**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**→0**m** _{sec}*= lim

*∆x**→0*

*f(c + ∆x)− f(c)*

*∆x* *.*

**Def (切線的定義)**

If*m = 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)*

## 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 =* ^{−1}_{4} ?

(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*
*a*^{2} *< 0.*

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

(b) When ^{−1}* _{a}*2

*= m =*

^{−1}_{4}

*, we see that a*

^{2}

*= 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/a*^{2}. So,
the tangent line becomes increasingly steep (陡峭的) as
*a→ 0 because*

*a*lim*→0*

*−1*

*a*^{2} =*−∞,*

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

*a**→−∞*lim

*−1*

*a*^{2} = 0 = lim

*a**→∞*

*−1*
*a*^{2} *.*

## 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* *= D**x**f(x) = D*_{1}*f(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 = ^{1}_{4},
*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)**

**⇒ 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*

*x*lim*→c**f(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**→0**f(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)* and*f*^{′}*(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*^{−}

*−h**h* =*−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}*[x*^{n}*] = n· x*^{n}* ^{−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)g}*2

^{′}*= (子微)(⺟不微) - (子不微)(⺟微)*

^{(x)}(⺟)2 .

## Example 1 (冪法則的例子)

(a) _{dx}^{d}*(x*^{3}*) = 3x*^{3}^{−1}*= 3x*^{2}.
(b) *d*

*dx(x*^{2/3}) = 2

3*x*^{(2/3)}* ^{−1}*= 2

3*x** ^{−1/3}*= 2
3

*√*

^{3}

*x*.
(c) _{dx}^{d}*(x*^{√}^{2}) =*√*

*2x*^{√}^{2}* ^{−1}*.
(d)

*d*

*dx*(1
*x*^{4}) = *d*

*dx(x** ^{−4}*) =

*−4x*

*=*

^{−4−1}*−4x*

*=*

^{−5}*−4*

*x*

^{5}.

## Example 3 (計算多項式的微分)

*Find the derivative of the polynomial y = x*^{3}+^{4}_{3}*x*^{2}*− 5x + 1.*

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

*dy*

*dx* = *d*
*dx*

(*x*^{3}+4

3*x*^{2}*− 5x + 1*)

= (x^{3})* ^{′}*+4

3(x^{2})^{′}*− 5(x)** ^{′}*+ (1)

^{′}*= 3x*^{3}* ^{−1}*+4

3 *· 2x*^{2}^{−1}*+ 5x*^{1}^{−1}*+ 0 = 3x*^{2}+8
3*x− 5.*

*Derivative of e*

^{x}**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:**

**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**

(a) *y =*1

*x(x*^{2}*+ e** ^{x}*) implies

*y*

*=*

^{′}*−1*

*x*^{2} *(x*^{2}*+ e** ^{x}*) +1

*x(2x + e*^{x}*) = 1 + (x− 1)e*^{x}*x*^{2}.
(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 =* *t*^{2}*− 1*

*t*^{3}+ 1 implies
*dy*

*dt* = *(2t)(t*^{3}+ 1)*− (t*^{2}*− 1)(3t*^{2})

*(t*^{3}+ 1)^{2} = *−t*^{4}*+ 3t*^{2}*+ 2t*
*(t*^{3}+ 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)(x*^{2}*− 2x)*

*x*^{4} .

**Sol: We first rewrite y as**

**Sol: We first rewrite y as**

*y =* *(x− 1)(x*^{2}*− 2x)*

*x*^{4} = *x*^{3}*− 3x*^{2}*+ 2x*

*x*^{4} *= 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*

*x*

^{2}+ 6

*x*^{3} *−* 6
*x*^{4}*.*

## 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}^{2}* ^{y}*2 =

_{dx}

^{d}^{2}2

*[f(x)]≡*

_{dx}

^{d}*(y*

*).*

^{′}*third derivative: y* ^{′′′}*= f*^{′′′}*(x) =*^{d}_{dx}^{3}* ^{y}*3 =

_{dx}

^{d}^{3}3

*[f(x)]≡*

_{dx}

^{d}*(y*

*).*

^{′′}*fourth derivative: y*^{(4)}*= f*^{(4)}*(x) =* ^{d}_{dx}^{4}* ^{y}*4 =

_{dx}

^{d}^{4}4

*[f(x)]≡*

_{dx}

^{d}*(y*

*).*

^{′′′}...

*nth derivative: y*^{(n)}*= f*^{(n)}*(x) =* ^{d}^{n}* ^{y}* =

^{d}

^{n}*[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)**

(1)

_{dx}

^{d}*[sin x] = cos x,*

_{dx}

^{d}*[cos x] =− sin x.*

(2) _{dx}^{d}*[tan x] = sec*^{2}*x,* _{dx}^{d}*[cot x] =− csc*^{2}*x.*

(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}*[ ln*

^{d}*|x| ] =*

^{1}

*for*

_{x}*x̸= 0*.

**Equivalent Def. of Euler number e**

**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*

]

= ^{cos}^{2}_{cos}* ^{x+sin}*2

*x*

^{2}

*= sec*

^{x}^{2}

*x.*

(3) _{dx}^{d}*[sec x] =* _{dx}* ^{d}*
[ 1

*cos x*

]

= ^{(0) cos x}*−1·(− sin x)*

cos^{2}*x* =

( 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 = x*^{2}*− sin x =⇒ y*^{′}*= 2x− cos x.*

(b) *y = e*^{x}*sin x =⇒ y* ^{′}*= (e** ^{x}*)

^{′}*sin x+e*

^{x}*(sin x)*

^{′}*= e*

^{x}*sin x+e*

^{x}*cos x.*

(c) *y =* *sin x*

*x* =*⇒ y* * ^{′}* =

*(sin x)*

^{′}*(x)− (sin x)(x)*

^{′}*x*^{2} = *x cos x− sin x*

*x*^{2} .

*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)** ^{′}* = cos

^{2}

*x− sin*

^{2}

*x.*

(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

*x*lim*→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 = (3x*^{2}+ 1)^{2}.

**Sol: Let**

*y = f(u) = u*

^{2}and

*u = g(x) = 3x*

^{2}+ 1. Applying the Chain Rule, we see that

*dy*
*dx* = *dy*

*du* *·du*

*dx* *= 2u· 6x = 12x · u*

*= 12x(3x*^{2}*+ 1) = 36x*^{3}*+ 12x.*

## Example 3 (連鎖律的例子)

*Differentiate y = sin(x*^{2}*+ e*^{x}*) w.r.t. x.*

**Sol: Let**

*y = f(u) = sin u*and

*u = g(x) = x*

^{2}

*+ e*

*. Then*

^{x}*y*

^{′}*= f*

^{′}*(g(x))· g*

^{′}*(x) = cos(x*

^{2}

*+ e*

*)*

^{x}*· (2x + e*

*)*

^{x}*= (2x + e*^{x}*) cos(x*^{2}*+ e** ^{x}*)
by the Chain Rule.

**Thm (Derivatives of e**

**Thm (Derivatives of e**

^{x}**and e**

**and e**

^{u}**)**

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

(1) _{dx}^{d}*e*^{x}*= e*^{x}*∀ x ∈ R.*

(2) _{dx}^{d}*e*^{u}*= e*^{u}*· u** ^{′}*. (切記!)

*Example 4 (利用連鎖律求 e*

^{u}## 的微分)

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

**Sol: Consider**

*y = f(u) = e*

*and*

^{u}*u = g(x) = cos x. It follows from*the Chain Rule that

*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)**

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

*u(x)*]*n*

is also a diff.

function for any*n∈ R*with
*y*^{′}*= n·*[

*u(x)*]*n**−1*

*· u*^{′}*(x).*

## Example 6 (Power Chain Rule 的例子)

(a) _{dx}^{d}*(5x*^{3}*− x*^{4})^{7} *= 7(5x*^{3}*− x*^{4})^{6}*·(5x*^{3}*− x*^{4})* ^{′}* =

*7(15x*

^{2}

*− 4x*

^{3}

*)(5x*

^{3}

*− x*

^{4})

^{6}.

(b) *y =* _{3x}^{1}_{−2}*= (3x− 2)** ^{−1}* =

*⇒ y*

*= (*

^{′}*−1)(3x − 2)*

^{−2}*· 3 =*

_{(3x}

^{−3}*2. (c)*

_{−2)}

_{dx}*sin*

^{d}^{5}

*x =*

_{dx}

^{d}*(sin x)*

^{5}= 5 sin

^{4}

*x cos x.*

(d) _{dx}^{d}*e*^{√}^{3x+1}*= e*^{√}^{3x+1}*·*[

*(3x + 1)*^{1/2}
]_{′}

=
*e*^{√}^{3x+1}*·*^{1}_{2}*(3x + 1)*^{−1/2}*· 3 =* _{2}^{√}_{3x+1}^{3} *e*^{√}* ^{3x+1}*.

**Section 3.7**

**Implicit Differentiation**

**(隱微分)**

**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**

**[Q]: How to fined y**

*=*

^{′}*under the implicit form?*

^{dy}## Example 2 (求圓上㇐點的斜率)

*Find the slope of the circle x*^{2}*+ y*^{2}*= 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*

(

*x*^{2}*+ [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 y*^{2} *= x*^{2}*+ 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(x*^{2}) + *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 *d*^{2}*y*

*dx*^{2} *if 2x*^{3}*− 3y*^{2} = 8.

**Sol:**

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

*6x*^{2}*− 6y · y* * ^{′}* = 0 =

*⇒y*

*=*

^{′}*6x*

^{2}

*6y*=

*x*

^{2}

*y.*
Furthermore, the second derivative is given by

*y** ^{′′}*=

*d*

*dx*

*(x*^{2}
*y*

)

= *2xy− x*^{2}*·y*^{′}*y*^{2}

= *2x*
*y* *−x*^{2}

*y*^{2} *·x*^{2}
*y* = *2x*

*y* *−x*^{4}

*y*^{3} *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 x*^{3}*+ y*^{3}*− 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*
*3x*^{2}*+ 3y*^{2}*y*^{′}*− 9(y + xy** ^{′}*) = 0 =

*⇒ y*

*=*

^{′}*9y− 3x*

^{2}

*3y*^{2}*− 9x* = *3y− x*^{2}
*y*^{2}*− 3x.*
*Since the slopes of the tangent line and normal line at P(2, 4)*
*are m*_{1} = ^{3y}* _{y}*2

^{−x}*−3x*

^{2}

*(2,4)*= ^{4}_{5} *and m*_{2} = ^{−5}_{4} , we know that
*y = 4 +*4

5*(x− 2) =* 4
5*x +*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*

^{′}*(x))*

^{−1}*.*

## Example 2 (Thm 3 的例子)

*Let f(x) = x*^{3}*− 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) = 3x*

^{2}

*x=2**= 12 and f** ^{−1}*(6) = 2.

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

**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*

*dx*ln*|x| =* 1

*x* *for x̸= 0,* *d*

*dx*ln*|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(x*^{2}+ 3) = *(x*^{2}+ 3)^{′}

*x*^{2}+ 3 = *2x*

*x*^{2}+ 3 for*x∈ 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}*ln*

^{d}*|x| =*

^{1}

_{x}*for x̸= 0.*

**Def (函數 a**

**Def (函數 a**

^{x}**和 log**

_{a}**x 的定義)**

**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_{a}*x :=* *ln x*

*ln a* *∀ x > 0.*

**Thm (函數 a**

**Thm (函數 a**

^{u}**和 log**

_{a}**u 的微分公式)**

**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*

*dx*log_{a}*x =* 1

*(ln a)x* *for x > 0,* *d*

*dx*log_{a}*u =* *u* ^{′}

*(ln a)u* *for u > 0.*

(3) *d*

*dx*log_{a}*|x| =* 1

*(ln a)x* for *x̸= 0,* *d*

*dx*log_{a}*|u| =* *u* ^{′}*(ln a)u* for
*u̸= 0*.

*Example 5 (求 a*

^{u}## 的導數)

(a) *d*

*dx*3* ^{x}*= 3

*ln 3.*

^{x}(b) *d*

*dx*3* ^{−x}*= 3

*(ln 3)*

^{−x}*· (−x)*

*=*

^{′}*−3*

*ln 3.*

^{−x}(c) *d*

*dx*3* ^{sin x}*= 3

*(ln 3)*

^{sin x}*· (sin x)*

*= 3*

^{′}

^{sin x}*(ln 3) cos x.*

## Example 6 (Logarithmic Differentiation)

Find *dy*

*dx* *if y =* *(x*^{2}*+ 1)(x + 3)*^{1/2}

*x− 1* for*x > 1.*

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

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

*ln y = ln(x*^{2}+ 1) + 1

2*ln(x + 3)− ln(x − 1).*

Differentiating this equation implicitly, we further obtain
*y*^{′}

*y* = *2x*
*x*^{2}+ 1+1

2 *·* 1

*x + 3−* 1
*x− 1.*

=*⇒y** ^{′}* =

*(x*

^{2}

*+ 1)(x + 3)*

^{1/2}

*x− 1*

( *2x*

*x*^{2}+ 1+ 1

*2x + 6−* 1
*x− 1*

)
*.*

## Example 7 (換底數後求微分)

*Find the derivative of y = x** ^{x}* for

*x > 0.*

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

**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}* for

*x > 0.*

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

**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*

*dx*sin^{−1}*u =* *u* ^{′}

*√*1*− u*^{2}*,* *d*

*dx*cos^{−1}*u =* *√−u*^{′}

1*− u*^{2} for *|u| < 1*.
(2) *d*

*dx*tan^{−1}*u =* *u* ^{′}

*1 + u*^{2}*,* *d*

*dx*cot^{−1}*u =* *−u*^{′}*1 + u*^{2}.
(3) *d*

*dx*sec^{−1}*u =* *u* ^{′}

*|u|√*

*u*^{2}*− 1,* *d*

*dx*csc^{−1}*u =* *−u*^{′}

*|u|√*

*u*^{2}*− 1* for

*|u| > 1*.

*Derivative of y = sec*

^{−1}*x*

*If y = sec*^{−1}*x for|x| > 1, then* *sec y = x*for*y∈ [0,*^{π}_{2})*∪ (*^{π}_{2}*, 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 =±*√

sec^{2}*y− 1 = ±√*

*x*^{2}*− 1 and hence*
*y** ^{′}* =

{ _{1}

*x**√*

*x*^{2}*−1**,* *x > 1,*

1
(*−x)**√*

*x*^{2}*−1**,* *x <−1* *or y* * ^{′}* = 1

*|x|√*
*x*^{2}*− 1*

## Examples 2 and 3

Applying the Chain Rule, we obtain the following derivatives.

(a) *d*

*dx*sin^{−1}*(x*^{2}) = *(x*^{2})^{′}

√1*− (x*^{2})^{2} = *2x*

*√*1*− x*^{4} for *−1 < x < 1*.

(b) *d*

*dx*sec^{−1}*(5x*^{4}) = *20x*^{3}

*|5x*^{4}*|*√

*(5x*^{4})^{2}*− 1* = 4
*x√*

*25x*^{8}*− 1* for
*25x*^{8} *> 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(x*^{3}*), then the differential dy is*
*dy = f* ^{′}*(x) dx = 3x*^{2}sec^{2}*(x*^{3}*) dx.*

## Linear Approximation of a Function

**Thm (利用 dy 估計 ∆y)**

**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.*