Thomas Calculus 13th [Solutions]

Chapter 3 Additional and Advanced Exercises 217
22. From Exercise 21 we have that fg is differentiable at 0 if f is differentiable at 0, f (0) 0 and g is continuous at 0.
(a) If f ( x)
sin x and g( x) | x |, then | x| sin x is differentiable because f (0) cos(0) 1, f (0) sin (0) 0
and g( x) | x | is continuous at x 0.
2/3 2/3
(b) If f ( x)
sin x and g( x) x , then x sin x is differentiable because
2/3
f (0) cos (0) 1, f (0) sin (0) 0 and g( x)
x is continuous at x 0.
(c) If f ( x) 1 cos x and g( x) 3 x , then 3 x (1 cos x ) is differentiable because f (0) sin (0) 0,
1/3
f (0) 1 cos (0) 0 and g( x)
x is continuous at x 0.
(d) If f ( x)
x and 1 2
g( x) x sin , then x sin 1
2
x
is differentiable because f (0) 1, f (0) 0 and
sin
1
lim x 1
x
sin lim lim sin t 0
1
x 0
x
x 0 x
t
(so g is continuous at x 0 ).
x
23. If f ( x)
x and ( ) sin 1 2
g x x , then x sin 1
x
x
is differentiable at x 0 because f (0) 1, f (0) 0 and
sin
1
lim x 1
x
sin lim lim sin t 0
1
x 0
x
x 0
t
(so g is continuous at x 0 ). In fact, from Exercise 21,
x t
h (0) g(0) f (0) 0. However, for x 2
0, h ( x ) x cos 1 1 2 sin 1 .
x
x 2
x
x
But
lim h ( x ) lim cos 1 2 sin 1
x 0 x 0
x
x x
does not exist because cos 1
x
has no limit as x 0. Therefore,
the derivative is not continuous at x 0 because it has no limit there.
24. From the given conditions we have f ( x h) f ( x) f ( h), f ( h) 1 hg( h ) and lim g( h ) 1. Therefore,
h 0
f ( x) lim f ( x h) f ( x) lim f ( x) f ( h) f ( x) lim ( ) f ( h) 1
( ) lim ( ) ( ) 1 ( )
0
h
0
h f x h h h 0
h
f x g h f x f x
h 0
f ( x) f ( x ) and f ( x ) exists at every value of x.
dy du1 du2
25. Step 1: The formula holds for n 2 (a single product) since y u1u 2 u
dx dx 2 u1
dx
Step 2: Assume the formula holds for n k:
dy du1 du2
du
y u1u 2 u 2 3 1 3 ... k
k u u uk u u uk u1u 2 uk
1 .
dx dx dx dx
dy d ( u1u 2 uk
)
duk
1
If y u1u 2 ukuk 1 u1u 2 uk u k 1, then uk
1 u
dx dx 1u2
uk
dx
du1 du2
duk
duk
1
u2u3 uk u1 u3 uk u1u dx dx 2 uk 1 u
dx k 1 u1u 2 uk
dx
du1 du2
duk
duk
1
u2u3 uk 1 u1 u3 uk 1 u1u 2 uk 1 uk 1 u1u 2 uk
.
dx dx dx dx
Thus the original formula holds for n ( k 1) whenever it holds for n k.
m
26. Recall m!
k k!( m k)! . m
Then m!
m m
1 m and
m! m!
m!( k 1) m!( m k)
1!( m 1)!
k k 1 k!( m k)! ( k 1)!( m k 1)! ( k 1)!( m k)!
m!( m 1)
( m 1)! m 1
( k 1)!( m k)!
( 1)!(( 1) ( 1))! k 1 . Now, we prove Leibnizs rule by mathematical induction.
k m k
d( uv)
Step 1: If n 1, then u dv v du . Assume that the statement is true for n k , that is:
dx dx dx
d k ( uv)
1 2 2 1
d k u v k d k u dv k d k u d v k
1 2 ... du d k v d k v
2 2 1
.
k k k dx
k k dv
u k 1 k
dx dx dx dx dx dx dx
Step 2: If n k 1, then
k 1 k k 1 k k k 1 2
d ( uv) d d ( uv)
d u v d u dv k d u dv k d u d v
k 1 dx k k 1 k dx k dx k 1 2
dx dx dx dx dx dx dx
k
k 1 2 k 2 3 2 k 1
k
d u d v k d u d v k
k
2 1 2 2 ... d u d v du d u
k k 2 3 k 1 2 k 1 k 1 dx
v k
dx dx dx dx dx dx dx
1 1 1 2
du d k v k k ( 1)
k k k
u d u d u v k d u dv d k u d v
dx k k 1 k 1 k dx 1 2 k 1 2
dx dx dx dx dx dx
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