1t?1?1) + 12(?4t?4?1) + 0
=
?14t?2
?
48t?5
or
?14 t2
?
48 t5
239
240
Chapter 4 CALCULATING THE DERIVATIVE
6 7 3p
14. y = x4 ? x3 + x + 5
=
6x?4
?
7x?3
+
3x?1
+
p 5
dy = 6(?4x?4?1) ? 7(?3x?3?1) + 3(?1x?1?1) + 0 dx
= ?24x?5 + 21x?4 ? 3x?2
or
?24 x5
+
21 x4
?
3 x2
317 15. y = x6 + x5 ? x2
= 3x?6 + x?5 ? 7x?2 dy = 3(?6x?7) + (?5x?6) ? 7(?2x?3) dx
= ?18x?7 ? 5x?6 + 14x?3
or
?18 x7
?
5 x6
+
14 x3
16. p(x) = ?10x?1=2 + 8x?3=2
?
?
p0(x) = ?10
?
1 2
x?3=2
+8
?
3 2
x?5=2
= 5x?3=2 ? 12x?5=2
or
5 x3=2
?
12 x5=2
17. h(x) = x?1=2 ? 14x?3=2
?
h0(x)
=
?
1 2
x?3=2
?
14
?
3 2
x?5=2
=
?x?3=2 2
+ 21x?5=2
or
?1 2x3=2
+
21 x5=2
18. y = 4p6 x = 6x?1=4
?
dy = 6 dx
?
1 4
x?5=4
=
?
3 2
x?5=4
?3 or 2x5=4
19. y = p?2 3x
=
?2 x1=3
=
?2x?1=3 ?
dy = ?2 dx
?
1 3
x?4=3
=
2x?4=3 3
or
2 3x4=3
20. f(x) = x3 + 5 x
= x2 + 5x?1 f 0(x) = 2x2?1 + 5(?1x?1?1)
= 2x ? 5x?2
or
2x
?
5 x2
21. g(x) = x3p? 4x x
=
x3 ? 4x x1=2
= x5=2 ? 4x1=2
?
g0(x)
=
5 2
x5=2?1
?
4
1 2
x1=2?1
=
5 2
x3=2
? 2x?1=2
or
5 2
x3=2
?
p2 x
22. g(x) = (8x2 ? 4x)2 = 64x4 ? 64x3 + 16x2
g0(x) = 64(4x4?1) ? 64(3x3?1) + 16(2x2?1) = 256x3 ? 192x2 + 32x
23. h(x) = (x2 ? 1)3 = x6 ? 3x4 + 3x2 ? 1
h0(x) = 6x6?1 ? 3(4x4?1) + 3(2x2?1) ? 0 = 6x5 ? 12x3 + 6x
24. A quadratic function has degree 2. When the derivative is taken, the power will decrease by 1 and the derivative function will be linear, so the correct choice is (b).
26. d (4x3 ? 6x?2) dx
= 4(3x2) ? 6(?2x?3) = 12x2 + 12x?3
=
12x2
+
12 x3
choice (c)
=
12x2(x3) x3
+
12
=
12x5 + x3
12
choice (b)
Neither choice (a) nor choice (d) equals
d (4x3 ? 6x?2): dx
Section 4.1 Techniques for Finding Derivatives
?
?
27. Dx
9x?1=2
+
2 x3=2
= Dx[9x?1=2 + 2x?3=2]
?
?
=9
?
1 2
x?3=2
+2
?
3 2
x?5=2
=
?
9 2
x?3=2
?
3x?5=2
or
?9 2x3=2
?
3 x5=2
?
?
28. Dx
p8 ? p3
4x
x3
= Dx [8x?1=4 ? 3x?3=2]
?
?
=8
?
1 4
x?5=4
?3
?
3 2
x?5=2
=
?2x?5=4
+
9x?5=2 2
or
?2 x5=4
+
9 2x5=2
29.
f (x)
=
x4 6
?
3x
=
1 6
x4
?
3x
f 0(x)
=
1 6
(4x3)
?
3
=
2 3
x3
?
3
f 0 (?2)
=
2 3
(?2)3
?
3
=
?
16 3
?
3
=
?
25 3
30.
f (x)
=
x3 9
?
7x2
=
1 9
x3
?
7x2
f 0 (x)
=
1 9
(3x2)
?
7(2x)
=
1 3
x2
?
14x
241
f 0(3)
=
1 (3)2 3
?
14(3)
= 3 ? 42 = ?39
31. y = x4 ? 5x3 + 2; x = 2 y0 = 4x3 ? 15x2
y0(2) = 4(2)3 ? 15(2)2 = ?28
The slope of tangent line at x = 2 is ?28: Use m = ?28 and (x1; y_1) = (2; ?22) to obtain the equation.
y ? (?22) = ?28(x ? 2) y = ?28x + 34
32. y = ?3x5 ? 8x3 + 4x2 y0 = ?3(5x4) ? 8(3x2) + 4(2x) = ?15x4 ? 24x2 + 8x
y0(1) = ?15(1)4 ? 24(1)2 + 8(1) = ?15(1)4 ? 24(1)2 + 8(1) = ?15 ? 24 + 8 = ?31
y(1) = ?3(1)5 ? 8(1)3 + 4(1)2 = ?7
The slope of the tangent line at x = 1 is ?31: Use m = ?31 and (x1; y1) = (1; ?7) to obtain the equation.
y ? (?7) = ?31(x ? 1) y + 7 = ?31x + 31 y = ?31x + 24
33.
y = ?2x1=2 + x3=2
?
y0 = ?2
1 2
x?1=2
+
3 2
x1=2
=
?x?1=2
+
3 2
x1=2
=
?
1 x1=2
+
3x1=2 2
y0(9)
=
?
1 (9)1=2
+
3(9)1=2 2
=
?
1 3
+
9 2
=
25 6
The
slope
of
the
tangent
line
at
x
=
9
is
25 6
:
242
34. y = ?x?3 + x?2 y0 = ?(?3x?4) + (?2x?3) = 3x?4 ? 2x?3
=
3 x4
?
2 x3
y0(2)
=
3 (2)4
?
2 (2)3
=
3 16
?
2 8
=
?
1 16
The
slope
of
the
tangent
line
at
x
=
2
is
?
1 16
.
35. f (x) = 9x2 ? 8x + 4 f0(x) = 18x ? 8
Let f 0(x) = 0 to ...nd the point where the slope of the tangent line is zero.
18x ? 8 = 0 18x = 8
x
=
8 18
=
4 9
Find the y-coordinate.
f (x) = 9x2 ? 8x + 4
f
4 9
?
=
9
4 9
?2
?
8
4 9
?
+
4
?
=9
16 81
?
32 9
+
4
=
16 9
?
32 9
+
36 9
=
20 9
T? 49h;e29s0l?o.pe of the tangent line is zero at one point,
36. f (x) = x3 + 9x2 + 19x ? 10 f0(x) = 3x2 + 18x + 19
If the slope of the tangent line is ?5, then the f 0(x) = ?5:
3x2 + 18x + 19 = ?5 3x2 + 18x + 24 = 0 3(x2 + 6x + 8) = 0 3(x + 2)(x + 4) = 0
x = ?2 or x = ?4
f(?2) = (?2)3 + 9(?2)2 + 19(?2) ? 10 = ?8 + 36 ? 38 ? 10 = ?20
f(?4) = (?4)3 + 9(?4)2 + 19(?4) ? 10 = ?64 + 144 ? 76 ? 10 = ?6
Chapter 4 CALCULATING THE DERIVATIVE
Thus, the points where the slope of the tangent line is ?5 are (?2; ?20) and (?4; ?6):
37. f(x) = 2x3 + 9x2 ? 60x + 4 f 0(x) = 6x2 + 18x ? 60
If the tangent line is horizontal, then its slope is zero and f0(x) = 0:
6x2 + 18x ? 60 = 0 6(x2 + 3x ? 10) = 0 6(x + 5)(x ? 2) = 0 x = ?5 or x = 2
Thus, the tangent line is horizontal at x = ?5 and x = 2:
38. f(x) = x3 + 15x2 + 63x ? 10 f 0(x) = 3x2 + 30x + 63
If the tangent line is horizontal, then its slope is zero and f0(x) = 0:
3x2 + 30x + 63 = 0 3(x2 + 10x + 21) = 0
3(x + 3)(x + 7) = 0 x = ?3 or x = ?7
Therefore, the tangent line is horizontal at x = ?3 or x = ?7:
39. f(x) = x3 ? 4x2 ? 7x + 8 f 0(x) = 3x2 ? 8x ? 7
If the tangent line is horizontal, then its slope is zero and f0(x) = 0:
3x2 ? 8x ? 7 = 0
p
x= 8?
64 + 84 6
p x = 8 ? 6 148
p
x
=
8
?
2 6
37
p
x
=
2(4
? 6
37)
p
x=
4? 3
37
p
Thus,
the
tangent
line
is
horizontal
at
x
=
4? 3
37 :
Section 4.1 Techniques for Finding Derivatives
40. f (x) = x3 ? 5x2 + 6x + 3 f0(x) = 3x2 ? 10x + 6
If the tangent line is horizontal, then its slope is zero and f 0(x) = 0:
3x2 ? 10x + 6 = 0
p
x
=
10
?
100 p6
?
72
x
=
10 ? 6
28 p
x
=
10
?2 p6
7
x
=
5
? 3
7
p
The
tangent
line
is
horizontal
at
x
=
5? 3
7:
41. f (x) = 6x2 + 4x ? 9 f0(x) = 12x + 4
If the slope of the tangent line is ?2; f 0(x) = ?2:
12x + 4 = ?2 12x = ?6
x ?
=
?
1 2
f
?
1 2
=
?
19 2
The
slope
of
the
tangent
line
is
?2
at
? ?
1 2
;
?
19 2
?
:
42. f (x) = 2x3 ? 9x2 ? 12x + 5 f0(x) = 6x2 ? 18x ? 12
If the slope of the tangent line is 12, f0(x) = 12:
6x2 ? 18x ? 12 = 12 6x2 ? 18x ? 24 = 0 6(x2 ? 3x ? 4) = 0 6(x ? 4)(x + 1) = 0
x = 4 or x = ?1
f (4) = ?59 and f (?1) = 6 The slope of the tangent line is ?12 at (4; ?59) and (?1; 6):
43. f (x) = x3 + 6x2 + 21x + 2 f0(x) = 3x2 + 12x + 21
If the slope of the tangent line is 9, f 0(x) = 9:
3x2 + 12x + 21 = 9 3x2 + 12x + 12 = 0 3(x2 + 4x + 4) = 0
3(x + 2)2 = 0 x = ?2
f (?2) = ?24
243
The slope of the tangent line is 9 at (?2; ?24):
44. f(x) = 3g(x) ? 2h(x) + 3 f 0(x) = 3g0(x) ? 2h0(x) f 0(5) = 3g0(5) ? 2h0(5)
= 3(12) ? 2(?3) = 42
45.
f (x)
=
1 2
g(x)
+
1 4
h(x)
f 0(x)
=
1 2
g0
(x)
+
1 4
h0(x)
f 0(2)
=
1 2
g0
(2)
+
1 4
h0(2)
=
1 2
(7)
+
1 4
(14)
=
7
46. (a) From the graph, f(1) = 2; because the curve goes through (1; 2):
(b) f0(1) gives the slope of the tangent line to f at 1. The line goes through (?1; 1) and (1; 2):
m
=
1
2?1 ? (?1)
=
1 2;
so
f 0(1)
=
1 2:
(c) The domain of f is [?1; 1) because the xcoordinates of the points of f start at x = ?1 and continue in...nitely through the positive real numbers.
(d) The range of f is [0; 1) because the y-coordinates of the points on f start at y = 0 and continue in...nitely through the positive real numbers.
49. f(x) = 1 ? f(x) kk
Use the rule for the derivative of a constant times
a function.
??
?
?
d f(x) = d 1 ? f(x)
dx k
dx k
= 1 f 0(x) k
= f0(x) k
50. Graph the numerical derivative of f (x) = 1:25x3 + 0:01x2 ? 2:9x + 1
for x ranging from ?5 to 5. (a) When x = 4, the derivative equals 57.18.
(b) The derivative crosses the x-axis at approximately ?0:88 and 0:88.
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