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The derivatives of complex functions, including the chain rule and product rule. It covers the derivatives of functions like ln(ln(ln x)), e^x, and ax^2 + bx + c. The document also includes examples of finding the derivatives of implicitly defined functions and circles.
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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
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 .
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 .
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 .
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
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
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
d
dx
e
n ln x = e
n ln x
d
dx
n ln x
= x
n
n
x
= nx
n− 1
y = x
n ⇒ ln y = n ln x ⇒
y ′
y
n
x
⇒ y
′ = x
n
n
x
= nx
n− 1
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 )
(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.
′ (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
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
2
; or
1 r
d^2 y dx^2 [ 1+( dy dx )
2
2
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
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
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).
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.
0 = 1, whose derivative is
(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
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.
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.