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Solutions to Supplementary Questions for HP Chapter 12
d dx (x 8 ln x) = 8x 7 ln x +
x^8 x = 8x 7 ln x + x 7
d 2
dx^2
(x 8 ln x) = d dx
(8x 7 ln x + x 7 ) =
8 · 7 x 6 ln x + 8 x 7
x
= 8 · 7 x 6 ln x + 8x 6
Now note that the power of x in all terms are the same, namely 8−k when we calculate dk dxk^ , so at the 8 th derivative, the power of x is zero in all terms. Also note that all terms
except for the first term are only a constant times a power of x, so at the eighth derivative,
all terms other than the first term become constant. Hence at the 9 th derivative, the only
term that is nonzero is the first term that includes ln x.
d 8 dx^8 (x 8 ln x) = 8 · 7 · 6 · 5 · 4 · 3 · 2 · 1 · x 0 ln x + constants
d 9 dx^9 (x 8 ln x) = 8 · 7 · 6 · 5 · 4 · 3 · 2 · 1 · x
8! x
d
dx
[ln(ln(ln(ln x)))]
ln(ln(ln x))
[
d
dx
ln(ln(ln x))
]
ln(ln(ln x))
ln(ln x)
d
dx
(ln(ln x))
ln(ln(ln x))
ln(ln x)
ln x
d
dx
(ln x)
ln(ln(ln x))
ln(ln x)
ln x
x
- (i)
d dx x
(aa) = a
a x
(aa−1)
(ii) Using the Chain Rule,
d dx
a f (x)
= a f (x) ln a(f ′ (x)), and hence,
d dx a (x a )
= a
(xa) ln a(ax
a− 1 )
(iii) Similarly,
d dx a
(ax) = a
(ax^ ) ln a(
d dx a
x ) = a
(ax^ ) ln a(a
x ln a) = a
(ax) a
x (ln a)
2 .
So
d dx
x
(aa)
(xa)
(ax)
= a
a x
(aa−1)
a
(xa)
x
a− 1
(ax) a
x (ln a)
2 .
- (a)
f
′ (x) = lim h→ 0
f (x + h) − f (x)
x
= lim h→ 0
f (x)f (h) − f (x)
x
= lim h→ 0
f (x)
[
f (h) − 1
h
]
= f (x) lim h→ 0
f (h) − f (0)
h
= f (x)f
′ (0)
= f (x)
(b) We have, for f (x) = e x ,
f ′ (0) = limh→ 0 e h − 1 h
f (0) = e 0 = 1
f (x + z) = e
x+z = e
x e
z = f (x)f (z) for all x and z.
Applying (a) shows f (x) = f ′ (x); i.e., d dx
e x = e x .
- Let g(x) = f (x)e −cx . By the product rule and chain rule we have
g
′ (x) = f
′ (x)e
−cx − f (x)ce
−cx
= cf (x)e
−cx − cf (x)e
−cx
Since the derivative of g is zero, and g is defined for all x, we must have g equal to a constant
K, g(x) = K, so K = f (x)e −cx and multiplying both sides by e cx gives f (x) = Ke cx .
- (a)
f (x) = ln(x) g(x) = ax
2
f ′ (x) =
1 x g ′ (x) = 2ax + b
f ′′ (x) = − 1 x^2
g ′′ (x) = 2a
so we have
g(1) = f (1) ⇒ a + b + c = 0
g ′ (1) = f ′ (1) ⇒ 2 a + b = 1
g ′′ (1) = f ′′ (1) ⇒ 2 a = − 1
Solving, we see that a = − 1 2
, b = 2, c = − 3 2
, so g(x) = − 1 2
x 2
(b)
f (x) = f ′ (x) = f ′′ (x) = e x
g(x) = ax
2
g
′ (x) = 2ax + b
g
′′ (x) = 2a
(c) Write q 1 = 3p − 5 , q 2 =
p− 5 6
. q 1 has point elasticity −5 by (a). If q 2 =
p− 5 6 , p = 6q 2 + 5
so q 2 has point elasticity
p p− 5 which is −
2 3 when p = 2.
From (b), we know that since q = q 1 q 2 , the point elasticity of q when p = 2 is
− 2 3
− 17 3
- By the chain rule,
dy dx
dy dp
dp dx
. For
dy dp , we have
dy
dp
d(ln f (p))
dp
f (p)
f
′ (p) =
q
dq
dp
For
dp dx we have x = ln p, p = e x ,
dp dx = e x = p. Therefore,
dy
dx
dy
dp
dp
dx
q
dq
dp
p =
p
q
dq
dp
= η
- (a) Taking natural logarithms, we have
ln q = a(ln p) − b(p + c)
and differentiating with respect to p, we have
q
dq
dp
a
p
− b
and
dq
dp
a
p
− b
p
a e
−b(p+c)
Since p >
a b
a p < b which shows that for all p >
a b
dq dp < 0. This means that the demand
decreases as the price increases or equivalently, that the demand increases as the price
decreases.
(b) Using the equation η =
p q
dq dp where η is the point elasticity of demand, we have
η =
p
q
a
p
− b
p a e −b(p+c)
p
q
a
p
− b
q
= a − bp
x n = e ln(x n ) = e n ln x
d
dx
x
n
d
dx
e
n ln x = e
n ln x
d
dx
n ln x
= x
n
n
x
= nx
n− 1
OR
y = x
n ⇒ ln y = n ln x ⇒
y ′
y
n
x
⇒ y
′ = x
n
n
x
= nx
n− 1
- Using implicit differentiation with respect to x,
x
y
dy
dx
dy
dx
y
x
y
x
(x 6 = 0)
An arbitrary point on the curve is (x 0 , (
c −
x 0 )
2 ), so the tangent line has the following
equation (note that x 0 6 = 0).
y − (
c −
x 0 )
2 = (x − x 0 )
c −
x 0 )^2
x 0
y − (
c −
x 0 )
2
(x − x 0 )(
x 0 −
c) √ x 0
When x = 0, then
y = −
x 0 √ x 0
x 0 −
c) + (
c −
x 0 )
2
x 0 (
c −
x 0 ) + (
c −
x 0 )
2
c −
x 0 )(
c −
x 0 +
x 0 )
c(
c −
x 0 )
So the y-intercept is
c(
c −
x 0 ) = c −
cx 0.
When y = 0 (note that y 0 6 = 0),
x − x 0 = −
c −
x 0 )
x 0 √ x 0 −
c
c −
x 0 )
x 0
x =
x 0 (
c −
x 0 ) + x 0 =
x 0 c − x 0 + x 0 =
x 0 c
so the x intercept is
x 0 c. The x-intercept and y-intercept have sum
x 0 c + c −
cx 0 = c.
- (a) f (g(x)) = x. Differentiating, we have f
′ (g(x))g
′ (x) = 1, so g
′ (x) =
1 f ′(g(x))
(b) g ′ (x) = 1 f ′(g(x))
1 ( 1 ex^ )
= e x f ′ (x) = 1 g′(f (x))
1 eln^ x^
1 x
- Differentiating implicitly with respect to x, we have
2 x − x
dy
dx
− y + 2y
dy
dx
so dy
dx
y − 2 x
2 y − x
Therefore,
1 r
d^2 y dx^2 [ 1+( dy dx )
2
] 3
2
; or
1 r
d^2 y dx^2 [ 1+( dy dx )
2
] 3
2
- Differentiating implicitly the first equation with respect to z,
y + z
dy
dz
2
3 dy
dz
(z + z
3 )
dy
dz
= −y(1 + 3z
2 )
dy
dz
y(1 + 3z
2 )
z + z^3
and also differentiating the second question with respect to x,
2 xz + x
2 dz
dx
2
dz
dx
= 6x
2 y + 2x
3 dy
dx
(x
2
dz
dx
= 6x
2 y − 2 xz − 3 z
2
3 dy
dx
dz
dx
6 x 2 y − 2 xz − 3 z 2
x^2 + 6xz
By the chain rule,
dy dx
dy dz
dz dx , so
dy
dx
= −y
(1 + 3z 2 )
z + z^3
6 x 2 y − 2 xz − 3 z 2
Solving for
dy dx we have
(z + z
3 )(x
2
dy
dx
= −y(1 + 3z
2 )(6x
2 y − 2 xz − 3 z
2
3 dy
dx
(z + z
3 )(x
2
−y(1 + 3z 2 )
dy
dx
= 6x
2 y − 2 xz − 3 z
2
3 dy
dx [
2 x
3
(z + z 3 )(x 2
y(1 + 3z 2 )
]
dy
dx
= 2xz + 3z
2 − 6 x
2 y
dy
dx
(2xz + 3z 2 − 6 x 2 y)y(1 + 3z 2 )
y(1 + 3z 2 )2x 3
- Set y =
ex
√ x^5 + (x+1)^4 (x^2 +3)^2 Then
ln y = ln
e x
x^5 + 2
(x + 1)^4 (x^2 + 3)^2
= ln(e
x ) + ln
x 5
4 − ln(x
2
2
= x +
ln(x
5
2
Differentiating both sides with respect to x gives
d ln y
dx
d
dx
[
x +
ln(x
5
2
]
y
′
y
d dx (x 5
2(x 5
d dx (x + 1))
x + 1
d dx (x 2
x 2
y ′
y
5 x 4
2(x 5
x + 1
4 x
x 2
Substituting back in y =
ex
√ x^5 + (x+1)^4 (x^2 +3)^2 and solving for y ′ gives
y
′
e
x
x 5
(x + 1)^4 (x^2 + 3)^2
5 x
4
2(x^5 + 2)
x + 1
4 x
x^2 + 3
- (a) First, set u = x x . Then
ln u = ln(x
x ) = x ln x
u ′
u
= ln x +
x
x
= ln x + 1
⇒ u
′ = x
x (ln x + 1)
Now, set y = x
(xx^ )
. Then ln y = ln x
(xx^ ) = x
x ln x. Taking the derivative of both sides
gives
y ′
y
= x
x (ln x + 1) ln x +
x x
x
⇒ y ′ = x (x x )
[
x x ((ln x) 2
]
(b) y = x
ln x ⇒ ln y = ln(x
ln x ) = ln x(ln x) = (ln x)
2
. Taking the derivative of both sides
gives
y ′
y
= 2(ln x)
[
d
dx
(ln x)
]
2(ln x)
x
⇒ y
′ = x
ln x
[
2 ln x
x
]
= 2 ln x(x
(ln x)− 1 )
(c) y = (ln x) x ⇒ ln y = ln(ln x) x = x ln(ln x). Taking the derivative of both sides give
y ′
y
= ln(ln x) +
x
ln x
[
d
dx
(ln x)
]
= ln(ln x) +
x
x ln x
= ln(ln x) +
ln(x)
⇒ y
′ = (ln x)
x
[
ln(ln x) +
ln x
]
With an initial guess of r= 0 .05, the iterates are
r 0 = 0. 05
r 1 = 0. 041247105
r 2 = 0. 03894101
r 3 = 0. 038779215
r 4 = 0. 038778432
r 3 and r 4 are the same to five decimal digits, so we may conclude r 4 is accurate to four
decimal digits, indicating the interest rate was 3.88% (rounded off).
Plugging in the values for P , L, p, g, we have 160z = e 70 z − e − 70 z where z =
4525 T
Therefore we wish to solve f (z) = e
70 z − e
− 70 z − 160 z = 0. f
′ (z) = 70e
70 z
− 70 z − 160.
The Newton’s method formula gives
xn+1 = xn −
e
70 xn − e
− 70 xn − 160 xn
70 e^70 xn^ + 70e−^70 xn^ − 160
We then have
x 0 = 0. 05 T 0 = 49. 05
x 1 = 0. 038387507 T 1 = 63. 8880
x 2 = 0. 028674915 T 2 = 85. 5277
x 3 = 0. 021330325 T 3 = 114. 9771
x 4 = 0. 016467949 T 4 = 148. 9256
x 5 = 0. 01388673 T 5 = 176. 6075
x 6 = 0. 01304748 T 6 = 187. 9673
x 7 = 0. 012958021 T 7 = 189. 2650
x 8 = 0. 012957042 T 8 = 189. 2793
x 9 = 0. 01295704 T 9 = 189. 2793
Since T 8 = T 9 up to four digits of accuracy, we may certainly conclude T = 189.3 is
accurate to one digit (rounded off).
- Since limx→ 0 + f (x) = limx→ 0 − f (x) = 0 = f (0), f is continuous at x = 0. Everywhere
else f is a polynomial, so f is continuous.
We use the definition of derivative to determine if f has a derivative at x = 0. Let
sgn (x) be the function defined by sgn (x) =
1; x > 0
0; x = 0
−1; x < 0
. Then
lim h→ 0
f (0 + h) − f (0)
h
= lim h→ 0
sgn (h)
1 2 (h) 2
h
= lim h→ 0
sgn (h)h
so f ′ (0) = 0.
Everywhere else f has the usual derivative of a polynomial, so we may write
f
′ (x) =
x; x ≥ 0
−x; x < 0
= |x|
As shown in the text, example 2, page 637, f ′ (x) is not differentiable at x = 0 so f (x)
does not have a second derivative at x = 0 but does have a derivative at x = 0.
- (a) Each derivative reduces the powers of x by 1, up until x
0 = 1, whose derivative is
- Taking ten derivatives will therefore give 0.
(b) By the above argument, we need only consider 3x 9
. The eighth derivative is
3 · 9 · 8 · 7 · 6 · 5 · 4 · 3 · 2 · x + 5 · 8 · 7 · 6 · 5 · 4 · 3 · 2
= 3(9!)x + 5(8!) = 1088640x + 201600
(c) g(t) = 3t 27
- 5t 24
- t 18
- 5t 15 − 4 t 9 − t 3
- By the same argument as (a), the 28 th
derivative will be zero.
(d) We need only consider 3t 27 since the other terms will have 26 th derivatives of 0. The
th derivative is 3(27!)t.
- The velocity of the particle is x ′ (t) = Ace ct − Bce −ct and the acceleration is x ′′ (t) =
Ac
2 e
ct
2 e
−ct = c
2 (Ae
ct
−ct ) = c
2 x(t) so the acceleration is given by the constant
c 2 multiplied with the position, i.e., the acceleration is proportional to the position.
