Homework 4: solutions
Section 2.3
x−(x+h)
h
1
− x(x+h)
−1
f (x + h) − f (x)
x(x+h)
55. f 0 (x) = lim
= lim x+h x = lim
= lim
h→0
h→0
h→0
h→0
h
h
h
h
1
1
1
= lim −
=−
= − 2.
h→0
x(x + h)
limh→0 x(x + h)
x
cos(x + h) − cos(x)
cos(x) cos(h) − sin(x) sin(h) − cos(x)
56. cos0 (x) = lim
= lim
h→0
h→0
h
h
µ
¶
cos(x)(cos(h) − 1) − sin(x) sin(h)
cos(h) − 1
sin(h)
= lim
= lim cos(x) ·
− sin(x) ·
h→0
h→0
h
h
h
sin(h)
cos(h) − 1
− sin(x) · lim
= cos(x) · 0 − sin(x) · 1 = − sin(x).
= cos(x) · lim
h→0
h→0
h
h
64. (a). The hyperbola is the graph of the function f (x) = c/x = cx−1 .
Let (a, c/a) the coordinates of P . We have to determine the x- and y-intercepts of the tangent line at a.
Equation of the tangent line: Y = f (a) + f 0 (a)(X − a) = ac − ac2 (X − a).
x-intercept. Set Y = 0: ac = ac2 (X − a) ⇒ X − a = a, so the x-intercept is the point (2a, 0).
2c
y-intercept. Set X = 0: Y = ac − ac2 (−a) = 2c
a , so the y-intercept is the point (0, a ).
c
Now you have to argue why the point (a, a ) is the midpoint of the segment joining (0, 2c
a ) and (2a, 0).
(b). Area = 12 · (2a) · ( 2c
)
=
2c
=
constant,
it
does
not
depend
on
the
choice
of
a
(choice
of P ).
a
65. This is f 0 (1), where f (x) = x1000 . Since f 0 (x) = 1000x999 , the answer is f 0 (1) = 1000.
Section 2.4
−2
2x
. Quotient rule: f 0 (x) = (x+1)
27. Let f (x) = x+1
2.
3
0
Tangent line at 1: Y = f (1) + f (1)(x − 1) = 1 − 21 (X − 1), so Y = − X
2 + 2.
(− sin(x)) · f (x) − cos(x) · f (x)
f (x) sin(x) + f 0 (x) cos(x)
=−
.
2
(f (x))
(f (x))2
√
√
√
f (π/3) · 23 + f 0 (π/3) · 12
4 · 23 + (−2) · 21
1−2 3
h0 (π/3) = −
=
−
=
(f (π/3))2
42
16
0
0
f (3)g(3) − f (3)g (3)
(−6) · 2 − 4 · 5
−32
42. (c). (f /g)0 (3) =
=
=
= −8.
(g(3))2
22
4
Section 2.5
40. h0 (x) =
3. y = f ◦ g(x), f (x) = x10 , g(x) = 1 − x2 .
−20 x (1 − x2 )9 .
dy
dx
= f 0 (g(x)) · g 0 (x) = 10(g(x))9 · g 0 (x) = 10(1 − x2 )9 · (−2x) =
30. The domain of this function is R−{0}. For x =
6 0, y = f (x)·g(x), with f (x) = x and g(x) = sin(1/x).
dy
Then dx
= f 0 (x)g(x) + f (x)g 0 (x) = g(x) + xg 0 (x) = sin(1/x) + xg 0 (x).
dy
By chain rule, g 0 (x) = cos(1/x)(−1/x2 ), so (for x 6= 0): dx
= sin( x1 ) − x1 cos( x1 ).
48. By differentiating (h(x))2 = 4 + 3f (x) we have: 2h(x)h0 (x) = 3f 0 (x). At x = 1: 2h(1)h0 (1) =
3f 0 (1)
3·4
√
3f 0 (1) ⇒ h0 (1) =
=
.
2h(1)
2· 4+3·7
√
One can also solve this problem by direct chain rule: h = g ◦ f , where g(x) = 4 + 3x. So h0 (1) =
g 0 (f (1)) · f 0 (1) = g 0 (7) · 4, so you need to compute g 0 (7).
52. (b). g(x) = f (x2 ) ⇒ g 0 (x) = f 0 (x2 ) · (2x). In particular g 0 (2) = 4f 0 (4). We can use the graph
to approximate f 0 (4): if you draw the tangent line at 4 it look like it passes through the point (4.5, 3).
3−2
= 2. So g 0 (3) ≈ 4 · 2 = 8. (Perhaps there are several
Therefore the slope of the tangent is f 0 (4) ≈ 4.5−4
possible answers to this problem.)
1