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Calculus: Derivatives of Complex Functions, Study notes of Calculus

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.

What you will learn

  • What is the derivative of ln(ln(ln x))?
  • What is the derivative of ax^2 + bx + c?
  • What is the derivative of a circle?
  • How do you find the derivative of implicitly defined functions?
  • How do you find the derivative of e^x?

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Solutions to Supplementary Questions for HP Chapter 12
1. 1) d
dx (x8ln x) = 8x7ln x+x8
x= 8x7ln x+x7
2) d2
dx2(x8ln x) = d
dx (8x7ln x+x7) = 8·7x6ln x+8x7
x+ 7x6
= 8 ·7x6ln x+ 8x6+ 7x6
Now note that the power of xin all terms are the same, namely 8kwhen we calculate
dk
dxk, so at the 8th derivative, the power of xis 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 9th derivative, the only
term that is nonzero is the first term that includes ln x.
8) d8
dx8(x8ln x) = 8 ·7·6·5·4·3·2·1·x0ln x+ constants
9) d9
dx9(x8ln x) = 8·7·6·5·4·3·2·1·
x=8!
x
2.
d
dx[ln(ln(ln(ln x)))]
=1
ln(ln(ln x)) d
dx ln(ln(ln x))
=1
ln(ln(ln x)) 1
ln(ln x)
d
dx (ln(ln x))
=1
ln(ln(ln x)) 1
ln(ln x)1
ln x
d
dx (ln x)
=1
ln(ln(ln x)) 1
ln(ln x)1
ln x
1
x
3. (i) d
dx x(aa)=aax(aa1)
(ii) Using the Chain Rule, d
dx af(x)=af(x)ln a(f0(x)), and hence, d
dx a(xa)
=a(xa)ln a(axa1)
(iii) Similarly, d
dx a(ax)=a(ax)ln a(d
dx ax) = a(ax)ln a(axln a) = a(ax)ax(ln a)2.
So d
dx x(aa)+a(xa)+a(ax)=aax(aa1) +aln aa(xa)xa1+a(ax)ax(ln a)2.
1
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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

  • 7x 6

= 8 · 7 x 6 ln x + 8x 6

  • 7x 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

  1. (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)

  • a

(xa)

  • a

(ax)

= a

a x

(aa−1)

  • a ln a

a

(xa)

x

a− 1

  • a

(ax) a

x (ln a)

2 .

  1. (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 .

  1. 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 .

  1. (a)

f (x) = ln(x) g(x) = ax

2

  • bx + c

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

  • 2x − 3 2

(b)

f (x) = f ′ (x) = f ′′ (x) = e x

g(x) = ax

2

  • bx + c

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

  1. 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

  1. (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

  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.

  1. (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

  1. 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

  1. Differentiating implicitly the first equation with respect to z,

y + z

dy

dz

  • y 3 z

2

  • z

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

  • 3z

2

  • 6xz

dz

dx

= 6x

2 y + 2x

3 dy

dx

(x

2

  • 6xz)

dz

dx

= 6x

2 y − 2 xz − 3 z

2

  • 2x

3 dy

dx

dz

dx

6 x 2 y − 2 xz − 3 z 2

  • 2x 3 dy dx

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

  • 2x 3 dy dx x^2 + 6xz

Solving for

dy dx we have

(z + z

3 )(x

2

  • 6xz)

dy

dx

= −y(1 + 3z

2 )(6x

2 y − 2 xz − 3 z

2

  • 2x

3 dy

dx

(z + z

3 )(x

2

  • 6xz)

−y(1 + 3z 2 )

dy

dx

= 6x

2 y − 2 xz − 3 z

2

  • 2x

3 dy

dx [

2 x

3

(z + z 3 )(x 2

  • 6xz)

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

  • (z + z 3 )(x 2
  • 6xz)
  1. 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

  • 2 − ln(x + 1)

4 − ln(x

2

2

= x +

ln(x

5

    1. − 4 ln(x + 1) − 2 ln(x

2

Differentiating both sides with respect to x gives

d ln y

dx

d

dx

[

x +

ln(x

5

    1. − 4 ln(x + 1) − 2 ln(x

2

]

y

y

d dx (x 5

2(x 5

d dx (x + 1))

x + 1

d dx (x 2

  • 3))

x 2

  • 3

y ′

y

5 x 4

2(x 5

x + 1

4 x

x 2

  • 3

Substituting back in y =

ex

√ x^5 + (x+1)^4 (x^2 +3)^2 and solving for y ′ gives

y

e

x

x 5

  • 2

(x + 1)^4 (x^2 + 3)^2

5 x

4

2(x^5 + 2)

x + 1

4 x

x^2 + 3

  1. (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

  • ln x) + x x− 1

]

(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).

  1. Plugging in the values for P , L, p, g, we have 160z = e 70 z − e − 70 z where z =

  2. 4525 T

Therefore we wish to solve f (z) = e

70 z − e

− 70 z − 160 z = 0. f

′ (z) = 70e

70 z

  • 70e

− 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).

  1. 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.

  1. (a) Each derivative reduces the powers of x by 1, up until x

0 = 1, whose derivative is

  1. Taking ten derivatives will therefore give 0.

(b) By the above argument, we need only consider 3x 9

  • 5x 8

. 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
    1. 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.

  1. The velocity of the particle is x ′ (t) = Ace ct − Bce −ct and the acceleration is x ′′ (t) =

Ac

2 e

ct

  • Bc

2 e

−ct = c

2 (Ae

ct

  • Be

−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.