Chapter 3 Derivatives (導數或是導函數)

Chapter 3 Derivatives (導數或是導函數)

Hung-Yuan Fan (范洪源)

Department of Mathematics, National Taiwan Normal University, Taiwan

Fall 2022

本章預定授課範圍

3.1 Tangent Lines and the Derivative at a Point 3.2 The Derivative as a Function

3.3 Differentiation Rules

3.5 Derivatives of Trigonometric Functions 3.6 The Chain Rule

3.7 Implicit Differentiation

3.8 Derivatives of Inverse Functions and Logarithms

3.9 Inverse Trigonometric Functions

Section 3.1

Tangent Lines and the Derivative at a Point

(切線與單點的導數)

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))?

示意圖 (承上頁)

示意圖 (承上頁)

Observation

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 .

Def (切線的定義)

∆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)

Example 1 (使用定義計算切線斜率)

(a) Find the slope of the curve y = ¹_x 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?

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− ¹_a

∆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 ⁻¹_a2 = 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).

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 .

Example 1 的示意圖

Section 3.2

The Derivative as a Function

(導函數)

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.

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).

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−1−_x−1^x 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.

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.

Example 2 的示意圖

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 ∃.

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) ∃.

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.

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).

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)@.

Example 4 的示意圖

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。

Sections 3.3 Differentiation Rules

(微分法則)

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^d[

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 .

Example 1 (冪法則的例子)

(a) _dx^d(x³) = 3x³⁻¹= 3x². (b) d

dx(x^2/3) = 2

3x^(2/3)−1= 2

3x^−1/3= 2 3√³

x. (c) _dx^d(x^√2) =√

2x^√2−1. (d) d

dx(1 x⁴) = d

dx(x⁻⁴) =−4x⁻⁴⁻¹=−4x⁻⁵ = −4 x⁵ .

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

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.

Derivative of e

Thm (自然指數函數的微分公式)

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

dx = f ^′(x) = e^x ∀ x ∈ R.

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

f^′(0) = lim

h→0

e^h− 1 h = 1.

Thus, we immediately obtain

( )

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

x(x²+ e^x) implies y^′ = −1

x² (x²+ e^x) +1

x(2x + e^x) = 1 + (x− 1)e^x x². (b) y = e^2x= e^x· e^x implies

y^′ = (e^x)^′e^x+ e^x(e^x)^′ = e^x· e^x+ e^x· e^x= 2e^2x.

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 e^x

)

= 0− 1 · e^x (e^x)² = −1

e^x =−e^−x.

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.

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_dx^2y2 = _dx^d22[f(x)]≡ _dx^d(y^′).

third derivative: y ^′′′ = f^′′′(x) =^d_dx^3y3 = _dx^d33[f(x)]≡ _dx^d(y^′′).

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

...

nth derivative: y⁽ⁿ⁾ = f⁽ⁿ⁾(x) = ^dny = ^dn [f(x)]≡ ^d(y⁽ⁿ⁻¹⁾)

Section 3.5

Derivatives of Trigonometric Functions

(三角函數的微分法則)

Thm (Derivatives of Elementary Functions)

(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

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

cos x

]

= ^cos2_cos^x+sin2x^2x = sec²x.

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

cos x

]

= ^{(0) cos x}−1·(− sin x)

cos²x =

( 1 cos x

)(sin x cos x

)

= sec x tan x.

Example 1 (與 sin x 微分有關的例子)

(a) y = x²− sin x =⇒ y^′ = 2x− cos x.

(b) y = e^xsin x =⇒ y ^′ = (e^x)^′sin x+e^x(sin x)^′ = e^xsin x+e^xcos x.

(c) y = sin x

x =⇒ y ^′ = (sin x)^′(x)− (sin x)(x)^′

x² = x cos x− sin x

x² .

Example 2 (與 cos x 微分有關的例子)

(a) y = 5e^x+ cos x =⇒ y ^′ = 5e^x− sin x.

(b) y = sin x cos x =⇒

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

(c) y = _{1−sin x}^{cos 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 x}2 = _{1−sin x}¹ .

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.

Section 3.6 The Chain Rule

(連鎖律)

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).

Example 1 (連鎖律的例子)

Find dy

dx if y = (3x²+ 1)².

Sol: Let

dy dx = dy

du ·du

dx = 2u· 6x = 12x · u

= 12x(3x²+ 1) = 36x³+ 12x.

Example 3 (連鎖律的例子)

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

Sol: Let

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

Thm (Derivatives of e

and e

)

(1) _dx^de^x= e^x ∀ x ∈ R.

(2) _dx^de^u= e^u· u^′. (切記!)

Example 4 (利用連鎖律求 e

的微分)

Differentiate the function y = e^{cos x}.

Sol: Consider

y^′ = e^u· u^′ = e^{cos x}· (cos x)^′ =−e^{cos x}sin x.

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

u(x)]n

is also a diff.

function for anyn∈ Rwith y^′ = n·[

u(x)]n−1

· u^′(x).

Example 6 (Power Chain Rule 的例子)

(a) _dx^d(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) _dx^d sin⁵x = _dx^d(sin x)⁵ = 5 sin⁴x cos x.

(d) _dx^de^√3x+1 = e^√3x+1·[

(3x + 1)^1/2 ]_′

= e^√3x+1·¹₂(3x + 1)^−1/2· 3 = ₂^√_3x+1³ e^√3x+1.

Section 3.7

Implicit Differentiation

(隱微分)

Representations of a function

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

Example 2 (求圓上㇐點的斜率)

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

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.

Example 3 (隱微分的例子)

Find dy

dx if y² = x²+ sin(xy).

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).

Example 4 (利用隱微分求二階導數)

Find d²y

dx² if 2x³− 3y² = 8.

Sol:

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.

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.

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 = ^3y_y2^−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.

Section 3.8

Derivatives of Inverse Functions and Logarithms

(反函數與對數函數的微分法則)

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)).

Example 2 (Thm 3 的例子)

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

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.

Thm (Derivatives of ln x and ln u)

(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.

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 ^′= ¹_x for x > 0 and

y^′ = ⁻¹_−x = ¹_x for x < 0, i.e., y^′ = _dx^d ln|x| = ¹_x for x̸= 0.

Def (函數 a

和 log

x 的定義)

(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.

Thm (函數 a

和 log

u 的微分公式)

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

(1) d

dxa^x= (ln a)a^x ∀ x ∈ R, d

dxa^u= (ln a)a^u· u^′. (2) d

dxlog_ax = 1

(ln a)x for x > 0, d

dxlog_au = u ^′

(ln a)u for u > 0.

(3) d

dxlog_a|x| = 1

(ln a)x for x̸= 0, d

dxlog_a|u| = u ^′ (ln a)u for u̸= 0.

Example 5 (求 a

的導數)

(a) d

dx3^x= 3^xln 3.

(b) d

dx3^−x= 3^−x(ln 3)· (−x)^′ =−3^−xln 3.

(c) d

dx3^{sin x}= 3^{sin x}(ln 3)· (sin x)^′ = 3^{sin x}(ln 3) cos x.

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

) .

Example 7 (換底數後求微分)

Find the derivative of y = x^x forx > 0.

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

y = x^x= e^{x ln x}.

From the Chain Rule, we immediately obtain y^′ = e^{x ln x}· (x ln x)^′

= e^{x ln x} (

1· ln x + x ·1 x )

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

Example 7 (利用對數微分法求導數)

Find the derivative of y = x^x 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^′ = x^x(ln x + 1), x > 0.

Section 3.9

Inverse Trigonometric Functions

(反三角函數的微分法則)

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.

Derivative of y = sec

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

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.

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.

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.

∆y 與 dy 的示意圖 (承上頁)

Thank you for your attention!