Calculus Made Ridiculusly Easy

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VCE

_coverage

Area of study

Units 3 & 4 • Calculus

5 In this

_chapter

5A The derivative of tan kx

5B Second derivatives

5C Analysing the behaviour

of functions using the

second derivative

5D Derivatives of inverse

circular functions

5E Antidifferentiation

involving inverse circular

functions

Differential

calculus

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172

S p e c i a l i s t M a t h e m a t i c s

Introduction

In this chapter we extend the functions which can be differentiated to include f(x) = tan kx and the inverse circular functions Sin−1x, Cos−1x and Tan−1x. It is assumed that the standard results for the differential calculus are familiar to the student as are the product rule, quotient rule and chain rule. These results are listed in the table below.

The derivative of tan

kx

Since tan , its derivative can be found using the quotient rule:

If f(x) = tan kx, =

where k is a constant, then using the quotient rule with u = sin kx and v = cos kx: f′(x) =

=

= (by factorising the numerator) = (since cos2kx + sin2kx = 1)

= k sec2_kx If f(x) = tan kx, then f′(x) = k sec2 kx. f(x) f′(x) axn anxn – 1 log_e kx ekx kekx sin kx k cos kx cos kx –k sin kx

u(x) × v(x) u′(x) × v(x) + u(x) × v′(x) (product rule) u′(x) × v(x) - u(x) × (quotient rule)

g[h(x)] h′(x) × g′[h(x)] (chain rule) 1 x ---u x( ) v x( ) --- v′ x( ) v2 ---kx sin kx cos kx ---= d dx --- u v ---    v du dx --- udv dx ---– v2 ---= sin kx cos kx

---k cos ---kx cos ---kx( ) sin kx k sin kx– (– ) cos2_kx ---k cos2_kx₊_{k sin}2_kx cos2_kx ---k cos₍ 2_kx₊_sin2_kx₎ cos2_kx ---k 1( ) cos2_kx

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---C h a p t e r 5 D i f f e r e n t i a l c a l c u l u s

173

If y = tan [f(x)] then = f ′(x) sec2 [f(x)].

Differentiate the following expressions with respect to x. a y = tan 6x b y = 2 tan

THINK WRITE

a Differentiate by rule: y = tan kx, then = ksec2kx where k = 6.

a y = tan 6x

= 6 sec2_6x

b Differentiate tan by rule where k = .

b y = 2 tan

=

Multiply the derivative by 2. =

4 x 3 ---dy dx --- dy dx ---1 4x 3 ---4 3 ---4x 3 ---dy dx --- 2 4 3 --- sec24x 3 ---    2 dy_dx--- 8₃--- sec24x---₃

1 WORKED

E

xample

If f(x) = tan (x2_{+ 5x), find f ′(x).} THINK WRITE

Let u = x2_{+ 5x to apply the chain rule.} _f(x)_{= tan(x}2_{+ 5x)}

Let u = x2+ 5x.

Find . = 2x + 5

Express f(x) in terms of u. So f(x) = y = tan u.

Find . = sec2u

Find using the chain rule. =

= (2x + 5) sec2_u

Replace u with the expression x2+ 5x. f′(x) = (2x + 5) sec2(x2+ 5x)

1 2 du dx --- du dx ---3 4 dy du --- dy du ---5 dy dx --- dy dx --- dy du --- du dx ---× 6

2 WORKED

E

xample

d y d x

---Find if y = loge [tan (−3x)].

THINK WRITE

Let u = tan (−3x) and consequently apply the chain rule.

y = log_e [tan(−3x)] Let u = tan(−3x). Find . = −3 sec2(−3x) d y d x ---1 2 du dx --- du dx

---3

WORKED

E

xample

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174

THINK WRITE

Express y in terms of u. y = log_eu

Find . =

Find using the chain rule. So = ×

= × −3 sec2_(–3x)

Replace u with the expression tan (−3x). ₌

Express in terms of sin (−3x) and cos (−3x) only in order to simplify the rational expression.

= ÷

Express the division of rational expressions as a multiplication.

= ×

Cancel out the factor of cos (−3x). = Use the double angle formula to

simplify sin (−3x) cos (−3x). =

State the answer in simplest form. and hence = – or −6 cosec(−6x).

3 4 dy du --- dy du --- 1 u ---5 dy dx --- dy dx --- dy du --- du dx ---1 u ---6 3sec 2 3x – ( ) – 3x – ( ) tan ---7 dy dx --- –3 cos2₍_–_3x₎ --- sin(–3x) 3x – ( ) cos ---8 –3 cos2₍_–_3x₎ --- cos(–3x) 3x – ( ) sin ---9 –3 3x – ( ) sin cos(–3x) ---10 –3 1 2 ---sin(–6x) ---11 dy dx --- 6 6x – ( ) sin

---Find the equation of the tangent to the curve y = 3x + cos 2x + tan x where x = .

THINK WRITE

To find the equation of a tangent line to a curve at a point, the coordinate of the point is needed as is the gradient of the curve at that point. Find y when x = to establish the coordinate.

y = 3x + cos 2x + tan x

If x = , y = + cos + tan = + 0 + 1 = 1 +

The coordinate is

(

, 1 +

)

Find to establish the gradient function. = 3 − 2 sin 2x + sec2_x π 4 ---1 π 4 ---π 4 --- 3π 4 --- π 2 --- π 4 ---3π 4 ---3π 4 ---π 4 --- 3π 4 ---2 dy dx --- dy dx

---4

WORKED

E

xample

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C h a p t e r 5 D i f f e r e n t i a l c a l c u l u s

175 The derivative of tan kx

1 Differentiate each of the following with respect to x.

a y = tan 4x b y = tan 5x c y = 3 tan 7x

d y = 4 tan 2x e y = tan(−3x) f y = tan(−12x)

g y = 3 tan(−4x) h y = −2 tan 3x i y = −5 tan(−2x)

j y = tan k y = tan l y = tan

m y = tan n y = 8 tan o y = 6 tan

p y = −3 tan q y = 5 tan r y = −4 tan

THINK WRITE

Evaluate when x = to find the gradient of the tangent at x = .

If x = , = 3 − 2 sin + sec2 = 3 − 2 +

= 1 + = 3

so the equation of the tangent is:

Substitute y₁= , x₁= and m = 3 into the equation for a straight line: y − y₁= m(x − x₁) where m is the gradient and (x₁, y₁) is a point on the line.

y −

(

1 +

)

= 3

(

x −

)

Simplify the equation and make y the subject. y = 3x − + 1 + y = 3x + 1 3 dy dx --- π 4 ---π 4 ---π 4 --- dy dx --- π 2 --- π 4 ---1 cos2π 4 --- ---1

(

1 2 ---

)

---4 1 3π 4 ---+ π 4 --- 3π 4 --- π 4 ---5 3π 4 --- 3π 4

---remember

1. If y = tan kx then = k sec2 kx.

2. If y = tan f(x) then = f ′(x) sec2 [f (x)]. dy dx ---dy dx

---remember

5A

WORKED Example 1 Ma_th cad Differentiator x 5 --- x 2 --- 3x 5 ---2x 7 --- 3x 4 --- 4x 9 ---5x 6 --- 8x 5 --- 7x 2

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---176

2 For each of the following find f′(x) if f(x) equals:

3

If y = etan 2x, then:

a can be found by using:

b is equal to:

4 For each of the following find if y equals:

5 For each of the following find f′(x) if f (x) equals

6 If f: [0, ] → R, f (x) = tan x, find the coordinates of the points on the graph where the gradient is equal to:

7 If f: [− , ] → R, f(x) = 4 tan , find the coordinates of the points on the graph where the gradient is equal to:

8 Find the equation of the tangent to the curve with equation y = 3 tan 2x at x = .

9 If f(x) = x tan x find the equation of i the tangent and ii the normal at the point on the curve where x = .

10 Show that there are no stationary points for the graph of y = tan x for all values of x.

11 Explain why the gradient of tan x is always positive.

12 If f:

[

− ,

]

→ R, f (x) = tan x − x:

a find any stationary points and state their nature

b if the domain is changed to R, show that the gradient can never be negative.

a tan(x2+ 3x) b tan(x + 2) c tan(5x − 4)

d tan(2x2− 3x) e tan(3x + 2) f tan 8x

g tan(7 − 4x) h tan(1 − 5x2)

A a direct rule B the product rule C the chain rule

D the quotient rule E graphical methods

A 2 sec22xetan 2x B 2 sec2xetan 2x C 4 sec2xetan 2x

D 2 tan2xesec 2x E ex sec22x

a log_e(tan 6x) b etan 3x

c tan(e3x) d sin(tan 5x)

e cos(tan 2x) f (tan 5x)

g sin 4x − tan 3x h log_e2x2+ 4 tan

i tan2x j tan42x

a x2 tan 3x b cos 2x tan c (5x3− 6x) tan 4x

d e4xtan(−3x) e e4x2tan 8x f g h a 1 b c 4 d 10 a 4 b 2 c 0 W WORKEDORKED E Examplexample 2 WORKED Example 2 m

multiple choiceultiple choice

dy dx ---dy dx ---WORKED Example 3 dy dx ---1 2 ---6x 7 ---4x 5 ---tan x 2x2 ---4x3 tan 2x --- sin 4x tan 4x ---π 4 3 ---π ---π ₂---x WORKED Example 4 π 6 ---π 4 ---π 2 --- π 2

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177 Second derivatives

Suppose that y = f (x).

Then the derivative, also known as the first derivative, is written as f′(x) or . By differentiating a second time, the second derivative, f′′(x) or is obtained. The process of obtaining f′′(x) or from y = f (x) is also called double differen-tiation.

The second derivative is commonly called ‘f double dash’ or ‘d squared y, dx squared’. dy dx ---d2y dx2 ---d2y dx2 ---Find if y = + 2x4− x−1. THINK WRITE Express as . y = + 2x4− x−1 y = + 2x4− x−1

Find the first derivative . = + 8x3+ x−2

Differentiate to obtain . _{= −} _{+ 24x}2_{− 2x}−3 or − + 24x2− d2_y d x2 --- x 1 x x 1 2 ---x x 1 2 ---2 dy dx --- dy dx --- 1 2 ---x 1 2 ---– 3 dy dx --- d2y dx2 --- d2y dx2 --- 1 4 --- x 3 2 ---– 1 4x 3 2 --- 2 x3

---5

WORKED

E

xample

Find i f′(x) and ii f′′(x) if f (x) is equal to: a ecos 2x+ logex b

.

THINK WRITE

a i Differentiate ecos 2x by the chain rule short cut and log_ex by rule to obtain f′(x).

a i f(x) = ecos 2x+ log_ex f′(x) = −2 sin 2x ecos 2x+

ii Express in index notation so it can be differentiated.

ii f′(x) = −2 sin 2x ecos 2x+ x−1

Differentiate f′(x) to obtain f ′′(x). Use the product rule to

differentiate −2 sin 2x ecos 2x.

f′′(x) = −4 cos 2x ecos 2x+ 4 sin2 2x ecos 2x−x−2 or = 4ecos 2x (sin2 2x − cos 2x) −

sin x x ---1 x ---1 1 x ---2 1 x2

---6

WORKED

E

xample

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178

THINK WRITE

b i Express in index notation. b i f(x) = f(x) =

Express f(x) as a product. = sin x

Differentiate f(x) using the product rule to obtain f′(x).

f′(x) = − sin x + cos x

ii Differentiate f′(x) using the product rule to obtain f′′(x).

ii f′′(x) = x sin x − x cos x

− x cos x − x sin x

Simplify by collecting like terms. = x sin x − x cos x − x sin x

Simplify f′′(x) by taking out the factor x .

= x (3 sin x − 4x cos x − 4x2_{sin x)}

or 1 x sin x x ---sin x x 1 2 --- ---2 x 1 2 ---– 3 1₂---x 3 2 ---– x 1 2 ---– 1 3₄ ---5 2 ---– ₁ 2 ---3 2 ---– 1 2 ---3 2 ---– 1 2 ---– 2 3 4 ---5 2 ---– 3 2 ---– 1 2 ---– 3 1 4 ---5 2 ---– 1 4 ---–5 2

---3 sin x–4x cos x_–4x2_{sin x}

4x

5 2

---

---If y = and , find the value of k.

THINK WRITE

Find from y = . y =

=

Find . ₌

Substitute and into the equation − 3 + 2y = 0.

so

Take out as a factor. (k2− 6k + 8) = 0

Factorise the quadratic function of k. (Note: cannot equal zero.)

(k − 4)(k − 2) = 0

State the solutions. Therefore k = 2 or k = 4.

e kx 2 --- _d2_y d x2 --- 3d y d x ---– +2 y = 0 1 dy dx --- e kx 2 ---e kx 2 ---dy dx --- ke kx 2 ---2 ---2 d 2_y dx2 --- d2y dx2 --- k2e kx 2 ---4 ---3 dy dx --- d2y dx2 ---d2_y dx2 --- dy dx ---k2_e kx 2 ---4 --- 3ke kx 2 ---2 ---– 2e kx 2 ---+ = 0 4 1₄---e kx 2 --- ₁ 4 ---e kx 2 ---5 1 4 ---e kx 2 ---1 4 ---e kx 2 ---6

7 WORKED

E

xample

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179 Second derivatives

1 Find if 2 If y = sin , then: a is equal to: b is equal to:

3 Findif′(x) andiif′′(x) if f(x) is equal to:

a y = 4x2+ 7x − 3 b y = 5x3− x2+ 3x c y = 6x d y = x4+ 2x3− 3x + 1 e y = x6+ 2x4− 3x f y = g y = 4x3+ h y = + 5 i y = x2+ 5x−3 j y = _k y = 2x−2+ x−1 A cos B C D sin E cos A x cos B C D sin _E

a log_e2x b x2− log_ex c sin 2x + 4 cos x

d 2 sin e 2e5x− 3ex + 1 f 4e−3x+ 3x3

g tan 2x h −3 tan (−4x) + 1 i

j _k e2 sin x l cos e5x

m n o log_e(cos x)

remember

1. is the second derivative of y with respect to x. It is found by differentiating y twice. 2. If y = f(x) then = f ′′(x). d2y dx2 ---d2y dx2

---remember

5B

WORKED Example 5 d2_y dx2 --- Ma_th cad Second derivatives x 7 2 ---x 3 2 ---+ 2 x–1 2x 9 2 ---6x 5 3 ---– 2 x ---₊3x–1 m

x dy dx ---x x cos x 2 x --- –cos x x ---1 2 x --- 1 2 x ---d2_y dx2 ---x x sin x–cos x x x --- x sin x–cos x 2x x ---1 4x x ---–     –cos x– x sin x 4x x ---WORKED Example 6 x 2 --- x 4 3 ---+ cos 2x x ---x 5 2 ---sin 2x ---sin 2x (cos 4x) 5 2

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---180

4 For each of the following functions f(x), match the graph which could represent f′′(x).

5 If y = cos x, show that .

6 If y = sin 3x, show that: + 9y = 0.

7 If y = x log_ex, show that: .

8 If y = xex, show that: .

9 Find the value of k if y = e−kx and .

10 If y = ekx and , find the value of k.

11 The position of a particle travelling in a straight line is given by the equation: x(t) = t3− t2− t + 7, where x has units in cm and t is in seconds. Find:

a , that is, the velocity at any time t

b , that is, the acceleration at any time t

c when and where the particle momentarily stops; that is, when = 0

d the minimum velocity.

a f (x) = x3+ 2x b f (x) = 4x2 c f (x) = x4− x2 d f (x) = sin x e f (x) = log_ex f f (x) = e2x g f (x) = A B C D E F G H I 4 x x 0 f"(x) y x 0 f"(x) y x 0 f"(x) y x 0 f"(x) y x 0 f"(x) y x 0 f"(x) y x 0 f"(x) y x 0 f"(x) y x 0 f"(x) y d2_y dx2 --- = –y d2_y dx2 ---dy dx --- (x+y) d 2_y dx2 ---× = d2_y dx2 --- dy dx --- y x --+ = WORKED Example 7 d2_y dx2 --- = 25y d2_y dx2 --- 3dy dx ---– –4y = 0 dx dt ---d2_x dt2 ---dx dt

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181 Analysing the behaviour of functions

using the second derivative

The second derivative of a function can be used for testing the nature of stationary points, but is not a requirement of this course. It is provided as an alternative, or extension, to the first derivative test which is a requirement to Mathematical Methods Units 3 and 4.

First derivative function

We have seen how the first derivative of a function, f′(x) or , can tell us where a function has a positive gradient, a negative gradient or a zero gradient (stationary point). For example, let us look at the functions f(x) = x2 and f(x) = x3.

Function f(x)

= x

2

Examine the graph at right. Since f (x) = x2, f′(x) = 2x

and f′(x) = 0 at x = 0

1. If f′(x) = 0

At x = 0, f (x) is a stationary point. 2. If f′(x) < 0

This occurs when 2x < 0. So if x < 0, then f (x) is a decreasing function, one with a negative gradient for all x < 0.

3. If f′(x) > 0

This occurs when 2x > 0. So if x > 0, then f (x) is an increasing function, one with a positive gradient for all x > 0.

Consequently, at x = 0, a minimum stationary point occurs.

Function f(x)

= x

3

Examine the graph at right. Since f (x) = x3, f′(x) = 3x2

and f′(x) = 0 at x = 0

1. If f′(x) = 0

At x = 0, f (x) is a stationary point.

2. Here, f′(x) > 0 for all values of x except zero.

Consequently, at x = 0, a stationary point of inflection occurs.

Second derivative function

Similarly the second derivative

(

f′′(x) or

)

can tell us where the gradient function

(

f′(x) or

)

is increasing or decreasing or neither (that is, changing from increasing to decreasing or vice-versa). Let us look at the situation when f′′(x) is greater than, less than and equal to zero.

dy dx ---x 0 x > 0 f (x) is increasing x < 0 f (x) is decreasing x = 0 f (x) is neither increasing nor decreasing f (x) = x2 y x 0 x > 0 f(x) is increasing x < 0 f (x) is increasing x = 0 f(x) is neither increasing nor decreasing f(x) = x3 y d2_y dx2 ---dy dx

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---182

Function f

′′(x) > 0

Examine the graph at right. When f′′(x) > 0 then the gradient function f (x) is increasing. That is, as x increases then f′(x) increases.

Again consider: f (x) = x2 f′(x) = 2x f′′(x) = 2

Function f

′′(x) < 0

Examine the graph at right. When f′′(x) < 0, then the gradient of f(x) is decreasing. That is, as x increases then f′(x) decreases.

For example, if f(x) = −x2

f′(x) = −2x (therefore there is a stationary point at x = 0)

f′′(x) = −2

that is, f′′(x) < 0 for all values of x.

This means that the gradient function f′(x) is always decreasing.

Function f

′′(x) = 0

The gradient of f (x) is changing from increasing to decreasing or vice-versa (providing the sign of f′′(x) changes at the point where f′′(x) = 0). Again consider:

f(x) = x3

f′(x) = 3x2 (therefore there is a stationary point at x = 0)

f′′(x) = 6x so f′′(x) = 0 at x = 0 and f′′(x) < 0 if x < 0, f′′(x) > 0 if x > 0.

Therefore the gradient is decreasing left of x = 0 and increasing right of x = 0. The point (0, 0) is called a point of inflection.

As f′(0) = 0, the point (0, 0) is also a stationary point, so it is called a stationary point of inflection.

Points of inflection

Points of inflection occur when the second derivative changes sign.

A tangent drawn at a point of inflection crosses the graph at that point (see figure 1). Sometimes points of inflection are also stationary points (see figure 2).

x 0 f (x) = x2 f ' (3) = 6 f ' (–3) = 6 f ' (2) = 4 f ' (–2) = –4 f ' (–1) = –2 f ' (1) = 2 y 1 –1 –2 –3 2 3

Gradient is always increasing from left to right

x 0 f(x) = –x2 f ' (3) = –6 f ' (–3) = 6 f ' (2) = –4 f ' (1) = –2 f ' (–1) = 2 f ' (–2) = 4 f ' (0) = 0 y 1 –1 –2 –3 2 3

Gradient is always decreasing from left to right

x < 0 f ' (x) is decreasing x > 0 f ' (x) is increasing f(x) = x3 x y f ' (2) = 12 f ' (–2) = 12 f ' (–1) = 3 f ' (1) = 3 0 1 2 –2 –1 Point of inflection [ f"(x) = 0] Tangent Tangent f"(x) < 0 f"(x) 0< Stationary points of inflection [f " (x) = 0 and f ' (x) = 0] Tangent Tangent f"(x) < 0 f " (x) 0< Figure 1 Figure 2

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183

The second derivative can tell us the nature of any stationary points or where points of inflection occur on a graph. In general:

1. If f′(a) = 0 and f ′′(a) > 0, then a local minimum stationary point occurs at x = a. 2. If f′(a) = 0 and f ′′(a) < 0, then a local maximum stationary point occurs at x = a. 3. If f′(a) = 0 and f ′′(a) = 0, and f ′′(x) changes sign at x = a, then a stationary point

of inflection occurs at x = a.

4. If f′′(a) = 0 and f ′′(x) changes sign at x = a, then a point of inflection occurs at x = a.

Notes:

1. Types of stationary points (1, 2, 3 above) can also be determined using the first derivative test either side of x = a.

2. The second derivative test is usually more efficient than the first derivative test in determining maximum or minimum stationary points, but not for stationary points of inflection.

3. Displaying a curve on a graphics calculator will assist in determining the nature of any stationary points.

a Find any stationary points, and their nature, if f (x) = x2(x − 1)(x + 1).

b Sketch the graph of f(x), clearly indicating all stationary points and axes intercepts.

THINK WRITE

a Expand f (x) so it can be differentiated easily. a f(x) = x2(x − 1)(x + 1) f(x) = x2(x2− 1) = x4_{− x}2 Find f′(x). f′(x) = 4x3− 2x Find f′′(x). f′′(x) = 12x2− 2

Solve for x where f′(x) = 0. For stationary points, f′(x) = 0 2x(2x2− 1) = 0

x = 0 or x2= x = 0, – or

Find f (0) and f′′(0) to determine one stationary point.

At x = 0, f (0) = 0, f ′(0) = 0 and f ′′(0) = −2. Therefore (0, 0) is a local maximum stationary point.

Find f

(

−

)

and f′′

(

−

)

for the second stationary point.

At x = – , f

(

–

)

= − = – and f′′

(

–

)

= 6 − 2

= 4

Therefore

(

– , −

)

is a local minimum stationary point.

Find f

( )

and f′′

( )

for the third stationary point.

At x = , f

( )

= − = −

and f′′

( )

= 6 − 2 = 4

Therefore

(

, −

)

is a local minimum stationary point. 1 2 3 4 1 2 ---1 2 --- 1 2 ---5 6 1 2 --- 1 2 ---1 2 --- 1 2 --- 1 4 --- 1 2 --- 1 4 ---1 2 ---1 2 --- 1 4 ---7 1 2 --- 1 2 ---1 2 --- 1 2 --- 1 4 --- 1 2 --- 1 4 ---1 2 ---1 2 --- 1 4

---8

WORKED

E

xample

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184

THINK WRITE

b Evaluate f (0) for the y-intercept. b f(0) = 0 so the y-intercept is 0.

Solve the factorised form of f (x) = 0 for the x-intercepts.

f(x) = x2(x − 1)(x + 1) = 0 has solutions x = 0, 1, −1

so the x-intercepts are −1, 0 and 1.

Sketch the graph of f (x).

Check the graph using a graphics calculator. 1 2 3 x 0 f (x) = x2_{(x – 1)(x + 1)} y 1 – 2 1 – 4 ( , – ) (– ,1– – ) 2 1 – 4 –1 1 1 –1 4 If y = x4− x3+ 2 find:

a any stationary points of inflection b any other points of inflection.

THINK WRITE

a Find . a y = x4− x3+ 2

= x3_{− 3x}2

Solve for x where = 0 (that is, the stationary points).

For stationary points, = 0.

x3− 3x2= 0 x2(x − 3) = 0 x = 0 or x = 3 Find . = 3x2− 6x Evaluate where x = 0. At x = 0, = 3(0)2− 6(0) = 0

Find y where x = 0. and y = (0)4− (0)3+ 2 = 2

Evaluate where x = 3. At x = 3, = 3(3)2− 6(3) = 27 − 18 = 9

State any stationary points of inflection. Therefore the point (0, 2) is a stationary point of inflection.

Check using a graphics calculator.

1 4 ---1 dy dx --- 1 4 ---dy dx ---2 dy dx --- dy dx ---3 d 2_y dx2 --- d2y dx2 ---4 d 2_y dx2 --- d2y dx2 ---5 1₄ ---6 d 2_y dx2 --- d2y dx2 ---7 8

9 WORKED

E

xample

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185

For worked example 9, the second derivative sign diagram verifies that at x = 0 and x = 2 there are indeed points of inflection as f′′(x) changes sign at these points.

This verification is not usually required but in rare cases it will show that what seems to be a point of inflection is in fact not.

For example, if f (x) = x4 f′(x) = 4x3

= 0 when x = 0. Hence a stationary point occurs at x = 0.

For f′′(x) = 12x2= 0, a solution occurs when x = 0.

Thus it appears, there is a stationary point of inflection at x = 0. But the sign diagram of the second derivative (see figure at right) shows that f′′(x) does not change sign at x = 0.

Therefore, there is not a stationary point of inflection at x = 0.

The first derivative test will verify that at x = 0 there is a local minimum stationary point as shown in the figure below (left).

The sketch graph of f (x) = x4 is shown in the figure below (right).

THINK WRITE

b Solve = 0 b For points of inflection, = 0 3x2− 6x = 0

3x(x − 2) = 0

x = 0 or x = 2

Find y at x = 2 only, since a stationary point has already been determined at the point (0, 2).

At x = 2, y = (2)4− 23+ 2 = 4 − 8 + 2 = −2

Check the sign of either side of x = 2.

If x = 1, = 3(1)2− 6(1) = −3

If x = 3, = 9 (from part a)

State the other point of inflection. Therefore the point (2, −2) is a point of inflection (not stationary.)

Check using a graphics calculator.

1 d 2_y dx2 --- d 2_y dx2 ---2 1₄ ---3 d 2_y dx2 --- d2y dx2 ---d2_y dx2 ---4 5 x 0 –1 1 2 – – sign ofd 2_y ––– dx2 + + x 0 –1 1 – – f "(x) + + x 0 –1 1 – – f ' (x) + + x 0 f (x) = x4 y

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186

For the function f(x) = 3 logex − x3− 5, find:

a all stationary points b any points of inflection c and sketch the graph of f (x).

THINK WRITE

a Find f′(x) and set it equal to zero. a f′(x) = − 3x2

For stationery points, f′(x) = 0 − 3x2 _{= 0}

Multiply the equation by x. or 3 − 3x3= 0, x ≠ 0

Solve f′(x) = 0. 3x3= 3 x3= 1 Thus x = 1. Evaluate f (1). f(1) = 3(0) − 13− 5 = −6 Find f′′(x). f′′(x) = −3x−2− 6x Evaluate f′′(1). _f_{′′(1) = − − 6} = −9

State any stationary point and its type. Therefore (1, −6) is a local maximum. b Set f′′(x) = 0 and multiply the equation

by x2.

b f′′(x) = −3x−2− 6x

For points of inflection, f′′(x) = 0 −3x−2_{− 6x = 0}

or −3 − 6x3= 0, x ≠ 0

Solve f′′(x) = 0. 6x3= −3

x3= −

Thus x = − or approximately −0.79.

Since x = −0.79 is the only solution there are no points of inflection because the implied domain is x > 0.

Therefore there are no points of inflection as −0.79 is outside the implied domain,

x > 0.

c Use a graphics calculator to determine intercepts.

Sketch the graph of f (x).

Check the graph using a graphics calculator. 1 3_x ---3 x ---2 3 4 5 6 3 1 ---7 1 2 1 2 ---1 2 ---   13 ---3 1 x 0 f (x) = 3 log_ex – x3_{– 5} y –6 1(1, –6) 2 3

10 WORKED

E

xample

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187

Different technology tools, including the graphics calculator, can be used to verify or obtain the coordinates of stationary points as well as draw the relevant graph.

The screen at right shows an example from the prepared Mathcad file ‘Stationary points’ found on the Maths Quest CD-ROM.

Find the minimum value of the gradient to the curve:

f(x) = x3+ 4x2− 7x.

THINK WRITE

Find the equation of the gradient function f′(x).

f′(x) = 3x2+ 8x − 7

Find the rate of change of the gradient function, f′′(x).

f′′(x) = 6x + 8

Set f′′(x) = 0. Let f′′(x) = 6x + 8 = 0

Solve f′′(x) = 0. 6x = −8

x = −

Evaluate f′

(

−

)

which is the minimum value of the gradient as f′(x) is a positive quadratic.

f′

(

−

)

= 3

( )

+ 8

(

−

)

− 7

= − − 7 = −12

which is a minimum as f′(x) is a positive quadratic equation.

State the minimum value of the gradient. Therefore the minimum gradient of f(x) is −12 .

Verify using a graphics calculator.

1 2 3 4 4 3 ---5 4 3 --- 4 3 --- 16 9 --- 4 3 ---16 3 --- 32 3 ---1 3 ---6 1₃ ---7

11 WORKED

E

xample

Ma_th cad Stationary points

remember

1. If f′(a) = 0 and f ′′(a) > 0, then a local minimum stationary point occurs at x = a. 2. If f′(a) = 0 and f ′′(a) < 0, then a local maximum stationary point occurs at x = a. 3. If f′(a) = 0 and f ′′(a) = 0, and f ′′(x) changes sign at x = a, then a stationary point

of inflection occurs at x = a.

4. If f′′(a) = 0 and f ′′(x) changes sign at x = a, then a point of inflection occurs at x = a.

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188 Analysing the behaviour of

functions using the second

derivative

Use a graphics calculator to assist you to draw the graphs.

1 For each of the following functions find the stationary points and the nature of the stationary points using the second derivative test.

2 Sketch the graph of each function in question 1, clearly indicating all stationary points and axes intercepts.

3

The function f(x) = x4+ 4x3+ 3:

a has a point or points of inflection when x is equal to:

b has a stationary point of inflection:

4

The minimum gradient of the curve with equation y = x3− 6x2− 8x is:

5

If f′(x) = 0 when x = 3 and x = −2, and f ′′(3) = 4 and f ′′(−2) = −5, then f(x) has:

A stationary points of inflection when x = −2 and x = 3

B a stationary point of inflection when x = 3 and a local maximum when x = −2

C a local maximum when x = −2 and a local minimum when x = 3

D a local minimum when x = −2 and a local maximum when x = 3

E a stationary point of inflection when x = −2 and a local minimum when x = 3

6 Give i any stationary points of inflection and ii any other points of inflection for each of the following functions.

7 Show that y = x log_ex does not have any points of inflection.

a f(x) = x2− 4x b g(x) = 12 − x2

c y = x2(x + 2) d y = x(x − 1)(x + 2)

e h(x) = (x − 3)(x + 3)(x + 1) _f _f(x)_{= x}3_{+ 4x}2_{− 4x − 16}

g g(x) = x3 h f(x) = x3+ 3

i h(x) = x3− 3x j y = x4+ x3

A 0 and −4 B 0 only C −2 only

D 0 and −2 E 0 and A (−4, 3) B (0, 3) C (−2, −13) D (−2, 32) E (−4, 0) A −20 B −32 C 2 D 0 E −12 a y = x3− 6x b f(x) = x2− c y = d g(x) = x3+ 2x2+ 1 e y = xex f y = 2x4− x3 g f(x) = x2ex h g(x) = 8x2− log_ex

5C

WORKED Example 8a Mathc ad Stationary points WORKED Example 8b m

1 2

---m

m

WORKED Example 9 1 x --- 1 x --- 1 x2 ---–

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189

8 Show that the graph of the function g(x) = ex2 has no points of inflection. Find its stationary point.

9 Find i stationary points and ii points of inflection, and iii sketch the graphs with the following rules:

Verify with a graphics calculator.

10 Show that the maximum value of the gradient to the curve f(x) = x3+ 3x2+ 2 is −3.

11 Find the maximum value of the gradient to the curve y = 10 + 3x2− 2x3.

12 Sketch the graph of the function y = x4− x2− 12, showing all intercepts with the axes, the stationary points and any points of inflection. (Verify with a graphics calculator.)

13 The downward displacement of a meteor t seconds after hitting the surface of the ocean is given by:

d = , where 0 ≤ t ≤ 20 and d is in metres.

a Find the depth of the meteor after 10 seconds.

b Find the velocity, v, at any time t.

c Find the maximum velocity of the meteor.

d If the depth of the ocean where the meteor strikes is 600 metres and the meteor disintegrates 20 seconds after hitting the ocean, does the meteor reach the ocean floor?

a f(x) = x3− x b y = x3+ 8x c f(x) = d g(x) = WORKED Example 10 x 1 x2 ---– x2 2 x ---– WORKED Example 11 40t t 2 4 ---- t3 24 ---– + Wor kSHEET

5.1

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190 Derivatives of inverse circular functions

The derivative of Sin

−1

, a > 0

If y = Sin−1 , −a < x < a and − < y < then = sin y, − < y <

x = a sin y So = a cos y

or = , cos y ≠ 0 and − < y < Since sin y =

cos y = (from diagram at right) or a cos y =

so = , −a < x < a.

Therefore, if f (x) = Sin–1

then f′(x) = , –a < x < a.

Note: Example 12(b) could also be done by the chain rule using the substitution u = 6x.

x

a

---x a --- π 2 --- π 2 ---x 0 y y = Sin–1x – a a –a – 2 – π – 2 π x a --- π 2 --- π 2 ---dx dy ---dy dx --- 1 acosy --- π 2 --- π 2 ---x a ---x 0 a x y a2_{– x}2 a2_–_x2 a ---a2_–_x2 dy dx --- 1 a2_–_x2 ---x a ---1 a2_–_x2

---Find the derivative of: a Sin−1 b Sin−1 6x.

THINK WRITE

a Differentiate by rule where a = 4. a If y = Sin−1 then =

b Express 6x as . _b _{If y}_{= Sin}−1_6x

then y = Sin−1

Differentiate by rule where a = . =

Take = out as a factor of the denominator to remove the fraction from the square root.

=

Simplify the derivative. =

= x 4 ---x 4 ---dy dx --- 1 16_–x2 ---1 x

(

1 6 ---

)

---x

(

1 6 ---

)

---2 1₆--- dy dx --- 1 1 36 ---_–x2 ---3 1 36 --- 1 6 --- 1 1 36 ---(1_–36x2) ---4 1 1 6 --- 1_–36x2 ---6 1_–36x2

---12

WORKED

E

xample

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191 The derivative of Cos

–1

, a > 0

If y = Cos−1 , −a ≤ x ≤ a and 0 ≤ y ≤ then = cos y, 0 ≤ y ≤

x = a cos y Thus, = −a sin y

or = , sin y ≠ 0 and 0 < y < . Since cos y =

sin y = (from diagram at right) or a sin y = So = – , −a < x < a. Therefore, if f (x) = Cos–1 then f′(x) = , –a < x < a

x

a

---x a --- π x 0 y y = Cos–1 x–_a a –a π – 2 π x a --- π dx dy ---dy dx --- –1 a sin y --- π x a --- a x y a 2_{– x}2 a2_–_x2 a ---a2_–_x2 dy dx --- 1 a2_–_x2 ---x a ---1 – a2_–_x2

---Find f′(x) if f(x) is equal to: a Cos−1 b Cos−1 .

THINK WRITE

a Differentiate by rule where a = . a f(x) = Cos−1

f′(x) =

b Express as . b f(x) = Cos−1

= Cos−1

Differentiate by rule where a = . f′(x) =

Take = out as a factor of the denominator. = = Simplify f′(x). = x 3 --- 2 x 5 ---3 ₃---x 1 – 3_–x2 ---1 2x 5 --- x

(

5 2 ---

)

--- 2x 5 ---x

(

5 2 ---

)

---2 5₂--- –1 25 4 ---_–x2 ---3 1₄--- 1 2 --- –1 1 4 ---(25_–4x2) ---–1 1 2 --- 25_–x2 ---4 –2 25_–x2

---13

WORKED

E

xample

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192 The derivative of Tan

−1

If y = Tan−1 , x ∈ R and − < y < then = tan y, – < y <

x = a tan y = a sec2_y = a(1 + tan2_y)

= (from the diagram at right) = = or = Therefore, if f(x) = Tan–1 then f′(x) = , x ∈ R

x

a

---x a --- π 2 --- π 2 ---x 0 y y = Tan–1x_– a – 2 π – 2 π – x a --- π 2 --- π 2 ---x 0 a x y a2 + x 2 dx dy ---a 1 x 2 a2 ---+     a a₍ 2₊_x2₎ a2 ---a2+_x2 a ---dy dx --- a a2+_x2 ---x a ---a a2₊_x2

---Find f′(x) if f(x) is equal to: a Tan−1 b Tan−1 .

THINK WRITE

a Differentiate by rule where a = 5. a f(x) = Tan−1

Thus, f′(x) =

b Express as . b f(x) = Tan−1

= Tan−1

Differentiate by rule where a = . f′(x) =

Take out as a factor of the denominator. =

Simplify the derivative by dividing by . = x 5 --- 8 x 3 ---x 5 ---5 25₊x2 ---1 8x 3 --- x

(

3 8 ---

)

--- 8x 3 ---x

(

3 8 ---

)

---2 3₈ ---3 8 ---9 64 ---₊x2 ---3 ₆₄---1 3 8 ---1 64 ---₍9₊64x2₎ ---4 3₈--- 1 64 --- 24 9₊64x2

---14

WORKED

E

xample

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193

Find if y is equal to: a Sin−1(4x + 7) b Cos−1(5 − 3x) c sin

(

Tan−1

)

.

THINK WRITE

a Let u = 4x + 7 so the chain rule can be applied. a y = Sin−1 (4x + 7) Let u = 4x + 7.

Find . = 4

Express y in terms of u. y = Sin−1 u

Find . =

Apply the chain rule. So = ×

=

Replace u with 4x + 7. =

b Let u = 5 − 3x so the chain rule can be applied. b y = Cos−1 (5 − 3x) Let u = 5 − 3x.

Find . = −3

Express y in terms of u. y = Cos−1 u

Find . =

=

Replace u with 5 − 3x. =

c Let u = Tan−1 so the chain rule can be applied. c y = sin

(

Tan−1

)

Let u = Tan−1 .

Find . =

Express y in terms of u. y = sin u

Find . = cos u d y d x --- x 5 ---1 2 du dx --- du dx ---3 4 dy du --- dy du --- 1 1_–u2 ---5 dy dx --- dy du --- du dx ---4 1_–u2 ---6 4 1_–(4x+7)2 ---1 2 du dx --- du dx ---3 4 dy du --- dy du --- –1 1_–u2 ---5 dy_dx--- _du---dy du_dx ---3 1_–u2 ---6 3 1_–(5–3x)2 ---1 x 5 --- x 5 ---x 5 ---2 du dx --- du dx --- 5 25+x2 ---3 4 dy du --- dy du

---15

WORKED

E

xample

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194

THINK WRITE

=

Replace u with Tan−1 .

= 5 dy dx --- dy du --- du dx ---5 cos u 25+x2 ---6 x 5 --- 5 Tan –1x 5 ---    cos 25₊x2

---Find the equation of the normal to the curve with equation:

y = 2 Cos−1 at the point where x = .

THINK WRITE

Find y when x = . y = 2 Cos−1

When x =

y = 2 Cos−1

= 2 =

Find . =

Substitute x = into to find the gradient of the tangent at x = .

When x = , the gradient of the tangent is =

= = −2

Find the gradient of the normal

.

The gradient of the normal = =

Substitute x₁= , y₁= and m = into the equation of a straight line rule: (y − y1) = m(x − x1) where m is the gradient and (x₁, y₁) is a point on the line.

Therefore, the equation of the normal is:

y − = (x − ) =

Simplify the equation. _y₌

(or 6y = 3x + 2 − 3 ) x 2 --- 3 1 3 x 2 ---3 3 2 ---π 6 ---    π 3 ---2 dy dx --- dy dx --- –2 4_–x2 ---3 3 dy dx ---3 3 dy dx --- –2 4–3 ---2 – 1 ---4 gradient of normal –1 gradient of tangent ---=     1 – 2 – ---1 2 ---5 3 π 3 --- 1 2 ---π 3 --- 1 2 --- 3 x 2 --- 3 2 ---– 6 x 2 --- π 3 --- 3 2 ---– + π 3

16 WORKED

E

xample

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195 Derivatives of inverse circular

functions

1 Find the derivative of each of the following expressions with respect to x:

2 Find if y is equal to:

3 Using the results of question 2, or otherwise, state the derivative of: where b is a real, positive constant.

4

Consider the function with the rule f (x) = Sin−1 .

a The maximal domain of f (x) is:

a Sin−1 b Sin−1 c Sin−1

d Cos−1 e Cos−1 f Cos−1

g Tan−1 h Tan−1 i Tan−1

j Tan−1 k Sin−1 l Sin−1

m Cos−1 n Cos−1 o Tan−1

p Tan−1 q Sin−1 r Tan−1

a Sin−1 2x b Sin−1 5x c Sin−1 3x d Sin−1 8x

e Cos−1 4x f Cos−1 6x g Cos−1 7x h Cos−1 10x

i Tan−1 3x j Tan−1 9x k Tan−1 4x l Tan−1 5x

a Sin−1 bx b Cos−1 bx c Tan−1 bx

A

[

− ,

]

B

[

− ,

]

C [−1, 1]

D [−3, 3] E

[

− ,

]

remember

1. If f (x) = Sin−1 then f′(x) = , −a < x < a

2. If f (x) = Cos−1 then f′(x) = , −a < x < a

3. If f (x) = Tan−1 then f′(x) = , x ∈ R x a --- 1 a2_–_x2 ---x a --- –1 a2_–_x2 ---x a --- a a2+_x2

---remember

5D

Ma_th cad Inverse trig. derivatives WORKED Example 12a x 2 --- x 5 --- x 8 ---WORKED Example 13a x 3 --- x 4 --- x 6 ---WORKED Example 14a x 2 --- x 4 --- x 7 ---x 3 --- x 5 --- x 0.2 ---x 2.5 --- x 7 --- x 3 ---x 0.8 --- x 6 --- x 10 ---dy dx ---WORKED Example 12b m

x 3 ---π 2 --- π 2 --- 3π 2 --- 3π 2 ---1 3 --- 1 3

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---196

b The maximal domain of f′(x) is:

5

Let f (x) = Cos−1 .

a When expressed in the form f (x) = Cos−1 , the value of a is:

b f′(x) is equal to:

6

The derivative of Tan−1 is equal to:

7 Find f′(x) if f (x) is equal to:

8 Using the results of question 7, or otherwise, state the derivative of y with respect to x:

where a and b are real, positive constants.

9 Find if y is equal to:

A

[

− ,

]

B [−3, 3] C (−3, 3) D [−1, 1] E (−1, 1) A B 7 C D E 3 A B C D E A B C D E

a Cos−1 b Cos−1 c Cos−1 d Cos−1

e Tan−1 f Tan−1 g Tan−1 h Tan−1

i Sin−1 j Sin−1 k Sin−1 l Sin−1

a y = Sin−1 b y = Cos−1 c y = Tan−1

a Sin−1(2x + 3) b Sin−1(3x − 5) c Cos−1(4x − 3)

d Cos−1(5x + 8) e Tan−1(3x + 2) f Tan−1(6x − 7)

g Sin−1 h Cos−1 i Tan−1

j Sin−1(4 − 3x) k Cos−1(7 − 2x) l Tan−1(8 − 5x)

m Sin−1 n Cos−1 o Tan−1 3π 2 --- 3π 2 ---m

3x 7 ---x a ---    7 3 --- 3 7 --- 1 7 ---7 – 49_–9x2 --- –7 9_–49x2 --- –3 9_–49x2 ---1 – 49_–x2 --- –3 49_–9x2 ---m

8x 5 ---40 25₊64x2 --- 64 64₊25x2 --- 40 64₊25x2 ---5 25+x2 --- 8 64+x2 ---WORKED Example 13b 3x 4 --- 7x 4 --- 9x 5 --- 5x 8 ---WORKED Example 14b 4x 5 --- 3x 8 --- 7x 2 --- 9x 5 ---2x 3 --- 5x 2 --- 6x 7 --- 8x 5 ---bx a --- bx a --- bx a ---WORKED Example 15 dy dx ---x+3 2 ---    2x+1 3 ---    4x–3 5 ---    3–4x 5 ---    6–3x 7 ---    2–3x 4 ---   

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197

10 Find the derivative of each of the following expressions with respect to x.

11 a Find the gradient of the graph of y = Sin−1 at the origin.

b Hence, find the equation of the tangent to this curve at the origin.

12 Find the equation of the normal to the curve y = Cos−1 2x at the point where it crosses the y-axis.

13 Find the equation of the tangent to the curve y = Sin−1 when x = 1.

14 a Find the coordinates of the point where the maximum gradient of f (x) = Tan−1 occurs (a is a constant).

b Find the maximum gradient of f (x).

1. To find the gradient of Tan–1x at x = 1, press , select 8:nDeriv(, enter

tan–1(X),X,1) and press . Check that

MODE is set to radians.

Alternatively, if the function is in Y1, then press , select 8:nDeriv(, enter Y1,X,1) and press . Remember: To enter Y1, press , select Y–VARS, 1:Function..., and 1:Y1

(similarly any Y variable).

2. To increase the accuracy, add an extra argument for h in nDeriv, for example h = 10–10. Press and (using the above method) set up

8:nDeriv(Y1,X,1,E–10) and press . To

enter E for E–10, press [EE].)

(Note: The inbuilt method actually calculates the gradient of a chord distance h either side of the point required. The default value of h is 0.001.)

a x3+ Cos−1 2x b 4x2− Sin−1

c sin 4x + Tan−1 d log_e 6x + x − Cos−1

e e7x + 3+ Tan−1 5x + 3 _f

g

(

Tan−1

)

4 _h 3x Cos−1 2x

i Sin−1 x + Cos−1 x j sin (Cos−1 x)

k tan

(

Sin−1

)

l cos

(

Tan−1

)

x 3 ---x 4 --- 92--- 2x 3 ---Sin–1x x 2 ---x 3 --- x 4 ---x 4 ---WORKED Example 16 x 2 ---x a

---Graphics Calculator

Graphics Calculator

tip!

_tip!

Finding numerical derivatives

MATH ENTER MATH ENTER VARS MATH ENTER 2nd

(28)

198 Antidifferentiation involving inverse

circular functions

We now know that:

(

Sin−1

)

= , −a < x < a

(

Cos−1

)

= , −a < x < a

(

Tan−1

)

= , x ∈ R It therefore follows that:

dx = Sin–1 + c, a > 0

dx = Cos–1 + c, a > 0

dx = Tan–1 + c

When finding antiderivatives of inverse circular functions, the integrand should be expressed in one of the standard forms above and then integrated.

d dx --- x a --- 1 a2_–_x2 ---d dx --- x a --- –1 a2_–_x2 ---d dx --- x a --- a a2+_x2 ---1 a2_–_x2

---∫

x a ---1 – a2_–_x2

---∫

x a ---a a2+_x2

---∫

x a

---Differentiate Tan−1 and hence find dx.

THINK WRITE

Write the equation. y = Tan−1

Differentiate Tan−1 by rule where a = 2. =

Express the result using integral notation. Therefore dx = Tan−1 + c.

x 2 --- 2 4+x2

---∫

1 x 2 ---2 x 2 --- dy dx --- 2 4+x2 ---3 2 4₊x2

---∫

x 2

---17

WORKED

E

xample

Find the antiderivative for each of the following expressions:

a b c .

THINK WRITE

a The antiderivative is an inverse sine function of the form Sin−1 where a = 5.

a dx = Sin−1 + c 1 25_–x2 --- –3 49_–x2 --- 20 16₊x2 ---x a ---1 25_–x2

---∫

x 5

---18

WORKED

E

xample

(29)

199

THINK WRITE

b Take 3 out as a factor. b dx = 3 dx

The antiderivative of

is an inverse cos function of the form Cos−1 where a = 7.

= 3 Cos−1 _{+ c}

(or −3 Sin−1 + c)

c Take 5 out as a factor. c dx = 5 dx

The antiderivative of

is an inverse tan function of the form Tan−1 where a = 4. = 5 Tan−1 _{+ c} 1 –3 49_–x2

---∫

–1 49_–x2

---∫

2 –1 49_–x2 ---x a ---x 7 ---x 7 ---1 20 16₊x2

---∫

4 16₊x2

---∫

2 4 16+x2 ---x a ---x 4

---Find each of the following indefinite integrals:

a dx b dx c dx.

THINK WRITE

a Use the substitution u = 5x. a dx Let u = 5x.

Find . = 5

Make dx the subject. dx =

Rewrite the integral in terms of u. dx =

= du

Antidifferentiate by rule. = Sin−1 u + c

Replace u with 5x. = Sin−1 5x + c

1 1_–25 x2

---∫

–1 9_–16 x2

---∫

8 9₊5 x2

---∫

1 1 1_–25x2

---∫

2 du dx --- du dx ---3 du---₅ 4 1 1_–25x2

---∫

1 1_–u2

---∫

du 5 ---1 5 --- 1 1_–u2

---∫

5 1₅ ---6 1₅

---19

WORKED

E

xample

(30)

200

THINK WRITE

b Use the substitution u = 4x. b dx Let u = 4x.

Find . = 4

Make dx the subject. or dx =

= du

Antidifferentiate by rule. = Cos−1 + c

Replace u with 4x. = Cos−1 + c

c Use the substitution u = x. c dx

Let u = .

Find . =

Make dx the subject. dx =

= du

Express the integral in standard form. = du

Antidifferentiate by rule. =

Replace u with x and simplify the surd factor. = 1 –1 9_–16x2

---∫

2 du dx --- du dx ---3 du 4 ---4 –1 9_–16x2

---∫

–1 9_–u2 --- du 4

---∫

1 4 --- –1 9_–u2

---∫

5 1₄--- u 3 ---6 1₄--- 4x 3 ---1 5 8 9₊5x2

---∫

5x 2 du dx --- du dx --- 5 3 du 5 ---4 8 9₊5x2

---∫

8 9+u2 --- du 5 ---×

∫

8 5 --- 1 9 + u2

---∫

5 8 3 5 --- 3 9+u2

---∫

6 8 3 5 --- Tan–1u 3 ---+c 7 5 8 5 15 --- Tan–1 5x 3 ---+c

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201

Antiderivatives involving all three inverse trigonometric functions can be obtained using the Mathcad file ‘Antidifferentiation involving trigonometric functions’. The screen at right shows the answer for part (c) of worked example 19.

Ma_th cad

Antiderivatives

Find the antiderivative of by first simplifying the rational expression.

THINK WRITE

Divide x2+ 4 into x3+ 4x – 16 to make the rational expression antidifferentiable.

Simplify by long division:

x3+ 4x

– 16

so = x –

Rewrite dx as two separate integrals.

dx = dx

Express dx in standard form. = x dx – 8 dx

Antidifferentiate both integrals. = x2 – 8 Tan–1 + c

x3+4 x–16 x2+4 ---1 x 3 4x–16 + x2+4 ---x x2+4 x3+4x–16 x3+4x–16 x2+4 --- 16 x2+4 ---2 x 3 4x–16 + x2+4

---∫

x3+4x–16 x2+4

---∫

x xd – 16 x2+4

---∫

∫

3 16 x2+4

---∫

∫

2 x2+4

---∫

4 1₂--- x 2

---20

WORKED

E

xample

remember

1. dx = Sin−1 + c, a > 0 2. dx = Cos−1 + c, a > 0 3. dx = Tan−1 + c 1 a2_–_x2

---∫

x a ---1 – a2_–_x2

---∫

x a ---a a2+_x2

---∫

x a