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3.1 Derivatives of Polynomials and Exponential Functions goo.gl/Vrcaa9 1

Chapter 3

Differentiation Rules

3.1

Derivatives of Polynomials and Exponential

Functions, page 172

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Property 1 (Derivative of a constant function, page 172). d

dx(c) = 0.

Proof. Let f (x) = c the constant function, then from the definition of a derivative, we have

f′_{(x) = lim} h→0 f (x + h) − f (x) h = limh→0 c − c h = limh→00 = 0.

Property 2 (The power rule, page 173). If n is any real number, then d

dx(x

n_{) = nx}n−1_.

Proof. Let f (x) = xn_{. Here we check the case n ∈ Z and show the general case in Section}

3.6. First, for n ∈ N, by the Binomial Theorem, we compute

f′_{(x) = lim} h→0 f (x + h) − f (x) h = limh→0 (x + h)n_{− x}n h = limh→0 Pn k=0Cknx(n−k)hk− xn h = lim h→0 n X k=1 C_knx(n−k)hk−1 = n X k=1  lim h→0C n kx(n−k)hk−1  = C₁nxn−1 = nxn−1.

Next, we check the case negative integer −n, n ∈ N. That is, let f (x) = x−n_{, and we will}

prove f′_{(x) = −nx}−n−1_: f′_{(x) = lim} h→0 f (x + h) − f (x) h = limh→0 1 h  1 (x + h)n − 1 xn  = lim h→0− n P k=0 Cn kx(n−k)hk− xn h(x + h)n_xn = lim_h→0− n X k=1 Cn kx(n−k)hk−1 (x + h)n_xn = − n X k=1 lim h→0 Cn kx(n−k)hk−1 (x + h)n_xn ! = −C1nx−n−1= −nx−n−1.





2 3.1 Derivatives of Polynomials and Exponential Functions goo.gl/Vrcaa9

Property 3(The constant multiple rule, page 175). If c is a constant and f is a differential function, then mVP9DEZRNBw d dx(cf (x)) = c d dxf (x). Proof. Let g(x) = cf (x). Then

g′_{(x) = lim} h→0 g(x + h) − g(x) h = limh→0 cf (x + h) − cf (x) h = limh→0c  f (x + h) − f(x) h  = c lim h→0 f (x + h) − f (x) h = cf ′_(x).

Property 4(The sum and difference rule, page 176). If f and g are both differentiable, then d dx(f (x) ± g(x)) = d dxf (x) ± d dxg(x). Proof. _{Let F (x) = f (x) ± g(x). Then}

F′_{(x) = lim} h→0 F (x + h) − F (x) h = limh→0 (f (x + h) ± g(x + h)) − (f (x) ± g(x)) h = lim h→0  f (x + h) − f(x) h ± g(x + h) − g(x) h  = lim h→0 f (x + h) − f (x) h ± limh→0 g(x + h) − g(x) h = f ′_{(x) ± g}′_(x). jIC1f3hWjsI

Example 5. Compute the derivative of the exponential function f (x) = ax_.

Solution. f′_{(x) = lim} h→0 f (x + h) − f (x) h = limh→0 ax+h_{− a}x h = limh→0 ax_ah_{− a}x h = lim h→0 ax_(ah_{− 1)} h =  lim h→0 ah_{− 1} h  ax= f′_{(0)f (x).}

Definition 6 (the number e, page 180). e is the number such that

lim

h→0

eh_{− 1}

h = 1.



n→∞(1 + 1 n)

n_{= e)。}

Property 7 (Derivative of the natural exponential function, page 178). d

dx(e

x_{) = e}x_.

Proof. Let f (x) = ex, then f′_{(x) = lim} h→0 f (x + h) − f (x) h = limh→0 ex_(eh_{− 1)} h =  lim h→0 eh_{− 1} h  ex = ex.

3.2 The Product and Quotient Rules goo.gl/KALD87 3

3.2

The Product and Quotient Rules, page 183

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Property 1 (The product rule, page 184). If f and g are both differentiable, then d dx(f (x)g(x)) = g(x) d dxf (x) + f (x) d dxg(x). Proof. Let F (x) = f (x)g(x), then

F′_{(x) = lim} h→0 F (x + h) − F (x) h = limh→0 f (x + h)g(x + h) − f (x)g(x) h = lim h→0 f (x + h)g(x + h) − f (x)g(x + h) + f (x)g(x + h) − f (x)g(x) h = lim h→0  f (x + h) − f(x) h  g(x + h) + f (x) g(x + h) − g(x) h  = lim h→0  f (x + h) − f(x) h  lim h→0g(x + h) + f (x) limh→0  g(x + h) − g(x) h  = f′_{(x)g(x) + f (x)g}′_(x).







Property 2 (The quotient rule, page 186). If f and g are both differentiable, then d dx  f (x) g(x)  = g(x) d dxf (x) − f (x)dxd g(x) (g(x))2 .

Proof. Let F (x) = f(x)_g(x). Then

F′_{(x) = lim} h→0 F (x + h) − F (x) h = limh→0 f(x+h) g(x+h) − f(x) g(x) h = limh→0 f (x + h)g(x) − g(x + h)f (x) hg(x + h)g(x) = lim h→0 f (x + h)g(x) − f (x)g(x) + f (x)g(x) − g(x + h)f (x) hg(x + h)g(x) = lim h→0 (f (x + h) − f (x))g(x) − f (x)(g(x + h) − g(x)) hg(x + h)g(x) = lim h→0 f(x+h)−f(x) h g(x) − f (x) g(x+h)−g(x) h g(x + h)g(x) = lim h→0 f(x+h)−f(x) h g(x) − f (x) lim_h→0 g(x+h)−g(x) h lim h→0g(x + h)g(x) = g(x)f′(x) − f (x)g′(x) (g(x))2 .



4 3.2 The Product and Quotient Rules goo.gl/KALD87 Example 3. Compute _dxd22  f(x) g(x)  .

HfjoNHEoqqA Solution. We compute

d2 dx2  f (x) g(x)  = d dx  d dx f (x) g(x)  = d dx  gf′_{− f g}′ g2  = g 2_(gf′_{− f g}′₎′_{− (gf}′_{− f g}′_{)(g · g)}′ g4 = g 2_(g′_f′_{+ gf}′′_{− f}′_g′_{− f g}′′_{) − (gf}′_{− f g}′_)(2g′_g) g4 = g(gf ′′_{− f g}′′_{) − 2g}′_(gf′_{− f g}′₎ g3 .



Example 4. The curve y = _1+x1 2 is called a witch of Maria Agnesi (箕舌線). Find an equation

of the tangent line to this curve at the point (−1,12).

Solution. Since y′_{(x) =} (1 + x2)(1)′ − 1(1 + x2)′ (1 + x2₎2 = −2x (1 + x2₎2, y′(−1) = (−2) · (−1) (1 + (−1)2₎2 = 1 2. we know the tangent line equation is y −12 = 12(x + 1).

3.3 Derivatives of Trigonometric Functions goo.gl/bFHgmZ 5

3.3

Derivatives of Trigonometric Functions, page

190

Goal: Find the derivative of six trigonometric functions.

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Example 1 (page 192–193). Calculate the derivative of f (θ) = sin θ and g(θ) = cos θ. Solution. We compute f′_{(θ) = lim} h→0 f (θ + h) − f (θ) h = limh→0 sin(θ + h) − sin θ h = limh→0 2 cos θ +h 2
sinh2 h = lim h→0cos  θ +h 2  lim h→0 sinh₂ h 2 = cos θ g′_{(θ) = lim} h→0 g(θ + h) − g(θ) h = limh→0 cos(θ + h) − cos θ h = limh→0 −2 sin θ + h 2
sinh2 h = lim h→0− sin  θ + h 2  lim h→0 sinh₂ h 2 = − sin θ.

Example 2 (page 193). Calculate the derivative of tan θ, cot θ, sec θ, and csc θ. Solution. Using the quotient rule, we get

d dθtan θ = d dθ  sin θ cos θ 

= cos θ(sin θ)′− sin θ(cos θ)′ cos2_θ =

cos θ cos θ − sin θ(− sin θ) cos2_θ = 1 cos2_θ = sec 2_θ d dθcot θ = d dθ  1 tan θ  = (tan θ)(1) ′_{− 1(tan θ)}′ tan2_θ = − sec2_θ tan2_θ = − csc 2_θ d dθsec θ = d dθ  1 cos θ  = (cos θ)(1)′− 1(cos θ)′ cos2_θ = sin θ

cos2_θ = sec θ tan θ

d dθcsc θ = d dθ  1 sin θ  = (sin θ)(1)′− 1(sin θ)′ sin2θ = − cos θ sin2θ = − csc θ cot θ.



θ = − csc 2_θ.



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Example 3. Calculate the derivative of sin2θ and cos2θ. Solution. We compute

(sin2θ)′ _{= (sin θ sin θ)}′ _{= cos θ sin θ + sin θ cos θ = 2 sin θ cos θ.}

(cos2θ)′ _{= (cos θ cos θ)}′ _{= − sin θ cos θ − cos θ sin θ = −2 sin θ cos θ.}

6 3.3 Derivatives of Trigonometric Functions goo.gl/bFHgmZ

Example 4. Compute _dθdnnsin θ and dn dθncos θ.

Solution. Direct computation gives

f (θ) = sin θ = f(4)(θ) = f(4k)(θ) g(θ) = cos θ = g(4)(θ) = g(4k)(θ) f′_{(θ) = cos θ = f}(5)_{(θ) = f}(4k+1)_{(θ) g}′_{(θ) = − sin θ = g}(5)_{(θ) = g}(4k+1)_(θ)

f′′_{(θ) = − sin θ = f}(6)_{(θ) = f}(4k+2)_{(θ) g}′′_{(θ) = − cos θ = g}(6)_{(θ) = g}(4k+2)_(θ)

f′′′_{(θ) = − cos θ = f}(7)_{(θ) = f}(4k+3)_{(θ) g}′′′_{(θ) = sin θ = g}(7)_{(θ) = g}(4k+3)_(θ).

Remark that we have another type of formula: d dθsin θ = sin  θ + π 2  ⇒ d n dθnsin θ = sin  θ + nπ 2  d dθcos θ = cos  θ +π 2  ⇒ d n dθncos θ = cos  θ + nπ 2  .

以下求導法則務必熟記

• (c)′ _{= 0. Derivative of a constant function}

• (xn₎′ _{= nx}n−1_{. The powe rule}

• (cf (x))′ _{= cf}′_{(x). The constant multiple rule}

• (f (x) + g(x))′ _{= f}′_{(x) + g}′_{(x). The sum rule}

• (f (x) − g(x))′ _{= f}′_{(x) − g}′_{(x). The difference rule}

• (f (x)g(x))′ _{= f}′_{(x)g(x) + f (x)g}′_{(x) The produce rule}

• fg(x)(x)

_′

= g(x)f′(x)−f(x)g_(g(x))2 ′(x). The quotient rule

⋆ (f (g(x)))′ _{= f}′_{(g(x)) · g}′_{(x). The chain rule}

三角函數的導函數

_,

務必熟記

(sin θ)′_{= cos θ (tan θ)}′ _{= sec}2_{θ (sec θ)}′ _{= sec θ tan θ}

(cos θ)′_{= − sin θ (cot θ)}′ _{= − csc}2_{θ (csc θ)}′ _{= − csc θ cot θ}

e

的定義與

_e

_的導函數

3.4 The Chain Rule goo.gl/6BEp2G 7

3.4

The Chain Rule, page 197

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Theorem 1 (The Chain Rule (鏈鎖律), page 198). If g is differentiable at x and f is differen-tiable at_{g(x), then the composite function F = f ◦g defined by F (x) = f (g(x)) is differentiable} at x and F′ _{is given by the product}

F′_{(x) = f}′_{(g(x)) · g}′_(x).

In Leibniz notation, if y = f (u) and u = g(x) are both differentiable functions, then dy dx= dy du· du dx.

Theorem 2(The power rule combined with the chain rule, page 200). If n is any real number and g(x) is differentiable, then

d dx(g(x))

n_{= n(g(x))}n−1_{· g}′_(x).

Proof. The relation is x_{→ u = g(x)}g _{→ y = u}f n_{. By the Chain Rule, we have}

dy dx = dy du     u=g(x) ·du_dx = nun−1 u=g(x)· g′(x) = n(g(x))n−1· g′(x).







Example 3. Find the derivatives. (a) (sin(ax))′ ₌ (b) (sin(x2₎₎′ ₌ (c) (sin2x)′ ₌ (d) (e2x)′ ₌ (e) (ex2 )′ ₌ jZcVk7bCXwo

Example 4 (page 202). Show that _dxdax_{= a}x_{ln a.}

8 3.4 The Chain Rule goo.gl/6BEp2G Example 5. Let f (x) = √3 x 3x+5. Find f′(1). Solution.





Example 7. _{Let f (x), g(x) ∈ C}1(R) and f (1) = 1, f′_{(1) = 0, g(0) = 0, g}′_{(0) = 1. Find the}

limit lim x→π 2 f (sin x)g(cos x) x2₋π 2x . Solution.



3.4 The Chain Rule goo.gl/6BEp2G 9 Example 8. Let 03m_ixBfAmo f (x) = ( x2_sin1 x x 6= 0 0 x = 0. (a) Find f′_(0). (b) When x 6= 0, find f′_(x).

(c) Does f′′_{(0) exist? If it exists, please find its value. If not, give reason to support your}

argument. Solution.









To prove the chain rule, one idea is the following:

F′_{(x) = lim} h→0 f (g(x + h)) − f (g(x)) h = lim h→0 f (g(x + h)) − f (g(x)) g(x + h) − g(x) · g(x + h) − g(x) h = lim h→0 f (g(x + h)) − f (g(x)) g(x + h) − g(x) · limh→0 g(x + h) − g(x) h = f′_{(g(x)) · g}′_(x).

This argument looks great, but it is not correct. What is the problem? How do we overcome the problem?

10 3.4 The Chain Rule goo.gl/6BEp2G

Appendix

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Proof of the Chain Rule. The function g(x) is differentiable at x. This means g′_{(x) exists and}

g(x + h) − g(x) h − g

′_{(x) → 0 as h → 0.}

Define a new variable v by

v = g(x + h) − g(x) h − g

′_{(x) ⇒ g(x + h) = g(x) + (g}′_{(x) + v)h. (1)}

Notice that v depends on the number h and that v → 0 as h → 0. Similarly, because the function f is differentiable at the point y = g(x), we have

f (y + k) − f (y) k − f

′_{(y) → 0 as k → 0.}

Define another variable w by

w = f (y + k) − f (y) k − f

′_{(y) ⇒ f (y + k) = f (y) + (f}′_{(y) + w)k. (2)}

Notice that w depends on the number k and that w → 0 as k → 0. From (1), we get

f (g(x + h)) = f (g(x) + (g′_{(x) + v)h).}

Use (2) applied to the right-hand-side with k = (g′_{(x) + v)h and y = g(x) to get.}

f (g(x) + (g′_{(x) + v)h) = f (g(x)) + (f}′_{(g(x)) + w)(g}′_{(x) + v)h.}

Note that k → 0 as h → 0, and so w → 0 as h → 0. So f (g(x + h)) − f (g(x)) h = f (g(x)) + (f′_{(g(x)) + w)(g}′_{(x) + v)h − f (g(x))} h = (f ′_{(g(x)) + w)(g}′_{(x) + v)h} h = (f ′_{(g(x)) + w)(g}′_{(x) + v).} Hence lim h→0 f (g(x + h)) − f (g(x)) h = limh→0(f ′_{(g(x)) + w)(g}′_{(x) + v)} =  lim h→0f ′_{(g(x)) + lim} h→0w   lim h→0g ′_{(x) + lim} h→0v  = f′_(g(x))g′_(x). since v → 0 as h → 0 and w → 0 as h → 0.

3.5 Implicit Differentiation goo.gl/6jtzyE 11

3.5

Implicit Differentiation, page 208

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The functions that we have met so far can be described by expressing one variable explicitly in terms of another variable y = f (x). However, there are a lot of functions are defined implicitly by a relation x and y and we formally write it as F (x, y) = 0. For example, F (x, y) = x2+ y2_{− 4 = 0, F (x, y) = x}3+ y3_{− 6xy = 0.}

x x

y y

Figure 1: (a) A circle x2_{+ y}2 _{= 4. (b) The folium of Descartes x}3_{+ y}3_{− 6xy = 0.}

Most of time, implicit functions are not “functions” (see the definition of a function in Section 1.1), but they are locally be expressed as functions.

x x y y Figure 2: A circle x2_{+ y}2 _{= 4.} x x x y y y

Figure 3: The folium of Descartes x3_{+ y}3_{− 6xy = 0.}

Furthermore, it’s not easy to solve implicit functions F (x, y) = 0 to explicit ones y = f (x). Fortunately, we can compute the derivative of implicit functions without solving implicit functions to explicit ones by .

12 3.5 Implicit Differentiation goo.gl/6jtzyE Example 1. If x2+ y2 = r2, find dy_dx. LlUA-nYD_A4 Solution. Solution 2.



(a) Find y′ _{if x}3_{+ y}3_{= 6xy.}

(b) Find the tangent to the folium of Descartes x3+ y3 = 6xy at the point (3, 3). (c) At what point in the first quadrant is the tangent line horizontal?

Solution.

If we solve the equation x3+y3 = 6xy for y in terms of x, we get three functions determined by the equation: y = f (x) = 3 s −1₂x3₊r 1 4x 6_{− 8x}3₊ 3 s −1₂x3₋r 1 4x 6_{− 8x}3 and y = 1 2  −f (x) ± √ −3   3 s −1₂x3₊r 1 4x 6_{− 8x}3₋ 3 s −1₂x3₋r 1 4x 6_{− 8x}3    .

It is very complicated to get the derivative by these formulae.

Implicit differentiation works for a lot of equations such as y5_{+ 3x}2_y2_{+ 5x}4_{= 12 for which}

3.5 Implicit Differentiation goo.gl/6jtzyE 13

An application of implicit differentiation is derivatives of inverse functions.

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Derivatives of Inverse Trigonometric Functions (page 214). d dxsin −1_{x = √} 1 1 − x2 d dxcos −1_{x = −√} 1 1 − x2 d dxtan −1_{x =} 1 1 + x2 d dxcot −1_{x = −} 1 1 + x2 d dxsec −1_{x =} 1 x√x2_{− 1} d dxcsc −1_{x = −} 1 x√x2_{− 1}.

Proof. Let y = y(x) = sin−1_{x, then sin y = x ⇒ cos y} dy

dx = 1. So dy dx = 1 cos y = 1 p 1 − sin2y = 1 √ 1 − x2.



Example 3 (Derivatives of inverse functions). Suppose f is a one-to-one differentiable func-tion and its inverse funcfunc-tion f−1 _{is also differentiable. Show that}

(f−1₎′_{(x) =} 1

f′_(f−1_(x))

provide that the denominator is not 0.

Solution.

LgLBwdDoWiI

Example 4. Find the tangent line of x23 + y 2 3 = a

2 3 at (x

0, y0) and the length between

x-intercept and y-intercept.

14 3.5 Implicit Differentiation goo.gl/6jtzyE

Example 5. _{Suppose that f (x) ∈ C}2(R) and f (x) satisfies x2+ xf (x) + (f (x))2 = k, where k is a constant, and f′_{(a) = f}′′_{(a) = 1. Find a and k.}

oA08sqVKgK4

Solution.

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Example 6. Two curves are orthogonal (正交) if their tangent lines are perpendicular at each point of intersection. Show that the given families of curves are orthogonal trajectories (正 交軌線) of each other; that is, every curve in one family is orthogonal to every curve in the other family.

(a) x2+ y2= r2, ax + by = 0. (b) x2+ y2= ax, x2+ y2= by. Solution.

(a)

(b) First, x2+ y2 _{= ax ⇒ 2x + 2yy}′ _{= a ⇒ y}′ ₌ a−2x

2y if y 6= 0. Next, x2+ y2 = by ⇒ 2x + 2yy′ _{= by}′ _{⇒ (b − 2y)y}′ _{= 2x ⇒ y}′ ₌ 2x b−2y if y 6= b 2. So if y 6= 0 and y 6= b 2, we have m1· m2 = a − 2x 2y · 2x b − 2y = 2ax − 4x2 2by − 4y2 = 2x2+ 2y2_{− 4x}2 2x2_{+ 2y}2_{− 4y}2 = 2y2_{− 2x}2 2x2_{− 2y}2 = −1.

If y = 0, then x2_{− ax = x(x − a) = 0 ⇒ x = 0 or x = a, so x}2_{+ y}2 _{= ax has vertical}

tangent line at x = 0 or x = a. If (x, y) = (0, 0), m2 = 0. If (x, y) = (a, 0), a 6= 0,

no curves in the family x2+ y2 = by passes through (a, 0). If y = b_{2, then x = ±2}b, so x2 + y2 = by has vertical tangent line at (b_{2, ±}b₂). At (₂b_{, ±2}b), we get a = b, and m1 = a − 2x = b − 2b₂ = 0, so x2+ y2 = ax has horizontal tangent line at (₂b, ±₂b).

3.6 Derivatives of Logarithmic Function goo.gl/QMmWh9 15

3.6

Derivatives of Logarithmic Function, page 218

Another application of implicit differentiation is getting the derivatives of logarithmic func-tions.

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Example 1 (page 218). Compute _dxd (log_ax) and _dxd(ln x).

Solution. Let y = log_ax. Then ay _{= x. Differentiating this equation implicit with respect}

to x, we get

In particular, we put a = e then _dxd(ln x) = . Example 2 (page 220). Find f′_{(x) if f (x) = ln |x|.}

Solution. Since f (x) = ( ln x if x > 0 ln(−x) if x < 0, it follows that f ′_{(x) =} ( 1 x if x > 0 if x < 0.

Thus f′_{(x) = for all x 6= 0.}

x y

Figure 1: f (x) = ln |x|.



Application: Logarithmic Differentiation (

對數微分法

₎

The calculation of derivatives of complicated functions involving products, quotients, or pow-ers can be simplified by the method of logarithmic differentiation.

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Example 3 (page 220). Differentiate y =x

3 4√x2+1 (3x+2)5 .

16 3.6 Derivatives of Logarithmic Function goo.gl/QMmWh9

The Power Rule _{(page 221). If n ∈ R and f (x) = x}n, then f′_{(x) = nx}n−1_.

pwlJ7ZGRGdE Proof. Let y = x

n _{and use logarithmic differentiation for x 6= 0:}

If x = 0, by the definition of derivative, we have

f′_{(0) =}

Example 4. Differentiate y = xx_{. (The function is defined on x > 0)}

Solution.

Solution 2.



The Number e as a Limit

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Example 5 (page 189). Show that e = lim

x→0(1 + x) 1 x = lim n→∞ 1 + 1 n
n .

Solution. We have shown that if f (x) = ln x, then f′_{(x) =} 1

x. Thus f′(1) = 1. From the

definition of a derivative as a limit and the continuity of the logarithmic function, we have

f′_{(1) = lim} x→0

f (1 + x) − f (1)

x =

=

Hence we have lim

x→0(1 + x) 1

x = e. If we put n = 1

x, then n → ∞ as x → 0

+_{, then we get an}

alternative expression for e:

lim n→∞  1 + 1 n n = e.

Example 6 (page 189). Show that lim

n→∞ 1 + x n
n = ex _{for any x > 0.} Solution.

3.8 Exponential Growth and Decay goo.gl/SdNZdK 17

3.8

Exponential Growth and Decay, page 237

In this section, we will show some examples of quantities grow or decay at a rate proportional to their size:

xB1bb_bh_Qo

(1) The number of individuals in a population of animals or bacteria.

(2) In nuclear physics, the mass of a radioactive substance decays at a rate proportional to the mass.

(3) In chemistry, the rate of unimolecular first-order reaction is proportional to the concen-tration of the substance.

(4) In finance, the value of a savings account with continuously compounded interest in-creases at a rate proportional to that value.

Definition 1 (page 237). If y(t) is the value of a quantity y at time t and if the rate of change of y with respect to t is proportional to its size y(t) at any time, then

dy

dt = ky, (3)

where k is a constant. It is called the law of natural growth (if k > 0) or the law of natural decay (if k < 0). The equation (3) is called a differential equation (微分方程) because it involves an unknown function y and its derivative dy_dt.

Theorem 2 (page 237). The only solutions of the differential equation dy_dt = ky are the exponential functions

y(t) = y(0)ekt.

Proof. Here we check any exponential function of the form y(t) = Cekt_{, where C is a constant,}

satisfies

y′_{(t) = C(ke}kt_{) = k(Ce}kt_{) = ky(t).}

We will prove in section 9.4 that any function that satisfies dy_dt = ky must be of the form y = Cekt_.

To see the significance of the constant C, we observe that

y(0) = Cek·0_{= C.}

Therefore C is the initial value of the function.



dt = ky。

18 3.8 Exponential Growth and Decay goo.gl/SdNZdK

Population Growth, page 237

In the context of population growth, where P (t) is the size of a population at time t, we can write nrKI3anQDhg dP dt = kP or 1 P dP dt = k.

The quantity _P1 dP_dt is the growth rate divided by the population size; it is called the relative growth rate(相對成長率).

Instead of saying “the growth rate is proportional to population size” we could say “the relative growth rate is constant.”

Example 3. The table gives the population of India, in millions, for the second half of the 20th century.

Year 1951 1961 1971 1981 1991 2001 Population 361 439 548 683 846 1029

(a) Use the exponential model and the census figure for 1951 and 1961 to predict the population in 2001. Compare with the actual population.

(b) Use the exponential model and the census figure for 1961 and 1981 to predict the population in 2001. Compare with the actual population. Then use this model to predict the population in the year 2010 and 2020.

Solution. (a)

(b) P (t) = P (0)ekt _{= 439e}kt_{, P (20) = 439e}20k _{= 683 ⇒ k =} 1 20ln683439

.

= 0.022099. P (40) = 439e40k _{= 439e}0.88396 _{= 1063, P (49) = 439e}. 49k _{= 439e}1.08289 _{= 1296.}.

P (59) = 439e59k = 439e1.30389 = 1617.. Exercise 4.

(a) Use the fact that the world population was 2560 million in 1950 and 3040 million in 1960 to model the population of the world in the second half of the 20th century. (b) What is the relative growth rate?

(c) Use the model to estimate the world population in 1993 and to predict the population in the year 2020.

3.8 Exponential Growth and Decay goo.gl/SdNZdK 19

Radioactive Decay, page 239

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Radioactive substances decay by spontaneously emitting radiation.

If m(t) is the mass remaining from an initial mass m0 of the substance after time t, then

the relative decay rate

−_m1 dm_dt

has been found experimentally to be constant. It follows that dm

dt = km,

where k is a negative constant. The solution is m(t) = m0ekt.

Physicist express the rate of decay in terms of half-life (半衰期), the time required for half of any given quantity to decay.

Example 5. Scientists can determine the age of ancient objects by the method of radiocarbon dating (放射性碳紀年). The bombardment of the upper atmosphere by cosmic rays converts nitrogen to a radioactive isotope of carbon, 14_{C, with a half-life of about 5730 years.}

Vege-tation absorbs carbon dioxide through the atmosphere and animal life assimilates 14C food chains. When a plant or animal dies, it stops replacing its carbon and the amount of 14C begins to decrease through radioactive decay. Therefore the level of radioactivity must also decay exponentially.

A parchment fragment was discovered that had about 74% as much 14C radioactivity as does plant material on the earth today. Estimate the age of the parchment.

Solution.

Exercise 6. Experiments show that if the chemical reaction N2O5→ 2NO2+

1 2O2

takes place at 45◦_{C, the rate of reaction of dinitrogen pentoxide is proportional to its}

concen-tration as follows:

−d[N_dt2O5] = 0.0005[N2O5]

(a) Find an expression for the concentration [N2O5] after t seconds if the initial

concentra-tion is C.

(b) How long will the reaction take to reduce the concentration of N2O5 to 90% of its

20 3.8 Exponential Growth and Decay goo.gl/SdNZdK

Newton’s Law of Cooling, page 240

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Newton’s Law of Cooling (牛頓冷卻定律) states that the rate of cooling of an object is propor-tional to the temperature difference between the object and its surroundings, provides that this difference is not too large.

If we let T (t) be the temperature of the object at time t and Ts be the temperature of the

surroundings, then we can formulate Newton’s Law of Cooling as a differential equation: dT

dt = k(T − Ts), where k is a constant.

When we make the change of variable y(t) = T (t) − Ts. Since Ts is constant, we have

y′_{(t) = T}′_{(t) and the equation becomes}

dy dt = ky. Hence we can solve y first and then find T .



Example 7. In a murder investigation, the temperature of the corpse was 32.5◦_{C at 1:30PM}

and 30.3◦_{C and hour later. Normal body temperature is 37.0}◦_{C and the temperature of the}

surroundings was 20.0◦_{C. When did the murder take place?}

Solution.

Exercise 8. A roast turkey is taken from an oven when its temperature has reached 85◦_C

and is placed on a table in a room where the temperature is 22◦_C.

(a) If the temperature of the turkey is 65◦_{C after half an hour, what is the temperature}

after 45 minutes?

(b) When will the turkey have cooled to 40◦_C?

Question 9. 兩杯熱咖啡, 一杯立馬加奶精, 然後放置五分鐘; 另一杯先放五分鐘後再加奶精, 哪一

3.8 Exponential Growth and Decay goo.gl/SdNZdK 21

Continuously Compounded Interest, page 241

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If an amount A0 is invested at an interest rate r, and interest is compounded n times a year,

then in each compounding period the interest rate is _nr and there are nt compounding periods in t years, so after t years the value of the investment is

A0  1 + r n nt .

The interest paid increases as the number of compounding periods n increases. If we let n → ∞, then we will be compounding the interest continuously (連續複利) and the value of the investment will be

A(t) = lim n→∞A0  1 +r n nt =

The above equation gives

dA

dt = rA(t),

which says that, with continuous compounding of interest, the rate of increase of an investment is proportional to its size.

Example 10.

(a) How long will it take an investment to double in value if the interest rate is 6% com-pounded continuously?

(b) What is the equivalent annual interest rate? Solution.

Exercise 11.

(a) If $3000 is invested at 5% interest, find the value of the investment at the end of 5 years if the interest is compounded (1) annually, (2) semiannually, (3) monthly, (4) weekly, (5) daily, and (6) continuously.

(b) If A(t) is the amount of the investment at time t for the case of continuous compounding, write a differential equation and an initial condition satisfied by A(t).

22 3.9 Related Rates goo.gl/r2aC5o

3.9

Related Rates, page 245

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Idea: Compute the rate of increase of one quantity in terms of the rate of change of another quantity (which may be more easily measured).

Procedure:

(1) Draw a picture or a diagram if possible.

(2) Introduce notation. Assign symbols to all quantities that are functions of time.

(3) Find an equation that relates the two quantities and then use the Chain Rule to differ-entiate both side with respect to time.

(4) Substitute the given information into the equation and get the unknown rate.

Example 1 (page 245). Air is being pumped into a spherical balloon so that its volume increases at a rate of 100 cm3_{/s. How fast is the radius of the balloon increasing when the}

diameter is 50 cm? Solution.

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Example 2 (page 246). A ladder 5 m long rests against a vertical wall. If the bottom of the ladder slides away from the wall at a rate of 1 m/s, how fast is the top of the ladder sliding down the wall when the bottom of the ladder is 3 m from the wall?

Solution.



3.9 Related Rates goo.gl/r2aC5o 23

Example 3 (page 246). A water tank has the shape of an inverted circular cone with base radius 2 m and height 4 m. If water is being pumped into the tank at a rate of 2 m3/min, find

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the rate at which level is rising the water is 3 m deep. Solution.

The water of level is rising at a rate of .

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Example 4 (page 247). Car A is traveling west at 90 km/h and car B is traveling north at 100 km/h. Both are headed for the intersection of the two roads. At what rate are the cars approaching each other when car A is 60 m and car B is 80 m from the intersection?

Solution.

The cars are approaching each other at a rate of .

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Example 5(page 248). A man walks along a straight path at a speed of 1.5 m/s. A searchlight is located on the ground 6 m from the path and is kept focused on the man. At what rate is the searchlight rotating when the man is 8 m from the point on the path closest to the searchlight?

Solution.

24 3.10 Linear Approximations and Differentials goo.gl/sq7mdV

3.10

Linear Approximations and Differentials, page

251

Recall two high school mathematics questions.

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Example 1. Let f (x) = 3x3_{− 22x}2_{+ 54x − 43. Find f (2.001) correct to three decimal place.} Solution.

Example 2. Find 1.0001100 _{correct to two decimal place.}

Solution.



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A curve lies very close to its tangent line near the point of tangency. In fact, by zooming in toward a point on the graph of a differentiable function, we noticed that the graph looks more and more like its tangent line. This observation is the basis for a method of finding approximate values of functions.

The idea is that it might be easy to calculate a value f (a) of a function, but difficult to compute nearby values of f . So we settle for the easily computed values of the linear function L whose graph is the tangent line of f at (a, f (a)).

Given a curve y = f (x), an equation of the tangent line at (a, f (a)) is y = f (a) + f′_{(a)(x − a).}

Definition 3 (page 252). The approximation

f (x) ≈ f (a) + f′_{(a)(x − a)}

is called the linear approximation or tangent line approximation of f at a (線性估計, 切線估 計). The linear function whose graph is this tangent line, that is,

L(x) = f (a) + f′_{(a)(x − a)}

3.10 Linear Approximations and Differentials goo.gl/sq7mdV 25

Example 4 (page 252). Approximate the numbers√3.98.

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Solution.



Example 5. Using a linear approximation to estimate cot 46◦_.

Solution.



Example 6. Go back to Example 1. and Example 2. to find out the calculation is in fact the linear approximation.

Solution. (1) f (x) = 3x3_{− 22x}2_{+ 54x − 43 = 1 + 2(x − 2) − 4(x − 2)}2_{+ 3(x − 2)}3. f′_{(x) = 2 − 8(x − 2) + 9(x − 2)}2_{. f (2.001) ≈ f (2) + f}′_{(2)(2.001 − 2) = 1.002.} (2) f (x) = (1 + x)100 = C₀1001100x0+ C₁100199x1+ C₂100198x2_{+ · · · + C100}10010x100. f′_{(x) = 100(1 + x)}99_{. f (0.0001) ≈ f (0) + f}′_{(0)(0.0001 − 0) = 1.01.}

Applications to Physics

Differentials

The ideas behind linear approximations are sometimes formulated in the terminology and notation of differentials (微分). If y = f (x), where f is a differentiable function, then the differential dx is an independent variable; that is, dx can be given the value of any real number. The differential dy is then defined in terms of dx by the equation

dydef.= f′_{(x) dx.}

26 3.10 Linear Approximations and Differentials goo.gl/sq7mdV

The geometric meaning of differentials is shown in Figure 1.

}

}

Figure 1: Geometric meaning of differentials.

Let P (x, f (x)) and Q(x + ∆x, f (x + ∆x)) be points on the graph of f and let dx = ∆x. The corresponding change in y is ∆y = f (x + ∆x) − f (x). The slope of the tangent line P R is the derivative f′_{(x). Thus the directed distance from S to R is f}′_{(x) dx = dy. Therefore}

dy represents the amount that the tangent line rises or falls (the change in the linearization), whereas ∆y represents the amount that the curve y = f (x) rises or falls when x changes by an amount dx.

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Example 7 (page 254). Compare the values of ∆y and dy if y = f (x) = x3_{+ x}2_{− 2x + 1 =}

9 + 14(x − 2) + 7(x − 2)2_{+ (x − 2)}3 _{and x changes from 2 to 2.05.}

Solution. _{Since f (2) = 9 and f (2.05) = 9 + 14 · 0.05 + 7 · (0.05)}2+ (0.05)3 = 9.717625, we have



Example 8 (page 255). The radius of a sphere was measured and found to be 21 cm with a possible error in measurement of at most 0.05 cm. What is the maximum error in using this value of the radius to compute the volume of the sphere.

Solution.

3.10 Linear Approximations and Differentials goo.gl/sq7mdV 27

Although the possible error in Example 8 may appear to be rather large, a better picture of the error is given by the relative error (相對誤差), which is computed by dividing the error by the total volume:

∆V V ∼ dV V = 4πr2_dr 4 3πr3 = 3dr r .

Thus the relative error in the volume is about three times the relative error in the radius. In Example 8 the relative error in the radius is approximately

dr r =

0.05

21 ∼ 0.0024

and it produces a relative error of about 0.007 in the volume. The error could also be expressed as percentage error (誤差百分比) of 0.24% in the radius and 0.7% in the volume.

28 3.11 Hyperbolic Functions goo.gl/gCgkTp

3.11

Hyperbolic Functions, page 259

Hyperbolic Functions, page 259

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Certain combinations of ex and e−x _{arise so frequently in mathematics and engineering. In}

many ways they are analogous to the trigonometric functions, and they have the same relation-ship to the hyperbola that the trigonometric functions have to the circle. For this reason they are called hyperbolic functions (雙曲函數) and individually called hyperbolic sine, hyperbolic cosine, and so on.

Definition 1 (Definition of the hyperbolic functions, page 259).

sinh x = e x_{− e}−x 2 cosh x = ex_{+ e}−x 2 tanh x = sinh x cosh x coth x = cosh x sinh x sech x = 1 cosh x csch x = 1 sinh x.

y= sinh x y= cosh x y= tanh x

1 1 1 1 1 1 1 − x x x y y y

Figure 1: Hyperbolic functions.

Hyperbolic Identities (page 260).

sinh(−x) = − sinh x cosh(−x) = cosh x cosh2_{x − sinh}2x = 1 _{1 − tanh}2x = sech2x sinh(x + y) = sinh x cosh y + cosh x sinh y cosh(x + y) = cosh x cosh y + sinh x sinh y

The identity cosh2_{x − sinh}2x = 1 indicates the curve (cosh x, sinh x) is hyperbola. Derivatives of Hyperbolic Functions (page 261).

d dxsinh x = cosh x d dxcosh x = sinh x d dxtanh x = sech 2_x d dxcoth x = −csch 2_x d

dxsech x = −sech x tanh x d

3.11 Hyperbolic Functions goo.gl/gCgkTp 29

Inverse Hyperbolic Functions

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Since the hyperbolic functions are defined in terms of exponential functions, the inverse hy-perbolic functions can be expressed in terms of logarithms:

sinh−1_{x = ln}_{x +}p_x2_{+ 1} cosh−1_{x = ln}_{x +}p_x2_{− 1} tanh−1_{x =} 1 2ln  1 + x 1 − x 

Derivatives of Inverse Hyperbolic Functions (page 263). d dxsinh −1_{x = √} 1 1 + x2 d dxcosh −1_{x = √} 1 x2_{− 1} d dxtanh −1_{x =} 1 1 − x2 d dxcoth −1_{x =} 1 1 − x2 d dxsech −1_{x = −} 1 x√_{1 − x}2 d dxcsch −1_{x = −} 1 |x|√x2_{+ 1}

Example 2 (page 265). Using principles from physics it can be shown that when a cable is hung between two poles, it takes the shape of a curve y = f (x) that satisfies the differential equation d2_y dx2 = ρg T s 1 + dy dx 2 ,

where ρ is the linear density of the cable, g is the acceleration due to gravity, T is the tension in the cable at its lowest point, and the coordinate system is chosen appropriately.

The function y = f (x) = T ρgcosh ρg T x 

is a solution of this differential equation.



a
is called a catenary (懸鏈線).

Example 3(page 265). Another application of hyperbolic functions occurs in the description of ocean waves: The velocity of a water wave with length L moving across a body of water with depth d is modeled by the function

v = s gL 2π tanh  2πd L  ,