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Chapter 2 Differentiation (微分)

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(微分)

Hung-Yuan Fan (范洪源)

Department of Mathematics, National Taiwan Normal University, Taiwan

Fall 2018

Hung-Yuan Fan (范洪源), Dep. of Math., NTNU, Taiwan Chapter 2, Calculus B 1/65

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本章預定授課範圍

2.6 Derivatives of Inverse Functions

Hung-Yuan Fan (范洪源), Dep. of Math., NTNU, Taiwan Chapter 2, Calculus B 2/65

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(導數與切線問題)

Hung-Yuan Fan (范洪源), Dep. of Math., NTNU, Taiwan Chapter 2, Calculus B 3/65

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Tangent Line Problem (切線問題)

Let f be a real-valued function defined on I = (a, b) with c∈ I.

What is the slope (斜率) m of the tangent line (切線) to the graph of f at the point P(c, f(c))?

Hung-Yuan Fan (范洪源), Dep. of Math., NTNU, Taiwan Chapter 2, Calculus B 4/65

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示意圖 (承上頁)

Hung-Yuan Fan (范洪源), Dep. of Math., NTNU, Taiwan Chapter 2, Calculus B 5/65

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示意圖 (承上頁)

Hung-Yuan Fan (范洪源), Dep. of Math., NTNU, Taiwan Chapter 2, Calculus B 5/65

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Obsevation

Since the slope of the secant line (割線) passing through P(c, f(c)) and Q(c + ∆x, f(c + ∆x)) is

msec = ∆y

∆x = f(c + ∆x)− f(c)

∆x ,

the slope m of the tangent line at P is determined by considering m = lim

∆x→0msec= lim

∆x→0

f(c + ∆x)− f(c)

∆x .

Hung-Yuan Fan (范洪源), Dep. of Math., NTNU, Taiwan Chapter 2, Calculus B 6/65

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Def (切線的定義)

Ifm = lim

∆x→0

f(c+∆x)−f(c)

∆x = lim

x→c

f(x)−f(c)

x−c , then the line passing through P(c, f(c))with slope m is called a tangent line to the graph of f at P.

Hung-Yuan Fan (范洪源), Dep. of Math., NTNU, Taiwan Chapter 2, Calculus B 7/65

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Vertical Tangent Lines

Def (垂直切線的定義)

A line x = c is called the vertical tangnet line (垂直切線) to the graph of f at the point (c, f(c)) if

lim

∆x→0+

f(c + ∆x)− f(c)

∆x = lim

x→c+

f(x)− f(c)

x− c =±∞

or

lim

∆x→0

f(c + ∆x)− f(c)

∆x = lim

x→c

f(x)− f(c)

x− c =±∞.

Hung-Yuan Fan (范洪源), Dep. of Math., NTNU, Taiwan Chapter 2, Calculus B 8/65

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垂直切線的示意圖

Hung-Yuan Fan (范洪源), Dep. of Math., NTNU, Taiwan Chapter 2, Calculus B 9/65

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Remark (切線方程式)

Equation of the tangent line to the graph of y = f(x) at the point (c, f(c)) is given by

y− f(c) = m(x − c). (Point-Slope Form; 點斜式)

Hung-Yuan Fan (范洪源), Dep. of Math., NTNU, Taiwan Chapter 2, Calculus B 10/65

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Example 4

Let f(x) =√

x for x≥ 0.

(a) Find the tangent line to the graph of f at (4, 2).

(b) Find the slopes of the graph of f at (1, 1) and (0, 0).

Sol: For c > 0, the slope of a tangent line passing through (c, f(c))

is given by

m = lim

∆x→0

f(c + ∆x)− f(c)

∆x = lim

∆x→0

√c + ∆x−√ c

∆x

= lim

∆x→0

∆x

∆x(

c + ∆x +√

c) = 1

√c + 0 +√

c = 1 2

c.

Hung-Yuan Fan (范洪源), Dep. of Math., NTNU, Taiwan Chapter 2, Calculus B 11/65

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(a) At the point (4, 2), m = 1

2

4 = 14 and the equation of the tangent line at (4, 2) is

y− 2 = 1

4(x− 4) or y = 1 4x + 1.

(b) The slope of the graph of f at (1, 1) is m = 1

2

1 = 12. In addition, the slope of a tangent line at (0, 0) is

lim

∆x→0+

f(0 + ∆x)− f(0)

∆x = lim

∆x→0+

1

∆x =∞, and hence the graph of f has a vertical tangent line at x = 0.

Hung-Yuan Fan (范洪源), Dep. of Math., NTNU, Taiwan Chapter 2, Calculus B 12/65

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示意圖 (承上例)

Hung-Yuan Fan (范洪源), Dep. of Math., NTNU, Taiwan Chapter 2, Calculus B 13/65

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Homework

Read Examples 1 and 2 by yourself.

Def (導數或導函數的定義)

(1) The derivative (導數) of f at x∈ dom(f) is f(x) = lim

∆x→0

f(x + ∆x)− f(x)

∆x = lim

z→x

f(z)− f(x) z− x .

(2) f is differentiable (可微分; 簡寫為 diff.) at x∈ dom(f) if the derivative f(x) .

(3) f is diff. on I = (a, b) if it is diff. at each x∈ I.

Hung-Yuan Fan (范洪源), Dep. of Math., NTNU, Taiwan Chapter 2, Calculus B 14/65

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Notes

If S ={x ∈ dom(f) | f (x)∃}, the first derivative f can be regarded as a function defined on S.

For any y = f(x), the derivative is often denoted by f (x) = y(x) = df(x)

dx = dy

dx = Dxf(x) = D1f(x).

Hung-Yuan Fan (范洪源), Dep. of Math., NTNU, Taiwan Chapter 2, Calculus B 15/65

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Recall

Sice the following equality

(a + b)3 = a3+ 3a2b + 3ab2+ b3 holds for any a, b∈ R, we immediately obtain

(x + ∆x)3= x3+ 3x2(∆x) + 3x(∆x)2+ (∆x)3.

Hung-Yuan Fan (范洪源), Dep. of Math., NTNU, Taiwan Chapter 2, Calculus B 17/65

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Differentiability v.s. Continuity

Thm 2.1 (可微分 =⇒ 連續)

Let f be a real-valued function defined on X⊆ R with c∈ X = dom(f). If f is diff. at c, then f is conti. at c.

Hung-Yuan Fan (范洪源), Dep. of Math., NTNU, Taiwan Chapter 2, Calculus B 18/65

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Hung-Yuan Fan (范洪源), Dep. of Math., NTNU, Taiwan Chapter 2, Calculus B 20/65

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Remarks

1 函數的可微分點必定是連續點，詳見 Thm 2.1。

2 但是，函數的連續點不一定是可微分點! 詳見 Example 6 和 Example 7。

Hung-Yuan Fan (范洪源), Dep. of Math., NTNU, Taiwan Chapter 2, Calculus B 21/65

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Basic Differentiation Rules andHigher-Order Derivatives(基本微分法則與高階導數)

Hung-Yuan Fan (范洪源), Dep. of Math., NTNU, Taiwan Chapter 2, Calculus B 22/65

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Thm (Basic Differentiation Rules)

Let f and g be diff. functions of x and c∈ R. Then (1) dxd[c] = 0.

(2) dxd[xn] = n· xn−1 for n∈ R. (3) dxd[f(x)± g(x)] = f(x)± g(x).

(4) dxd[c· f(x)] = c · ˙f(x).

(5) dxd[f(x)g(x)] = f(x)g(x) + f(x)g(x) =(前微)(後不微)+(前不微)(後微). (6) dxdh

f(x) g(x)

i

= f(x)g(x)[g(x)]−f(x)g2 (x) = (子微)(母不微) - (子不微)(母微)

(母)2 .

Hung-Yuan Fan (范洪源), Dep. of Math., NTNU, Taiwan Chapter 2, Calculus B 23/65

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Thm (Derivatives of Elementary Functions)

(1) dxd[sin x] = cos x, dxd[cos x] =− sin x.

(2) dxd[tan x] = sec2x, dxd[cot x] =− csc2x.

(3) dxd[sec x] = sec x tan x, dxd[csc x] =− csc x cot x.

(4) dxd[ex] = ex, dxd[ ln|x| ] = 1x forx̸= 0.

Equivalent Def. of Euler number e

The number e is the base number of an exponential function s.t.

the slope of the tangent line at (0, 1) is 1, i.e., it satisfies

lim

∆x→0

e∆x− e0

∆x = lim

∆x→0

e∆x− 1

∆x = 1.

Hung-Yuan Fan (范洪源), Dep. of Math., NTNU, Taiwan Chapter 2, Calculus B 28/65

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Sketch of the Proof

(1)

d

dx[sin x] = lim

∆x→0

sin(x + ∆x)− sin x

∆x

= lim

∆x→0

sin x cos(∆x) + cos x sin(∆x)− sin x

∆x

= sin x

 lim

∆x→0

cos(∆x)− 1

∆x



+ cos x

 lim

∆x→0

sin(∆x)

∆x



= (sin x)(0) + cos x· 1 = cos x.

(2) dxd[tan x] = dxd hsin x

cos x

i

= cos2cosx+sin2x2x = sec2x.

(3) dxd[sec x] = dxd h 1

cos x

i

= (0) cos x−1·(− sin x)

cos2x =

 1 cos x

sin x cos x



= sec x tan x.

Hung-Yuan Fan (范洪源), Dep. of Math., NTNU, Taiwan Chapter 2, Calculus B 29/65

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Higher-Order Derivatives (高階導函數)

Let y = f(x) be a diff. function of x. Then first derivative: y = f (x) = dydx = dxd[f(x)].

second derivative: y ′′= f ′′(x) = ddx2y2 = dxd22[f(x)]≡ dxd(y).

third derivative: y ′′′ = f′′′(x) =ddx3y3 = dxd33[f(x)]≡ dxd(y′′).

fourth derivative: y(4)= f(4)(x) = ddx4y4 = dxd44[f(x)]≡ dxd(y ′′′).

...

nth derivative: y(n) = f(n)(x) = ddxnyn = dxdnn[f(x)]≡ dxd(y(n−1)) for n∈ N. Here, we denote y(0)= y = f(x).

Hung-Yuan Fan (范洪源), Dep. of Math., NTNU, Taiwan Chapter 2, Calculus B 32/65

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(連鎖律)

Hung-Yuan Fan (范洪源), Dep. of Math., NTNU, Taiwan Chapter 2, Calculus B 33/65

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Thm 2.11 (The Chain Rule; 連鎖律)

If y = f(u) is a diff. function of u and u = g(x) is a diff. function of x, then y = f(g(x)) is a diff. function of x and

dy dx = dy

du du

dx or d dx

h

f(g(x))i

= f (g(x))· g(x).

Hung-Yuan Fan (范洪源), Dep. of Math., NTNU, Taiwan Chapter 2, Calculus B 34/65

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Thm 2.12 (General Power Rule; 廣義冪法則)

If u(x) is a diff. function of x, then y =h

u(x)in

is also a diff.

function for anyn∈ Rwith y = n·h

u(x)in−1

· u(x).

Hung-Yuan Fan (范洪源), Dep. of Math., NTNU, Taiwan Chapter 2, Calculus B 36/65

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Example 8

Find f(x) if f(x) = x p3

x2+ 4 = x (x2+ 4)1/3.

Sol: From the Quotient Rule of Differentiation and the Chain

Rule, we see that

f (x) = 1· (x2+ 4)1/3− x ·(1/3)(x2+ 4)−2/3(2x) (x2+ 4)2/3

= 1

3(x2+ 4)−2/3·h3(x2+ 4)− 2x2 (x2+ 4)2/3

i

= 1

3(x2+ 4)2/3 · x2+ 12

(x2+ 4)2/3 = x2+ 12 3(x2+ 4)4/3.

Hung-Yuan Fan (范洪源), Dep. of Math., NTNU, Taiwan Chapter 2, Calculus B 38/65

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Example 11 (1/2)

Fid the derivative y.

(a) y = cos 3x2 = cos(3x2).

(b) y = (cos 3)x2.

(c) y = cos(3x)2 = cos(9x2).

Sol: By the Chain Rule, we have

(a) y = (− sin 3x2)· (6x) = −6x sin(3x2).

(b) y = (cos 3)· (2x) = (2 cos 3)x.

(c) y = (− sin 9x2)· (18x) = −18x sin(9x2).

Hung-Yuan Fan (范洪源), Dep. of Math., NTNU, Taiwan Chapter 2, Calculus B 39/65

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Example 11 (2/2)

(d) y = cos2x = (cos x)2. (e) y =√

cos x = (cos x)1/2.

Sol: From the Chain Rule, we see that

(d) y = 2(cos x)(− sin x) = −2sixx cos x.

(e) y = 12(cos x)−1/2(− sin x) = − sin x 2

cos x.

Hung-Yuan Fan (范洪源), Dep. of Math., NTNU, Taiwan Chapter 2, Calculus B 40/65

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u

and ln|u|)

If u = u(x) is a diff. function of x, then

(1) dxd[ex] = ex ∀ x ∈ R, dxd[eu] = eu· u.

(2) dxd[ ln x ] = 1x for x > 0, dxd[ ln u ] = uu for u > 0.

(3) dxd[ ln|x| ] = 1x for x̸= 0, dxd[ ln|u| ] = uu for u̸= 0.

Hung-Yuan Fan (范洪源), Dep. of Math., NTNU, Taiwan Chapter 2, Calculus B 41/65

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Example 13

(b) Find y = f(x) if y = f(x) = ln(x2+ 1).

Sol: If we let

u = u(x) = x2+ 1 > 0for x∈ R, then f (x) = d

dx h

ln(x2+ 1) i

= u

u = 2x

x2+ 1 ∀ x ∈ R.

Homework

Hung-Yuan Fan (范洪源), Dep. of Math., NTNU, Taiwan Chapter 2, Calculus B 42/65

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Example 15

Differentiate the following function f(x) = lnx(x2+ 1)2

√2x3− 1 ∀ x ∈ ( 1

3

2,∞).

Sol: Firstly, it follows from Laws of Logarithm that

f(x) = lnh

x(x2+ 1)2

i− lnh

(2x3− 1)1/2i

= ln x + 2 ln(x2+ 1)1

2ln(2x3− 1).

Thus, the first derivative of f is given by f(x) = 1

x+ 2

 2x x2+ 1

1 2

 6x2 2x3− 1



= 1

x+ 4x

x2+ 1 3x2

2x3− 1 ∀ x ∈ ( 1

3

2,∞).

Hung-Yuan Fan (范洪源), Dep. of Math., NTNU, Taiwan Chapter 2, Calculus B 44/65

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x

a

x 的定義)

Let 0 < a̸= 1.

(1) The exponential function to the base a (以 a 為底的指數函 數) is defined by

ax≡ ex ln a ∀ x ∈ R.

(2) The logarithmic function to the base a (以 a 為底的對數函 數) is defined by

logax≡ ln x

ln a ∀ x > 0.

Hung-Yuan Fan (范洪源), Dep. of Math., NTNU, Taiwan Chapter 2, Calculus B 45/65

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u

a

|u| 的微分公式)

Let u = u(x) be a diff. function of x and 0 < a̸= 1.

(1) dxd[ax] = (ln a)ax ∀ x ∈ R, dxd[au] = (ln a)au· u. (2) dxdh

logaxi

= (ln a)x1 for x > 0, dxdh logaui

= (ln a)uu for u > 0.

(3) dxdh

loga|x|i

= (ln a)x1 for x̸= 0. (4) dxdh

loga|u|i

= (ln a)uu foru̸= 0.

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Implicit Differentiation(隱微分)

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Representations of a function

Explicit Form (顯式):

y = f(x),

where x is the independent variable (自變量) and y is the dependent variable (應變量).

Implicit Form (隱式):

F(x, y) = 0,

where we assume that y = y(x) is a diff. function of x.

[Q]: How to fined y

= dydx under the implicit form?

Ans: apply the technique of Implicit Differentiation!

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Example (補充題)

Find dydx if xy = 1.

Sol:

Assume that y = y(x) is a diff. function of x. Thaen x and y satisfy the following equation

F(x, y)≡ xy − 1 = x ·y(x)− 1 = 0.

Differentiating w.r.t. x on both sides of equality gives that d

dx[x· y(x) − 1] = y + x ·dy dx = d

dx(0) = 0.

So, we have dydx = −yx for x̸= 0.

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Example 8 (利用隱微分求切線方程式)

Find the tangent line to the graph of

x2(x2+ y2) = y2 or F(x, y)≡ x4+ x2y2− y2 = 0 at the point (

2/2,√ 2/2).

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Solution of Example 8

Sol:

Assume that y = y(x) is a diff. function of x. It follows from the Chain Rule that

4x3+ 2xy2+ 2x2y·dy

dx− 2y · dy dx = 0.

Thus, the first derivative dydx is given by dy

dx = −2x(2x2+ y2)

2y(x2− 1) = x(2x2+ y2) y(1− x2) . At the given point, m = dydx

x=y=22 = 3/21/2 = 3, and hence the equation of the tangent line at this point is

y−

2 2 = 3

 x−

2 2



or y = 3x−√ 2.

Hung-Yuan Fan (范洪源), Dep. of Math., NTNU, Taiwan Chapter 2, Calculus B 52/65

for each level – (1) the number of harvested nodes (2) the diameter of the largest node (3) the macroscopic suspicion of tumour (4) extracapsular spread. Harvesting and

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Lecture 1: Introduction and overview of supergravity Lecture 2: Conditions for unbroken supersymmetry Lecture 3: BPS black holes and branes. Lecture 4: The LLM bubbling

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11.4 Differentials and the Chain Rules 11.5 Directional Derivatives and Gradients 11.6 Tangent Planes and Normal Lines 11.7 Extrema of Functions of Two Variables 11.8

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