## 導函數

### 牙醫系第 2 題

Use the definition of the derivative  \displaystyle g' \left( t \right) = \lim_{h \to 0} \frac {g \left( t + h \right) - g \left( t \right)} h $\displaystyle g' \left( t \right) = \lim_{h \to 0} \frac {g \left( t + h \right) - g \left( t \right)} h$ given that to find the derivative of g(t) = 1/t2.


\begin{align*}
g' \left ( t \right)
&= \lim_{h \to 0} \frac {g \left( t + h \right) - g \left( t \right)} h \\
&= \lim_{h \to 0} \frac {\left( t + h \right)^{-2} - t^{-2}} h \\
&= \lim_{h \to 0} \frac {t^2 - \left( t + h \right)^2} {ht^2 \left( t + h \right)^2} \\
&= -\lim_{h \to 0} \frac {h \left( 2t + h \right)} {ht^2 \left( t + h \right)^2}
\end{align*}


 \frac {2t + h} {t \left( t + h \right)} = \frac1t + \frac 1 {t + h} 


\begin{align*}
g' \left( t \right)
&= -\frac1t \lim_{h \to 0} \frac 1 {t + h} \left( \frac1t + \frac 1 {t + h}\right) \\
&= -\frac1t \left( \frac1t + \lim_{h \to 0} \frac 1 {t + h}\right) \lim_{h \to 0} \frac 1 {t + h}
\end{align*}


\lim_{h \to 0} \frac 1 {t + h} = \frac1t 


\begin{align*}
g' \left( t \right)
&= -\frac1t \left( \frac1t + \frac1t \right) \frac1t \\
&= -\frac 2 {t^3}
\end{align*}


### 牙醫系第 3 題

Given  \displaystyle f \left( x \right) = \frac {x \sin x} {1 + \cos x} $\displaystyle f \left( x \right) = \frac {x \sin x} {1 + \cos x}$, find f′(x).

 f' \left( x \right) = \frac {\sin x + x \cos x} {1 + \cos x} + \frac {x \sin^2 x} {\left( 1 + \cos x \right)^2} 

## 導函數

### 牙醫系第 2 題

Use the definition of the derivative  \displaystyle f' \left( x \right) = \lim_{h \to 0} \frac {f \left( x + h \right) - f \left( x \right)} h $\displaystyle f' \left( x \right) = \lim_{h \to 0} \frac {f \left( x + h \right) - f \left( x \right)} h$ given that to find the derivative of f(x) = cos(x).

### 其他系第 2 題

given that to find the derivative of f(x) = x3 + 7x.

 f' \left( x \right) = \lim_{h \to 0} \frac {\left( x + h \right)^3 + 7 \left( x + h \right) - \left( x^3 + 7x \right)} h 

(x + h)3x3 = 3x2h + 3xh2 + h3

(x + h)3 + 7(x + h) − (x3 + 7x) = (3x2 + 7)h + 3xh2 + h3

### 牙醫系第 5 題

Logarithmic differentiation. Let $\displaystyle f \left( x \right) = \frac {\left( x^3 - 1 \right)^4 \sqrt{3x - 1}} {x^2 + 4}$ and find f′(x)

## 導數

### 第 3 題

Given $\displaystyle f \left( x \right) = \frac {x^2 \left( 1-x \right)^3} {1+x}$, find f′(3).

## 切線與法線

### 牙醫系第 4 題

Find the equation of the tangent line of the family of curves y3 + xy = sec(xy2) + c at (1, 1).

##### Hint

3y2y′ + y + xy′ = sec(xy2) tan(xy2) (y2 + 2xyy′)

(3y2 + x) y′ + y = sec(xy2) tan(xy2) (y2 + 2xyy′)

4y′ + 1 = sec 1 tan 1 (1 + 2y′)

cos2 1 (4y′ + 1) = sin 1 (2y′ + 1)

cos2 1 − sin 1= (2 sin 1 − 4 cos2 1) y

### 其他系第 4 題

Find the equation of the normal line of the family of curves y3 + xy = xy2 + c at (3/16, 2).

3y2y′ + y + xy′ = y2 + 2xyy

(3y2 − 2xy + x) y′ = y2y

(12 − 9/16) y′ = 2

y′ = 32/183

## 應用題

### 牙醫系第 6 題

Electron speed. An electron with a whose mass of is 9.1 × 10-31 kg and a charge of is -1.6 × 10-19 C travels in a circular path with no loss of energy in a magnetic field of 0.05 T that is orthogonal to the path of the electron. If the radius of the path is 0.002 m, what is the speed of the electron?

mv2 / r = q |v × B|

|v × B| = vB

mv2 / r = qvB

v = qrB / m

v = (1.6 × 10-19 C) (0.002 m) (0.05 T) / (9.1 × 10-31 kg)

v ≈ 1.76 × 107 m/s

### 其他系第 6 題

In the late 1830s, French physiologist Jean Poiseuille discovered the formula we use today to predict how much the radius of a particular clogged artery decreases the normal volume of flow. His formula,

V = kr4

says that volume of fluid flowing through a small pipe or tube in a unit of time at a fixed pressure is a constant times the fourth power of the tube’s radius r. How dose does a 10% decrease in r affect V?

#### 官方解法

Δr = 0.05r

ΔV ≈ 0.2kr4 = 0.2V

V 增加 20%

#### 直接解法

V0 = kr04

r = 1.05r0

V = k (1.05r0)4 = 1.054 V0

## 線性回歸

### 其他系第 1 題

In a study of 12 subjects, a clinical researcher obtained these data relating the ages of a subject pool (in years) to that of their systolic blood pressure (in mmHg):

No. Age (years) BPsystolic (mmHg)
125120
237139
320118
449150
527125
623115
734146
856170
924128
1036137
1122126
1242148

x y x2 xy
25120 6253000
3713913695143
20118 4002360
4915024017350
27125 7293375
23115 5292645
3414611564964
5617031369520
24128 5763072
3613712964932
22126 4842772
4214817646216
39516221446555349

## 有理函數的積分

c積分常數，則

### 例題 1

#### 傳統作法

A 算是最容易算出來的。這是二種作法的共同步驟，所以不分析成本。

• 展開 57 (x2 + 4)2 + x2 + 16x
• −57x4 − 287x2 + 2704x − 912 除以 x − 3
• −57x3 − 171x2 − 800x + 304 除以 x2 + 4

θ = arctan(x/2)，使得 x = 2 tan θ

#### Hermite reduction

U = x − 3，V = x2 + 4。我們開始進行輾轉相除法。取 UV′ 除以 V

UV′ = 2V − 6x − 8

6x + 8 = 2VUV(*)

18V = (3x − 4) (2VUV′) + 104

104 = (3x − 4) UV′ − (6x − 26) V(†)

### 例題 2

Hermite reduction 專門處理這種題目。設 U = (x − 1) (x − 2)2V = x − 3。

UV′ = (x2 − 2x + 2) V + 2

UV′ = 2V + 3x − 7

9V = (3x − 8) (UV′ − 2V) − 2

2 = (3x − 8) UV′ + (7 − 6x) V

369x − 664 = 123 (3x − 7) + 197

## 無理函數的積分

y = arsinh x. 此時 sinh y = x, $\cosh y = \sqrt {x^2 + 1}$, dx = cosh y dy.

c 為積分常數。

## 線性回歸

### 第 1 題

In a study of five industrial areas, a researcher obtained these data relating the average number of units of a certain pollutant in the air and the incidence (per 100,000 people) of a certain disease:

 Units of pollutant Incidence of disease 3 4 5 8 10 48 52 58 70 96

Find the equation of the least-squares line y = Ax + B (to two decimal points places.)

Given that A = \frac {n \sum xy - \sum x \sum y} {n \sum x^2 - \left( \sum x \right)^2}$A = \frac {n \sum xy - \sum x \sum y} {n \sum x^2 - \left( \sum x \right)^2}$ and B = \frac {\sum y - A \sum x} n$B = \frac {\sum y - A \sum x} n$.

x y x2 xy
3 48 9 144
4 52 16 208
5 58 25 290
8 70 64 560
10 96100 960
303242142162


A = \frac {5 \cdot 2162 - 30 \cdot 324} {5 \cdot 214 - 30^2} = \frac{109}{17} \approx 6.411764705882353


B = \frac {324 - \left( 109/17 \right) \cdot 30} {5} = \frac{2238}{85} \approx 26.32941176470588


## 導函數

### 第 2 題，A 卷

Use the definition of the derivative \displaystyle f' \left( x \right) = \lim_{h \to 0} \frac {f \left( x + h \right) - f \left( x \right)} h $\displaystyle f' \left( x \right) = \lim_{h \to 0} \frac {f \left( x + h \right) - f \left( x \right)} h$ to find \displaystyle \frac d{dx} \sin x $\displaystyle \frac d{dx} \sin x$.


\frac d{dx} \sin x = \lim_{h \to 0} \frac {\sin \left( x + h \right) - \sin x} h



\sin \left( x + h \right) = \sin x \cos h + \cos x \sin h



\begin{align*}
\frac d{dx} \sin x &= \lim_{h \to 0} \frac {\sin x \cos h + \cos x \sin h - \sin x} h \\
&= \left( \sin x \right) \left( \lim_{h \to 0} \frac {\cos h - 1} h \right) + \left( \cos x \right) \lim_{h \to 0} \frac {\sin h} h
\end{align*}



\frac d{dx} \sin x = \cos x


## 切線與法線

### 第 3 題，A 卷

Find the equation of the tangent line to the curve y3xy2 + xy = 14 3 at (1, −2).

### 第 3 題，B 卷

… the normal line ….

y3xy2 + xy = 3

3y2y′ − (2xyy′ + y2) + (xy′ + y) = 0

(3y2 − 2xy + x) y′ = y2y

 y' = \frac{y^2 - y}{3y^2 - 2xy + x} 
 y' \left( 1, -2 \right) = \frac6{17} 

 y + 2 = \frac6{17} \left( x - 1 \right) 

 y + 2 = \frac{-17}6 \left( x - 1 \right) 

## 導數

### 第 4 題

Given \displaystyle f \left( x \right) = \frac {x^2 \left( 1-x \right)^3} {1+x} $\displaystyle f \left( x \right) = \frac {x^2 \left( 1-x \right)^3} {1+x}$, find f′(2)


f' \left( x \right) = \frac {-2x \left( x-1 \right)^3 - 3x^2 \left( x-1 \right)^2} {x+1} + \frac {x^2 \left( x-1 \right)^3} {\left( x+1 \right)^2}


f' \left( 2 \right) = \frac{-44}9


### 第 6 題，B 卷

Given f(x) = (x2 + 1)sin x, find f′(π/2)

f(x) = exp(sin x ln(x2 + 1))


\frac d{dx} \sin x \ln \left( x^2 + 1 \right) = \cos x \ln \left( x^2 + 1 \right) + \frac {2x \sin x} {x^2 + 1}


f' \left( x \right) = \left( x^2 + 1 \right)^{\sin x} \left( \cos x \ln \left( x^2 + 1 \right) + \frac {2x \sin x} {x^2 + 1} \right)


f' \left( \frac \pi 2 \right) = \left( \frac {\pi^2} 4 + 1 \right) \left( \frac \pi {\pi^2/4 + 1} \right) = \pi


## 微分

### 第 7 題

In the late 1830s, French physiologist Jean Poiseuille discovered the formula we use today to predict how much the radius of a particular clogged artery decreases the normal volume of flow. His formula,

V = kr4

says that volume of fluid flowing through a small pipe or tube in a unit of time at a fixed pressure is a constant times the fourth power of the tube’s radius r. How dose does a 10% decrease in r affect V?

#### 官方解法


\frac{dV}{dr} = 4kr^3


Δr = -0.1r


\Delta V \approx \frac{dV}{dr} \Delta r


ΔV ≈ -0.4kr4 = -0.4V

V 減少 40%

#### 直接解法

V0 = kr04

r = 0.9r0

V = k (0.9r0)4 = 0.94 V0


\frac {V - V_0} {V_0} = 0.9^4 - 1 = -0.3439


## 線性近似

### 第 8 題

Use the differentials to approximate the quantity \sqrt{4.6}$\sqrt{4.6}$ to four decimal points places.

f(x) ≈ f(a) + f′(a) (xa)


\sqrt x \approx \sqrt a + \frac {x - a} {2 \sqrt a}


x = 4 為起點，


\begin{align*}
\sqrt{4.6} &\approx 2 + \frac {4.6 - 4} {2 \cdot 2} = 2.15 \\
\sqrt{4.6} &\approx 2.15 + \frac {4.6 - 2.15^2} {2 \cdot 2.15} = \frac{3689}{1720} \approx 2.144767441860465 \\
\sqrt{4.6} &\approx \frac{3689}{1720} + \frac {4.6 - \left( 3689/1720 \right)^2} {2 \cdot \left( 3689/1720 \right)} \approx 2.144761058962219
\end{align*}



\sqrt{4.6} \approx 2.1448


## 應用題

### 第 9 題

Water boils at 212℉ at sea level and 200℉ at an elevation of 6000 ft. Assume that the boiling point B varies linearly with altitude α. Find the function B = f(α) that describes the dependence. Comment on whether a linear function gives a realistic model.

f(α) = mα + k

f(0) = k = 212

f(6000) = 6000m + 212 = 200

m 為 -0.002，所以

f(α) = -0.002α + 212

## 導函數

#### 第 2 題

Given f(x) = x3 − 3x2 + 3, use the definition $\displaystyle f' \left( x \right) = \lim_{h \to 0} \frac {f \left( x + h \right) - f \left( x \right)} h$ to find f′(x).

(x + h)3x3 = 3x2h + 3xh2 + h3

(x + h)2x2 = 2xh + h2

## 切線與法線

#### 公、保、管第 3 題

Find the equation to the tangent of the curve (x2 + 3) (x − 3)1/2 at x = 4.

#### 牙醫系第 3 題

… at x = 5.

f(x) = (x2 + 3) (x − 3)1/2，則

#### 牙醫系第 7 題

Find the equation of the tangent line to the curve x2 + 3xy + y2 = 5 at (1, 1).

#### 公、保、管第 7 題

… the normal line ….

2x + 3y + 3xy′ + 2yy′ = 0

5 + 5y′ = 0

y − 1 = -(x − 1)

y − 1 = x − 1

## 導數、二階導數

#### 公、保、管第 4 題

Consider a curve $\displaystyle f \left( x \right) = \frac {x + 2} {\left( x - 3 \right)^{0.5}}$. Find f′(5).

… and f″(5).

#### 公、保、管第 6 題

Given f(x) = 2x2+1, find f′(2).

y = x2 + 1，由鏈式法則

f′(x) = (ln 2) x 2y+1 = (ln 2) x 2x2+2

f′(2) = 128 ln 2

#### 牙醫系第 6 題

Given f(x) = ln(sec4(x) tan2(x)), find f′(π/4).

f′(x) = y′ / y

y′ = 4 sec4(x) tan3(x) + 2 sec6(x) tan(x)

f′(π/4) = 8

## 函數圖形

#### 第 5 題

Sketch the graph of $\displaystyle f \left( x \right) = \frac {3x^5 - 20x^3} {32}$ and also find the relative extreme points and inflection points at the interval of [-1, 1].

## 線性近似

#### 公、保、管第 8 題

Use the differentials to approximate the quantity $\sqrt{5.6}$ to two decimal points places.

#### 牙醫系第 8 題

$\sqrt{5.5}$ ….

f(x) ≈ f(a) + f′(a) (xa)

f(5.6) ≈ 2 + f′(4) (5.6 − 4)

## 應用題

#### 牙、公、保第 9 題

When a person coughs, the trachea (windpope windpipe) contracts, allowing air to be expelled at a maximum velocity. It can shown that during a cough the velocity v of airflow is given by the function v = f(r) = kr2(Rr) , where r is the trachea’s radius (in centimeters) during a cough, R is the trachea’s normal radius (in centimeters), and k is a positive constant that depends on the length of the trachea. Find the radius r for which the velocity of airflow is greatest.

f(0) = f(R) = 0

f(r) = kRr2kr3

f′(r) = 2kRr − 3kr2

f(2R/3) = k (2R/3)2 (R/3) > 0

#### 醫管系第 9 題

Suppose that during a nationwide program to immunize the population against certain from of influenza, public health officials found that the cost of inoculating x% of the population was approximately $\displaystyle C \left( x \right) = \frac {150x} {200 - x}$ million dollars.

1. What was the cost of inoculating the first 50% of the population?
2. What was the cost of inoculating the second 50% of the population?
3. What percentage of the population had been inoculated by the time 37.5 million dollars had been spent?

1. C(50) = 50
2. C(100) − C(50) = 100
3. 這實質上是解 解得 x 為 40，答案為 40%。

#### 牙醫系第 10 題

A rain gutter is made from sheets of metal 9 in wide. The gutters have a 3-in base and two 3-in sides, folded up at an angle θ (see figure). What angle θ maximizes the cross-sectional area of the gutter?

[本圖受著作權保護，請勿轉載。]

A = 9 (sin θ) (1 + cos θ)

#### 公、保第 10 題

Several mathematical stories originated with the second wedding of the mathematician and astronomer Johannes Kepler. Here is one: While shopping for wine for his wedding, Kepler noticed that the price of a barrel of wine (here assumed to be a cylinder) was determined solely by the length d of a dipstick that was inserted diagonally through a hole in the top of the barrel to the edge of the base of the barrel (see figure). Kepler realized that this measurement does not determine the volume of the barrel and that for a fixed value of d. The volume varies with the radius r and height h of the barrel. For a fixed value of d, what is the ratio r/h that maximizes the volume of the barrel?

[本圖受著作權保護，請勿轉載。]

V = πr2h

r2 = d2h2

V = π (d2hh3)

#### 醫管系第 10 題

Based on a study conducted in 1997, the percentage of the U.S. population by age affilcted with Alzheimer’s disease is given by the function

P(x) = 0.0726x2 + 0.7902x + 4.9623　where　0 ≤ x ≤ 25

where x is measured in years, with x = 0 corresponding to age 65 yr. Show that P is an increasing function of x on the interval (0, 25). What does your result tell you about the relationship between Alzheimer’s disease and age for the population that is aged 65 year and or older?

P 在 [0, 25] 等於多項式，所以在 (0, 25) 可微。我們觀察它的導函數。

P′(x) = 0.1452x + 0.7902　where　0 ≤ x ≤ 25

## 線性回歸

#### 第 1 題

In a study of five industrial areas, a researcher obtained these data relating the average number of units of a certain pollutant in the air and the incidence (per 100,000 people) of a certain disease:

 Units of pollutant Incidence of disease 3.4 4.6 5.2 8 10.7 48 52 58 70 96

Find the equation of the least-squares line y = Ax + B (to two decimal points places.)

Given that $A = \frac {n \sum xy - \sum x \sum y} {n \sum x^2 - \left( \sum x \right)^2}$ and $B = \frac {\sum y - A \sum x} n$.

x y x2 xy
3.4 48 11.56 163.2
4.6 52 21.16 239.2
5.2 58 27.04 301.6
8 70 64 560
10.7 96114.491027.2
31.9324238.252291.2

## 導函數

#### 第 2 題

Use the definition of the derivative $\displaystyle f' \left( x \right) = \lim_{h \to 0} \frac {f \left( x + h \right) - f \left( x \right)} h$ to find $\displaystyle \frac d{dx} \cos x$.

## 切線與法線

#### 第 3 題

Find the equation of the tangent line and the normal line of the curve x2y3 + xy = 10 at (1, 2).

3x2y2y′ + 2xy3 + xy′ + y = 0

(3x2y2 + x) y′ + 2xy3 + y = 0

13 y′ + 18 = 0

## 函數圖形

#### 第 4 題

Sketch the graph of f(x) = 6x5 − 5x3 and also find the relative extreme points and inflection points.

f′(x) = 30x4 − 15x2

f″(x) = 120x3 − 30x

f‴(x) = 360x2 − 30

6x5 − 5x3 = x3 (6x2 − 5)

## 線性近似

#### 第 5 題

Use the differentials to approximate the quantity $\sqrt{0.089}$ to four decimal points places.

f(x) ≈ f(a) + f′(a) (xa)

f(0.089) ≈ 0.3 + f′(0.09) (0.089 − 0.09)

## 牛頓法

#### 第 6 題

Use Newton’s method to find the real root r of f(x) = x3x − 1 to two decimal points places, given that initial point x0 = 1.5.

nxn
01.5
11.347826086956522
21.325200398950907
31.324718173999054

x ≈ 1.32