# 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

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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) 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).

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## 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.

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## 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).

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## 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.

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## 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.

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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) 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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## References

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