Homework 7
Exercise 18. For each n ∈ N,
d n
x = nxn−1 .
dx
Proof. We proceed by induction on n ∈ N. Let n ∈ N. For the case n = 1, we have
d
(x + h) − x
h
x = lim
= lim = lim 1 = 1.
h→0
h→0 h
h→0
dx
h
Assume
d n
x = nxn−1 . By the product rule, we have
dx
d n+1
d n
x
=
(x · x)
dx
dx
= xn
d
d
x + x xn
dx
dx
= xn (1) + x(nxn−1 )
= xn + nxn
= (n + 1)xn .
Exercise 20. For each n ∈ N, there is a polynomial fn of degree n + 1 such that the n-th
derivative of tan x is given by
dn
tan x = fn (tan x).
dxn
Proof. We proceed by induction on n ∈ N. Let n ∈ N. First assume n = 1. Pick f1 (x) =
d
tan x = sec2 x = tan2 x + 1 = f1 (tan x).
dx
dn
Suppose there is a polynomial fn of degree n + 1 such that
tan x = fn (tan x). Pick
dxn
d
d
fn+1 (x) = (x2 + 1)( fn )(x). Since fn is a polynomial of degree n+1,
fn (x) is a polynodx
dx
d
mial of degree n. Since the degree of
fn (x) and x2 + 1 are n and 2, respectively, fn+1 (x)
dx
x2 + 1. Then f1 is a polynomial of degree 2 and
1
is a polynomial of degree n + 2. By the chain rule, we have
dn+1
d
tan x =
n+1
dx
dx
=
dn
tan x
dxn
d
fn (tan x)
dx
=(
d
d
fn )(tan x) · (tan x)
dx
dx
=(
d
fn )(tan x) · sec2 x
dx
= (tan2 x + 1)(
d
fn )(tan x)
dx
= fn+1 (tan x).
Exercise 22 For each natural number n, for all n times differentiable functions f and g
on R,
(n)
(f g)
=
n X
n
k
k=0
f (n−k) g (k) .
Proof. We proceed by induction on n ∈ N. Let n ∈ N. First assume n = 1. Let f and g be
differentiable functions on R. By the product rule,
1 X
1 (1−k) (k)
1 0
1
0
(f g) = f g + f g =
f g+
fg =
f
g .
0
1
k
k=0
0
0
0
Assume for all n times differentiable functions f and g on R,
(n)
(f g)
=
n X
n
k
k=0
2
f (n−k) g (k) .
Let f and g be n + 1 times differentiable functions on R. Then f and g are n times
differentiable. Therefore, by the inductive hypothesis and the product rule,
d
(f g)(n)
dx
!
n d X n (n−k) (k)
=
f
g
dx k=0 k
n X
n d
f (n−k) g (k)
=
k dx
k=0
n X
n
f (n+1−k) g (k) + f (n−k) g (k+1)
=
k
k=0
n n X
n (n+1−k) (k) X n (n−k) (k+1)
f
g +
f
g
=
k
k
k=0
k=0
(f g)(n+1) =
n n−1 n (n+1) (0)
n (0) (n+1) X n (n+1−k) (k) X n (n−k) (k+1)
=
f
g +
f g
+
f
g +
f
g
0
n
k
k
k=1
k=0
n
n n (n+1) (0)
n (0) (n+1) X n (n+1−k) (k) X
n
=
f
g +
f g
+
f
g +
f (n+1−k) g (k)
0
n
k
k
−
1
k=1
k=1
n
n
n
n (n+1) (0)
n (0) (n+1) X
=
f
g +
f g
+
+
f (n+1−k) g (k)
k
k
−
1
0
n
k=1
n n + 1 (n+1) (0)
n + 1 (0) (n+1) X n + 1 (n+1−k) (k)
=
f
g +
f g
+
f
g
0
n+1
k
k=1
n+1 X
n + 1 (n−k) (k)
=
f
g .
k
k=0
3