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# Chapter 3 Derivatives (導數或是導函數)

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## Chapter 3Derivatives(導數或是導函數)

### Hung-Yuan Fan (范洪源)

Department of Mathematics, National Taiwan Normal University, Taiwan

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## (切線與單點的導數)

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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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### Observation

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 .

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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.

### 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 1 (使用定義計算切線斜率)

(a) Find the slope of the curve y = 1x at x = a̸= 0. What is the slope at x =−1?

(b) Where does the slope m = −14 ?

(c) What happens to the tangent lime to the curve at (a, 1/a) as a changes?

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## Solution of Example 1 (1/2)

(a) The slope of the curve y = f(x) = 1/x at x = a̸= 0 is

m = lim

∆x→0

f(a + ∆x)− f(a)

∆x = lim

∆x→0 1 a+∆x 1a

∆x

= lim

∆x→0

−∆x

(∆x)a(a + ∆x) = −1

a(a + 0) = −1 a2 < 0.

So, the slope at x =−1 is m = −1/(−1)2=−1.

(b) When −1a2 = m = −14 , we see that a2= 4 or a =±2. That is, the curve has slope m =−1/4 at (−2, −1/2) and (2, 1/2).

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## Example 1 (2/2)

(c) Note that the slope of a tangent line at (a, 1/a) is −1/a2. So, the tangent line becomes increasingly steep (陡峭的) as a→ 0 because

alim→0

−1

a2 =−∞,

and it becomes more and more horizontal as a→ ±∞ because

a→−∞lim

−1

a2 = 0 = lim

a→∞

−1 a2 .

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## (導函數)

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### Def (導函數的定義)

(1) The derivative (導函數) of f w.r.t. x is a function f whose value at x∈ dom(f) is

f (x) = lim

h→0

f(x + h)− f(x)

h = 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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## Example 1 (利用定義求導函數)

Differentiate the fuction f(x) = x− 1 for x ̸= 1.x

### Sol: Applying the Def. of the derivative, we see that

f(x) = lim

z→x

f(z)− f(x) z− x = lim

z→x z

z−1x−1x z− x

= lim

z→x

−(z−x) (z−1)(x−1)

z− x = lim

z→x

−1 (z− 1)(x − 1)

= −1

(x− 1)2 for x̸= 1.

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## Example 2 (利用定義求切線的斜率與方程式)

(a) Find the derivative of f(x) =√

x for x > 0.

(b) Find the tangent line to the curve y = f(x) at x = 4.

### Sol:

(a) For x > 0, the derivative of f w.r.t. x is f(x) = lim

h→0

f(x + h)− f(x)

h = lim

h→0

√x + h−√ x h

= lim

h→0

h h(√

x + h +√

x) = 1

√x + 0 +√

x = 1 2

x. (b) Since the slope of the tangent line at x = 4 is m = 1

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

y− 2 = 1

(x− 4) or y = 1 x + 1.

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## Differentiability on a Closed Interval

### Def (閉區間上的可微分性)

f is diff. onI = [a, b]if it is diff. on(a, b), the right-hand derivative (右導數) at x = a

f+(a) = lim

h→0+

f(a + h)− f(a)

h = lim

x→a+

f(x)− f(a) x− a ∃, and the left-hand derivative (左導數) at x = b

f(b) = lim

h→0

f(b + h)− f(b)

h = lim

x→b

f(x)− f(b) x− b ∃.

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### Remark (內點可微分性的等價條件)

Let f be a real-valued function defined onI = (a, b)with c∈ I. Then f is diff. at x = c⇐⇒f+(c) = f(c) = f (c) .

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## Differentiability v.s. Continuity

### Thm 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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## Proof of Thm 1

Since f is diff. at c∈ dom(f), we know that f (c) = lim

x→c

f(x)− f(c) x− c ∃.

Then the continuity of f at x = c follows, since

xlim→cf(x) =lim

x→c

[f(c) + f(x)− f(c)

x− c · (x − c)]

= f(c) + f(c)· 0 = f(c).

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## Example 4 (Thm 1 的反例)

f(x) =|x| =

{ x, x≥ 0

−x, x < 0 is conti. at x = 0, but not diff. there.

### Sol: Since it is easily seen that lim

x→0f(x) = 0 = f(0), f is conti. at x = 0.

Because the slope of a line y = mx + n is m, we see that f(x) =−1 ∀ x ∈ (−∞, 0) andf(x) = 1 ∀ x ∈ (0, ∞).

But, f is NOT diff. at x = 0 because the one-sided derivatives of f at x = 0 are

f(0) = lim

h→0

f(0+h)−f(0)

h = lim

h→0

|0+h|−|0|

h = lim

h→0

−hh =−1 and f+(0) = lim

h→0+

f(0+h)−f(0)

h = lim

h→0+

|0+h|−|0|

h = lim

h→0+ h

h = 1. In this case,f (0)@.

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## Example 4 的示意圖

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### Remarks

1 函數的可微分點必定是連續點，詳見 Thm 1。

2 但是，函數的連續點不㇐定是可微分點! 詳見 Example 4。

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## (微分法則)

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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) dxd[

f(x) g(x)

]

= 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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## Example 1 (冪法則的例子)

(a) dxd(x3) = 3x3−1= 3x2. (b) d

dx(x2/3) = 2

3x(2/3)−1= 2

3x−1/3= 2 33

x. (c) dxd(x2) =

2x2−1. (d) d

dx(1 x4) = d

dx(x−4) =−4x−4−1=−4x−5 = −4 x5 .

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## Example 3 (計算多項式的微分)

Find the derivative of the polynomial y = x3+43x2− 5x + 1.

### Sol: Applying the differentiation rules (1)–(4), we obtain

dy

dx = d dx

(x3+4

3x2− 5x + 1)

= (x3)+4

3(x2)− 5(x)+ (1)

= 3x3−1+4

3 · 2x2−1+ 5x1−1+ 0 = 3x2+8 3x− 5.

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## Derivative of e

x

### Thm (自然指數函數的微分公式)

The derivative of f(x) = ex is the function f itself, i.e., dy

dx = f (x) = ex ∀ x ∈ R.

### pf: In Section 1.5, we know that f satisfies the following property:

f(0) = lim

h→0

eh− 1 h = 1.

Thus, we immediately obtain

( )

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## Example 6 (乘法法則的例子)

Find the derivative of the given functions.

### Sol: Applying the Product Rule (5), we see that

(a) y = 1

x(x2+ ex) implies y = −1

x2 (x2+ ex) +1

x(2x + ex) = 1 + (x− 1)ex x2. (b) y = e2x= ex· ex implies

y = (ex)ex+ ex(ex) = ex· ex+ ex· ex= 2e2x.

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## Example 7 (商法則的例子)

Applying the Quotient Rule (6), we obtain (a) y = t2− 1

t3+ 1 implies dy

dt = (2t)(t3+ 1)− (t2− 1)(3t2)

(t3+ 1)2 = −t4+ 3t2+ 2t (t3+ 1)2 . (b) y = e−x implies dy

dx = d dx

(1 ex

)

= 0− 1 · ex (ex)2 = −1

ex =−e−x.

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## Example 8 (化簡後使用冪法則求導數)

Find the derivative of y = (x− 1)(x2− 2x)

x4 .

### Sol: We first rewrite y as

y = (x− 1)(x2− 2x)

x4 = x3− 3x2+ 2x

x4 = x−1− 3x−2+ 2x−3. Applying the Power Rule (2), the derivative of y is given by

y=−x−2+ 6x−3− 6x−4 = −1 x2 + 6

x3 6 x4.

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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) = dny = dn [f(x)]≡ d(y(n−1))

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## (三角函數的微分法則)

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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

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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 [sin x

cos x

]

= cos2cosx+sin2x2x = sec2x.

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

cos x

]

= (0) cos x−1·(− sin x)

cos2x =

( 1 cos x

)(sin x cos x

)

= sec x tan x.

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## Example 1 (與 sin x 微分有關的例子)

(a) y = x2− sin x =⇒ y = 2x− cos x.

(b) y = exsin x =⇒ y = (ex)sin x+ex(sin x) = exsin x+excos x.

(c) y = sin x

x =⇒ y = (sin x)(x)− (sin x)(x)

x2 = x cos x− sin x

x2 .

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## Example 2 (與 cos x 微分有關的例子)

(a) y = 5ex+ cos x =⇒ y = 5ex− sin x.

(b) y = sin x cos x =⇒

y = (sin x)(cos x) + (sin x)(cos x) = cos2x− sin2x.

(c) y = 1−sin xcos x =⇒ y = (cos x)(1−sin x)−(cos x)(1−sin x)

(1−sin x)2 =

(− sin x)(1−sin x)−(cos x)(− cos x)

(1−sin x)2 = (1−sin x)1−sin x2 = 1−sin x1 .

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## Example 6 (利用連續性求三角函數的極限)

Since all of six trigonometric functions are conti. at every point in their domains, we immediately obtain

xlim→0

√2 + sec x cos(π− tan x) =

2 + 1 cos(π− 0) =

3

−1 =−√ 3.

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## (連鎖律)

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### Thm 2 (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 [

f(g(x))]

= f (g(x))· g(x).

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## Example 1 (連鎖律的例子)

Find dy

dx if y = (3x2+ 1)2.

### Sol: Let

y = f(u) = u2 and u = g(x) = 3x2+ 1. Applying the Chain Rule, we see that

dy dx = dy

du ·du

dx = 2u· 6x = 12x · u

= 12x(3x2+ 1) = 36x3+ 12x.

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## Example 3 (連鎖律的例子)

Differentiate y = sin(x2+ ex) w.r.t. x.

### Sol: Let

y = f(u) = sin uandu = g(x) = x2+ ex. Then y = f(g(x))· g(x) = cos(x2+ ex)· (2x + ex)

= (2x + ex) cos(x2+ ex) by the Chain Rule.

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x

u

### )

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

(1) dxdex= ex ∀ x ∈ R.

(2) dxdeu= eu· u. (切記!)

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u

## 的微分)

Differentiate the function y = ecos x.

### Sol: Consider

y = f(u) = eu andu = g(x) = cos x. It follows from the Chain Rule that

y = eu· u = ecos x· (cos x) =−ecos xsin x.

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### Thm (Power Chain Rule or General Power Rule; 廣義冪法則)

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

u(x)]n

is also a diff.

function for anyn∈ Rwith y = n·[

u(x)]n−1

· u(x).

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## Example 6 (Power Chain Rule 的例子)

(a) dxd(5x3− x4)7 = 7(5x3− x4)6·(5x3− x4) = 7(15x2− 4x3)(5x3− x4)6.

(b) y = 3x1−2 = (3x− 2)−1 =⇒ y= (−1)(3x − 2)−2· 3 = (3x−3−2)2. (c) dxd sin5x = dxd(sin x)5 = 5 sin4x cos x.

(d) dxde3x+1 = e3x+1·[

(3x + 1)1/2 ]

= e3x+1·12(3x + 1)−1/2· 3 = 23x+13 e3x+1.

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## (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 under the implicit form?

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## Example 2 (求圓上㇐點的斜率)

Find the slope of the circle x2+ y2= 25 at the point (3,−4).

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## Solution of Example 2

Assume that y = y(x) is a diff. function of x. Differetiating w.r.t. x on both sides of the equation, we have

d dx

(

x2+ [y(x)]2 )

= d

dx(25) =⇒ 2x + 2ydy dx = 0.

Then dy

dx = −2x 2y = −x

y and hence the slope of the circle at (3,−4) is m = dy

dx

(3,−4)= −x y

(3,−4)= −3

−4 = 3 4.

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## Example 3 (隱微分的例子)

Find dy

dx if y2 = x2+ sin(xy).

(54)

## Solution of Example 3

Assume that y = y(x) is a diff. function of x. Differentiating the given equation implicitly, we obtain

d

dx[y(x)]2 = d

dx(x2) + d

dx[sin(x· y(x))]

=⇒2ydy

dx = 2x + cos(xy)· (y + xdy dx).

=⇒[2y − x cos(xy)]dy

dx = 2x + y cos(xy).

=⇒dy

dx = 2x + y cos(xy) 2y− x cos(xy).

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## Example 4 (利用隱微分求二階導數)

Find d2y

dx2 if 2x3− 3y2 = 8.

### Sol:

Assume that y = y(x) is a diff. function of x. Then

6x2− 6y · y = 0 =⇒y= 6x2 6y = x2

y. Furthermore, the second derivative is given by

y′′= d dx

(x2 y

)

= 2xy− x2·y y2

= 2x y −x2

y2 ·x2 y = 2x

y −x4

y3 for y̸= 0.

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## Example 5 (利用隱微分求切線與法線)

(a) Show that P(2, 4) lies on the curve x3+ y3− 9xy = 0.

(b) Find the tangent line and normal line (法線) to the curve at the point P.

(57)

## Solution of Example 5

(a) It is true because (2)3+ (4)3− 9(2)(4) = 72 − 72 = 0.

(b) Assume that y = y(x) is a diff. function of x. Then 3x2+ 3y2y− 9(y + xy) = 0 =⇒ y = 9y− 3x2

3y2− 9x = 3y− x2 y2− 3x. Since the slopes of the tangent line and normal line at P(2, 4) are m1 = 3yy2−x−3x2

(2,4)= 45 and m2 = −54 , we know that y = 4 +4

5(x− 2) = 4 5x +12

5 and y = −5 4 x +13

2 are the equations of the tangent line and normal line at P.

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## The Derivative Rule for Inverses

### Thm 3 (反函數的可微分性)

Let f be diff. on an open interval I. If f has an inverse function f−1, then f−1 is diff. at any x∈ range(f) for whichf (f−1(x))̸= 0, with the derivative

(f−1)(x) = 1 f(f−1(x)).

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## Example 2 (Thm 3 的例子)

Let f(x) = x3− 2 for x > 0. Find the value of dfdx−1 at x = 6 = f(2) without finding the formula for f−1(x).

(61)

## Solution of Example 2

Since f : (0,∞) → (−2, ∞) is one-to-one (Check!), its inverse function f−1: (−2, ∞) → (0, ∞) ∃.

Moreover, f−1 is diff. at x = 6 = f(2) by Thm 3 and the derivative of f−1 at x = 6 is given by

(f−1)(6) = 1

f (f−1(6)) = 1

f(2) = 1 12, since f(2) = 3x2

x=2= 12 and f−1(6) = 2.

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### Thm (Derivatives of ln x and ln u)

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

(1) d

dxln x = 1

x for x > 0, d

dxln u = u

u for u > 0.

(2) d

dxln|x| = 1

x for x̸= 0, d

dxln|u| = u

u for u̸= 0.

(63)

## Example 3 (計算 ln u 的微分)

(a) d

dxln(2x) = (2x) 2x = 2

2x = 1

x for x > 0.

(b) d

dxln(x2+ 3) = (x2+ 3)

x2+ 3 = 2x

x2+ 3 forx∈ R. (c) Let y = ln|x| for x ̸= 0. Then y =

{ ln x, x > 0,

ln(−x), x < 0. Thus, it follows from Chain Rule that y = 1x for x > 0 and

y = −1−x = 1x for x < 0, i.e., y = dxd ln|x| = 1x for x̸= 0.

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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.

(65)

u

a

### u 的微分公式)

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

(1) d

dxax= (ln a)ax ∀ x ∈ R, d

dxau= (ln a)au· u. (2) d

dxlogax = 1

(ln a)x for x > 0, d

dxlogau = u

(ln a)u for u > 0.

(3) d

dxloga|x| = 1

(ln a)x for x̸= 0, d

dxloga|u| = u (ln a)u for u̸= 0.

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u

## 的導數)

(a) d

dx3x= 3xln 3.

(b) d

dx3−x= 3−x(ln 3)· (−x) =−3−xln 3.

(c) d

dx3sin x= 3sin x(ln 3)· (sin x) = 3sin x(ln 3) cos x.

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## Example 6 (Logarithmic Differentiation)

Find dy

dx if y = (x2+ 1)(x + 3)1/2

x− 1 forx > 1.

### Sol: Note that y > 0 if x > 1. Then we see that

ln y = ln(x2+ 1) + 1

2ln(x + 3)− ln(x − 1).

Differentiating this equation implicitly, we further obtain y

y = 2x x2+ 1+1

2 · 1

x + 3− 1 x− 1.

=⇒y = (x2+ 1)(x + 3)1/2 x− 1

( 2x

x2+ 1+ 1

2x + 6− 1 x− 1

) .

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## Example 7 (換底數後求微分)

Find the derivative of y = xx forx > 0.

### Sol: For x > 0, we can rewrite y as

y = xx= ex ln x.

From the Chain Rule, we immediately obtain y = ex ln x· (x ln x)

= ex ln x (

1· ln x + x ·1 x )

= xx(ln x + 1), x > 0.

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## Example 7 (利用對數微分法求導數)

Find the derivative of y = xx forx > 0.

### Sol: Note that y > 0 if x > 0 and assume that y = y(x) is a diff.

function of x. Using the logarithmic differentiation forln y = x ln x, we see that

y

y = 1· ln x + x ·1 x

=⇒y = xx(ln x + 1), x > 0.

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## (反三角函數的微分法則)

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### Thm (反三角函數的微分公式)

Let u = u(x) be a diff. function of x. Then (1) d

dxsin−1u = u

1− u2, d

dxcos−1u = √−u

1− u2 for |u| < 1. (2) d

dxtan−1u = u

1 + u2, d

dxcot−1u = −u 1 + u2. (3) d

dxsec−1u = u

|u|√

u2− 1, d

dxcsc−1u = −u

|u|√

u2− 1 for

|u| > 1.

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## Derivative of y = sec

−1

### x

If y = sec−1x for|x| > 1, then sec y = xfory∈ [0,π2)∪ (π2, π].

Using the implicit differentiation, we see that (sec y tan y)y = 1 =⇒ y = 1

sec y tan y. Since sec y = x, tan y =±

sec2y− 1 = ±√

x2− 1 and hence y =

{ 1

x

x2−1, x > 1,

1 (−x)

x2−1, x <−1 or y = 1

|x|√ x2− 1

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## Examples 2 and 3

Applying the Chain Rule, we obtain the following derivatives.

(a) d

dxsin−1(x2) = (x2)

√1− (x2)2 = 2x

1− x4 for −1 < x < 1.

(b) d

dxsec−1(5x4) = 20x3

|5x4|

(5x4)2− 1 = 4 x√

25x8− 1 for 25x8 > 1 orx > 1

8

25.

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## Differential 的定義

### Def of Differentials

Let f be diff. on an open interval I with x∈ I.

(1) The differential dx is an independent variable of any nonzero real number.

(2) The differential of y = f(x) is defined by dy = f(x)dx.

For example, if y = f(x) = tan(x3), then the differential dy is dy = f (x) dx = 3x2sec2(x3) dx.

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## Linear Approximation of a Function

### Thm (利用 dy 估計 ∆y)

If dx = ∆x≈ 0 is sufficiently small, then (1) f(x + ∆x)− f(x) =∆y≈ dy= f (x)dx.

(2) f(x + dx) = f(x + ∆x)≈ f(x) + dy = f(x) + f(x)dx.

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## Thank you for your attention!

Project 1.3 Use parametric bootstrap and nonparametric bootstrap to approximate the dis- tribution of median based on a data with sam- ple size 20 from a standard normal

Department of Mathematics, National Taiwan Normal University, Taiwan..

Department of Mathematics, National Taiwan Normal University, Taiwan..

Feng-Jui Hsieh (Department of Mathematics, National Taiwan Normal University) Hak-Ping Tam (Graduate Institute of Science Education,. National Taiwan

2 Department of Educational Psychology and Counseling / Institute for Research Excellence in Learning Science, National Taiwan Normal University. Research on embodied cognition

Department of Mathematics, National Taiwan Normal University,

4.1 Extreme Values of Functions on Closed Intervals 4.2 The Mean Value Theorem.. 4.3 Monotonic Functions and the First Derivative Test 4.4 Concavity and

Department of Mathematics, National Taiwan Normal University, Taiwan..