Chapter 2 Differentiation (微分)

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Chapter 2 Differentiation

(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

2.1 The Derivative and the Tangent Line Problem 2.2 Basic Differentiation Rules and Rates of Change 2.3 Product and Quotient Rules and Higher-Order Derivatives

2.4 The Chain Rule

2.5 Implicit Differentiation

2.6 Derivatives of Inverse Functions

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Section 2.1

The Derivative and the Tangent Line Problem

(導數與切線問題)

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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))?

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示意圖 (承上頁)

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示意圖 (承上頁)

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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

m_sec = ∆y

∆x = f(c + ∆x)− f(c)

∆x ,

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

∆x→0m_sec= lim

∆x→0

f(c + ∆x)− f(c)

∆x .

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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.

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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 =±∞.

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垂直切線的示意圖

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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)

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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.

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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.

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示意圖 (承上例)

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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.

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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).

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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.

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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.

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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。

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Sections 2.2 & 2.3

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

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Thm (Basic Differentiation Rules)

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

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

(4) _dx^d[c· f(x)] = c · ˙f^′(x).

(5) _dx^d[f(x)g(x)] = f^′(x)g(x) + f(x)g^′(x) =^{(前微)(後不微)}+^{(前不微)(後微)}. (6) _dx^dh

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 .

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Thm (Derivatives of Elementary Functions)

(1) _dx^d[sin x] = cos x, _dx^d[cos x] =− sin x.

(2) _dx^d[tan x] = sec²x, _dx^d[cot x] =− csc²x.

(3) _dx^d[sec x] = sec x tan x, _dx^d[csc x] =− csc x cot x.

(4) _dx^d[e^x] = e^x, _dx^d[ ln|x| ] = ¹_x 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.

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Sketch of the Proof

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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) _dx^d[tan x] = _dx^d hsin x

cos x

i

= ^cos²_cos^x+sin2x²^x = sec²x.

(3) _dx^d[sec x] = _dx^d h 1

cos x

i

= ^{(0) cos x}−1·(− sin x)

cos²x =

1 cos x

sin x cos x

= sec x tan x.

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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) = ^dy_dx = _dx^d[f(x)].

second derivative: y ^′′= f ^′′(x) = ^d_dx²^y2 = _dx^d²2[f(x)]≡ _dx^d(y^′).

third derivative: y ^′′′ = f^′′′(x) =^d_dx³^y3 = _dx^d³3[f(x)]≡ _dx^d(y^′′).

fourth derivative: y⁽⁴⁾= f⁽⁴⁾(x) = ^d_dx⁴^y4 = _dx^d⁴4[f(x)]≡ _dx^d(y ^′′′).

...

nth derivative: y⁽ⁿ⁾ = f⁽ⁿ⁾(x) = ^d_dxⁿ^yn = _dx^dⁿn[f(x)]≡ _dx^d(y⁽ⁿ⁻¹⁾) for n∈ N. Here, we denote y⁽⁰⁾= y = f(x).

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Section 2.4 The Chain Rule

(連鎖律)

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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).

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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).

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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.

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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).

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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.

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Thm (Derivatives of e

^u

and ln |u|)

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

(1) _dx^d[e^x] = e^x ∀ x ∈ R, _dx^d[e^u] = e^u· u^′.

(2) _dx^d[ ln x ] = ¹_x for x > 0, _dx^d[ ln u ] = ^u_u^′ for u > 0.

(3) _dx^d[ ln|x| ] = ¹_x for x̸= 0, _dx^d[ ln|u| ] = ^u_u^′ for u̸= 0.

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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

Please read Example 13 of Section 2.4 by yourself.

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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,∞).

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Def (函數 a

^x

和 log

_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

a^x≡ e^{x ln a} ∀ x ∈ R.

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

log_ax≡ ln x

ln a ∀ x > 0.

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Thm (函數 a

^u

和 log

_a

|u| 的微分公式)

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

(1) _dx^d[a^x] = (ln a)a^x ∀ x ∈ R, _dx^d[a^u] = (ln a)a^u· u^′. (2) _dx^dh

log_axi

= _{(ln a)x}¹ for x > 0, _dx^dh log_aui

= _{(ln a)u}^u^′ for u > 0.

(3) _dx^dh

log_a|x|i

= _{(ln a)x}¹ for x̸= 0. (4) _dx^dh

log_a|u|i

= _{(ln a)u}^u^′ foru̸= 0.

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Section 2.5

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

^′ = ^dy_dx under the implicit form?

Ans: apply the technique of Implicit Differentiation!

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Example (補充題)

Find ^dy_dx 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 ^dy_dx = ^−y_x 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 ^dy_dx is given by dy

dx = −2x(2x²+ y²)

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

x=y=^√₂² = ^3/2_1/2 = 3, and hence the equation of the tangent line at this point is

y−

√2 2 = 3

x−

√2 2

or y = 3x−√ 2.