Chapter 7

Trigonometric Functions

This chapter starts with graphs of sine functions. Then graphs of cosine, tangent and other trigonometric functions are shown. After that, the inverse of those functions and their graphs come. The essential part of this chapter is the differentiation of trigonometric functions.

7.1 Graphs of Sine Functions

Graph of the Sine Function y = sin x

Domain: The set ℝ of all real numbers. Range: {y| -1 ≤ y ≤ 1}

$$\scriptsize{ \text{Five key points:} \begin{cases}x\text{-intercepts:} (0,0), (\pi ,0) (2\pi , 0)\\ \text{maximum point:} (\dfrac{\pi}{2},1) \quad \text{minimum point:} (\dfrac{3\pi}{2},-1) \end{cases} } \,$$

Periodic function: If f(x) = f(x + p) where p is a positive real number, then the function F is called a periodic function. If p is the smallest such number, then p is the period of function f. The sine function y = sin x is a periodic function with period 2π, since 2π is the smallest positive real number such that sin x = sin (x + 2x). The amplitude of a periodic function is half the difference between the maximum and the minimum values. So the amplitude of the sine function y = sin x is ^{1 - (-1)}⁄₂ = 1.

Graph of the Sine Function y = a sin bx, a > 0, b > 0

In the following example, we will see how to obtain the graph of y = a sin bx, where a > 0 and b > 0, from the graph of y = sin x.

Example 1.
From the graph of y = sin x, draw step-by-step transformation graph to get the graph of y = 2 sin ^π⁄₂ x.
Solution
Method 1
$$y = \text{sin}\, x \xrightarrow[\text{scale factor 2}] {\text{vertical scaling}} \, y = \text{2 sin}\, x $$ fig 7.1-2

$$y = \text{sin}\, x \, \xrightarrow[\text{scale factor 2}]{\text{vertical scaling}} \, y = \text{2\, sin} \, x \, \xrightarrow[\text{scale factor}\, \tiny{\dfrac{1}{\dfrac{\pi}{2}} = \dfrac{2}{\pi} }]{\text{horizontal scaling}} y = \text{2 sin} \dfrac{\pi}{2} x \, $$

Method 2
$$y = \text{sin}\, x \, \xrightarrow[\text{scale factor}\, \tiny{\dfrac{1}{\dfrac{\pi}{2}} = \dfrac{2}{\pi} }]{\text{horizontal scaling}} y = \text{sin} \dfrac{\pi}{2} x $$ fig 7.1-4

$$y = \text{sin}\, x \, \xrightarrow[\text{scale factor}\, \tiny{\dfrac{1}{\dfrac{\pi}{2}} = \dfrac{2}{\pi} }]{\text{horizontal scaling}} y = \text{sin}\, \dfrac{\pi}{2} \, x = \xrightarrow[\text{scale factor 2}]{\text{vertical scaling}} \, \text{2 sin} \dfrac{\pi}{2} x \,$$

From the graph of y = sin x, the graph of y = a sin bx, a > 0, b > 0, can be obtained as

$$y = \text{sin}\, x \, \xrightarrow[\text{scale factor }a]{\text{vertical scaling}} \, y = a\, \text{sin} \, x \, \xrightarrow[\text{scale factor}\, \tiny{\dfrac{1}{b}} ] {\text{horizontal scaling}} y = a\, \text{sin} \, bx \, \, $$

$$y = \text{sin}\, x \, \xrightarrow[\text{scale factor}\, \tiny{\dfrac{1}{b}} ] {\text{horizontal scaling}} \, y = \text{sin} \, bx \, \xrightarrow[\text{scale factor }a]{\text{vertical scaling}} y = a\, \text{sin} \, bx \, $$

y = sin x → y = a sin bx
(x,y) → (^x⁄_b, ay)

$$\scriptsize{ \text{Five key points:} \begin{cases}x\text{-intercepts:} (0,0), (\dfrac{\pi}{b} ,0), (\dfrac{2\pi}{b} , 0)\\ \text{maximum point:} (\dfrac{\pi}{2b},a) \quad \text{minimum point:} (\dfrac{3\pi}{2b},-a) \end{cases} } \,$$

Graph of the Sine Function y = -a sin bx, a > 0, b > 0

$$y = a\, \text{sin}\, bx\, \xrightarrow{\text{Reflection on the}\, x\text{-axis}}\, y = -a\, \text{sin}\, bx \,$$

Example 2
Draw the graph of y = 2 sin ^π⁄₂ x and y = -2 sin ^π⁄₂ x.
Solution
fig 7.1-8

Note that y = a sin (-b)x = -a sin bx, so no need to consider b < 0.

Graph of the Sine Function y = a sin b(x - h) + k

$$\small{y = a \, \text{sin}\, bx\, \xrightarrow[\text{vertical translation}\, k\, \text{units}] {\text{horizontal translation}\, h\, \text{units}} \, y}$$ $$\small{= a \, \text{sin} \, b(x - h) + k }$$

y = sin x → y = a sin bx → y = a sin b(x - h) + k
(x, y) → (^x⁄_b, ay) → ( ^x⁄_b + h, ay + k)

Key points on the midline y = k

(0,0) → (0, 0) → (h, k)
(π, 0) → (^π⁄_b, 0) → (^π⁄_b + h, k)
(2π, 0) → (^2π⁄_b, 0) → (^2π⁄_b+h, k)

Maximum and minimum points

(^π⁄₂,1) → (^π⁄_2b, a) → (^π⁄_2b+h, a+k)
(^3π⁄₂, -1) → (^3π⁄_2b, -a) → (^3π⁄_2b+h, -a+k)

Example 3.
Draw the graph of y = 3 sin 2 (x - 1) + 4.
Solution
Midline: y = 4,
Amplitude: 3,
Period: π
Five key points: (1, 4), (^π⁄₄+1, 7), (^π⁄₂+1, 4), (^3π⁄₄+1, 1), (π+1, 4)
fig 7.1-10

Example 4.
Draw the graph of y = -3 sin π(x + 1) - 2.
Solution
Midline: y = -2,
Amplitude: 3,
Period: 2
Five key points: (-1, -2), (- ¹⁄₂, -5), (0, -2), (¹⁄₂, 1), (1, -2)
fig 7.1-11

Exercise 7.1
1. From the graph of y = sin x, draw step-by-step transformation graphs to get the graph of y = -3 sin ^π⁄₃ x.
2. Draw the graph of (a) y = ¹⁄₂ sin x
(b) y = sin 4x.
3. Draw the graph of (a) y = 2 sin ^π⁄₄ x
(b) y = -sin πx.
4. Draw the graph of (a) y = 2 sin ^π⁄₃(x-2) + 1
(b) y = -2 sin ¹⁄₂(x+1) + 2.
5. Show that y = a sin bx is an odd function.

7.2 Graphs of Cosine Functions

Graph of the Coine Function y = cos x

Domain: The set ℝ of all real numbers. Range: {y| -1 ≤ y ≤ 1}

$$\scriptsize{ \text{Five key points:} \begin{cases}x\text{-intercepts:} (\dfrac{\pi}{2},0), (\dfrac{3\pi}{2} , 0)\\ \text{maximum point:} (0,1), (2\pi, 1) \quad \text{minimum point:} (\pi,-1) \end{cases} } $$

$$\text{Note\ that\ sin} (x + \dfrac{\pi}{2}) = cos \ x ,$$ $$y = \text{sin}\, x \xrightarrow[- \dfrac{\pi}{2} \, \text{units}] {\text{horizontal translation}} \, y = \text{cos}\, x $$

Graph of the Cosine Function y = a cos bx, a > 0, b > 0

From the graph of y = cos x, the graph of y = a cos bx can be obtained as

$$\scriptsize{y = a \, \text{sin}\, bx\, \xrightarrow[\text{horizontal scaling, scale factor}\, \tiny{\dfrac{1}{b}}] {\text{vertical scaling, scale factor}\, a} \, y = a \, \text{cos}\, bx}$$ $$\scriptsize{(x,y) \rarr \, (\dfrac{x}{b}, ay) }$$

$$\scriptsize{ \text{Five key points:} \begin{cases}x\text{-intercepts:} (\dfrac{\pi}{2b},0), (\dfrac{3\pi}{2b} , 0)\\ \text{maximum point:} (0,a), (\dfrac{2\pi}{b}, a) \quad \text{minimum point:} (\dfrac{\pi}{b},-a) \end{cases} } $$

Graph of the Cosine Function y = -a cos bx, a > 0, b > 0

$$y = a\, \text{cos}\, bx\, \xrightarrow{\text{Reflection on the}\, x\text{-axis}}\, y = -a\, \text{cos}\, bx $$

Example 5.
Draw the graphs of y = ¹⁄₂ cos 2x and y = - ¹⁄₂ cos 2x
Solution
fig 7.2-3

Note that y = a cos (-b)x = a cos bx, so no need to consider b < 0.

Graph of the Cosine Function y = a cos b(x - h) + k , a > 0, b > 0

$$\small{y = a \, \text{cos}\, bx\, \xrightarrow[\text{vertical translation}\, k\, \text{units}] {\text{horizontal translation}\, h\, \text{units}} \, y}$$ $$\small{= a \, \text{cos} \, b(x - h) + k }$$

y = cos x → y = a cos bx → y = a cos b(x - h) + k
(x, y) → (^x⁄_b, ay) → ( ^x⁄_b + h, ay + k)

Five key points: points on the midline : (^π⁄_2b + h, k), (^3π⁄_2b + h, k)
maximum points: (h, a + k), (^2π⁄_b + h, a+k) minimum point: (^π⁄_b + h, -a+k)

Example 6.
Draw the graph of y = 2 cos ^π⁄₃ (x + 1) + 3 and y = -2 cos ^π⁄₃ (x + 1) + 3.
Solution
Midline: y = 3,
Amplitude: 2,
Period: 6
Five key points for y = 2 cos ^π⁄₃ (x + 1) + 3: (-1, 5), (0.5, 3), (2, 1), (3.5, 3), (5, 5)
Five key points for y = - 2 cos ^π⁄₃(x + 1) + 3: (-1, 1), (0.5, 3), (2, 5), (3.5, 3), (5, 1)
fig 7.2-5

Exercise 7.2
1. From the graph of y = cos x, draw step-by-step transformation graphs to get the graph of y = -3 cos ^π⁄₂ x.

2. Draw the graph of (a) y = ¹⁄₂ cos x
(b) y = cos 4x.

3. Draw the graph of (a) y = 2 cos ^π⁄₄ x
(b) y = -sin πx.

4. Draw the graph of (a) y = 2 cos π(x-2) + 1
(b) y = -2 cos ¹⁄₂(x+1) + 2.

5. Show that y = cos bx is an even function.

7.3 Graphs of Other Trigonometric Functions

Graph of y = 10 x and y = cot x

Domain: x ≠ ± ^π⁄₂, ± ^3π⁄₂, ...
Range: ℝ
Period: π
x-intercepts: x = 0, ±&pi, ±2π, ...
Asymptotes: x = ±^π⁄₂, ± ^3π⁄₂, ...
fig 7.3-2

Domain: x ≠ 0, ±π, ±2π, ...
Range: ℝ Period: π
x-intercepts: x = ±^π⁄₂, ± ^3π⁄₂, ...
Asymptotes: x = 0, ±π, ±2π, ...
Example 7.
Draw the graphs of y = a tan bx and y = a cot bx for a, b > 0.

Solution
y = tan x → y = a tan bx,
(x, y) → (^x ⁄_b, ay)
fig 7.3-3

y = cot x → y = a cot bx
(x, y) → (^x ⁄_b, ay)
fig 7.3-4

Graph of y = sec x and y = csc x

Domain: x ≠ ± ^π⁄₂, ± ^3π⁄₂, ...
Range: y ≤ -1 or y ≥ 1
Period: 2π
x-intercepts: x = 0, ±π, ±2π, ...
Asymptotes: x = ±^π⁄₂, ± ^3π⁄₂, ...
fig 7.3-6

Domain: x ≠ 0, ±π, ±2π, ...
Range: y ≤ -1 or y ≥ 1
Period: 2π
x-intercepts: x = ±^π⁄₂, ± ^3π⁄₂, ...
Asymptotes: x = 0, ±π, ±2π, ...
Note that transform y = sec x to y = a sec bx and y = csc x to y = a csc bx with (x, y) → (^x⁄_b, ay), we get the graph of y = a sec bx and y = a csc bx.

Exercise 7.3
1. Draw the graph of (a) y = ¹⁄₂ tan πx
(b) y = 2 tan ¹⁄₂ x.

2. Draw the graph of (a) y = 2 cot ^π⁄₃ x
(b) y = 3 cot 2x.

3. Draw the graph of (a) y = 2 sec πx

(b) y = 2 csc ¹⁄₂ x.

4. Show that y = a tan bx , y = a cot bx, and y = a csc bx are odd functions.

5. Show that y = a sec bx is an even function.

7.4 Inverse Trigonometric Functions

Trigonometric functions are not one-to-one functions. So we need to restrict the domain of each function to be a one-to-one function to get the inverse trigonometric functions. The following graphs shows that y = sin x, y = cos x, and y = tan x are one-to-one in the restricted domains.

In these restricted domains, we can define the inverse functions y = sin^-1x, y = cos^-1x and y = tan^-1x.
fig 7.4-1

y = sin^-1x if sin y = x for -1 ≤ x ≤ 1, - ^π⁄₂ ≤ y ≤ ^π⁄₂
y = cos^-1 x if cos y = x for -1 ≤ x ≤ 1, 0 ≤ y ≤ π
y = tan^-1 x if tan y = x for x ∈ R

We can define the other trigonometric functions in the same manner.

y = cot ^-1x if cot y = x for x ∈ R, 0 < y < π
y = sec^-1 x if sec y = x for |x| ≥ 1, 0 ≤ y ≤ π, y ≠ ^π⁄₂
y = csc^-1 x if csc y = x for |x| ≥ 1, - ^π⁄₂ ≤ y ≤ ^π⁄₂, y ≠ 0

Example 8.
Evaluate (a) sin^-1 ¹⁄₂
(b) cos^-1 - (¹⁄₂)
$$\small{\text{(c) tan}^{-1} (-\sqrt{3})}$$ (d) sin^-1 (sin ^5π⁄₆ ).
Solution
(a) Since sin ^π⁄₆ = ¹⁄₂ and - ^π⁄₂ ≤ ^π⁄₆ ≤ ^π⁄₂,
we have sin^-1 ¹⁄₂ = ^π⁄₆.

(b) Since cos ^2π⁄₃ = - ¹⁄₂ and
0 ≤ ^2π⁄₃ ≤ π, we have
cos^-1 (- ¹⁄₂) = ^2π⁄₃.

(c) Since tan(- ^π⁄₃) = - √ 3 and
- ^π⁄₂ ≤ - ^π⁄₃ ≤ ^π⁄₂, we have
tan^-1 (- √ 3 ) = - ^π⁄₃.

(d) sin^-1(sin ^5π⁄₆) = sin^-1 (¹⁄₂) = ^π⁄₆.

From y = sin^-1 x if sin y = x, we have sin(sin^-1 x) = x for the domain -1 ≤ x ≤ 1 of y = sin^-1 x, but from Example 8(d) one can see that
sin^-1(sin x) = x is not true in general.
Since sin(-sin^-1x) = -sin(sin^-1x) = -x, we have

sin^-1(-x) = -sin^-1x

so y = sin^-1x is an odd function.

From y = cos^-1x if cos y = x, we have cos(cos^-1x) = x for the domain -1 ≤ x ≤ 1 of y = cos^-1x. Since cos(π - cos^-1x) = - cos(cos^-1x) = -x,

π - cos^-1x = cos^-1(-x)

cos^-1x + cos^-1(-x) = π

Exercise 7.4
1. Evaluate (a) csc(sin^-10.3)
(b) cot(tan^-1 5)
(c) sec(cos^-1(-0.75)).

2. Evaluate (a) sin^-1 ^{√ 3}⁄₂
(b) cos^-1 ^{√ 2}⁄₂
(c) tan^-1 ^{√ 3}⁄₃

3. Evaluate (a) sin(cos^-1 ^{√ 3}⁄₂ )
(b) tan(sin^-1 ^{√ 3}⁄₂)
(c) csc(tan^-1(- ^{√ 3}⁄₃ ))

4. Evaluate (a) sin^-1(sin ^2π⁄₃)
(b) cos^-1(cos(- ¹⁄₂))
(c) tan^-1(tan(- √ 3 ))

5. Prove that sin^-1 x + cos^-1 x = ^π⁄₂

7.5 Differentiation of Trigonometric Functions

Before we study the differentiation of trigonometric functions, we first evaluate an important limit, $$\small{\lim_{x\to 0} \frac{sin\, x}{x}.} $$

Consider the unit circle in the following figure with radii OA and OB with OA = OB = 1 and ∠AOB = x radians.
Obviously, BD < arc AB < AC. In right ▵OBD, BD = OB sin x = sin x.

fig 7.5-1

In right ∆OAC, AC = OA tan x = tan x.
Length of arc AB = OB ⋅ x = x, we have
sin x < x < tan x
^{sin x}⁄_{sin x} < ^x⁄_{sin x} < ^{tan x}⁄_{sin x}
1 < ^x⁄_{sin x} < ¹⁄_{cos x}
1 > ^{sin x}⁄_x > cos x

When x → 0, cos x → 1.
$$\small{ \text{Since}\, \lim_{x\to 0}1 = 1 \, \text{and} \, \lim_{x\to0}\, cos \, x\, = \, 1, } $$ $$\small{ \lim_{x\to0} \frac{sin \, x}{x} = 1 } $$ Derivative of sin x
Let y = sin x.
y + δy = sin(x + δx)
∴ δy = sin(x + δx) - sin x
= 2 cos(x + ^δx⁄₂) ⋅ sin ^δx⁄₂ (sin α - sin β = 2cos ^{α + β}⁄₂ ⋅ sin ^{α - β}⁄₂)
$$\small{ \frac{dy}{dx} = \lim_{\delta x\to0} \frac{\delta y}{\delta x} } $$ $$\small{ = \lim_{\delta x\to0} \frac{2 \text{cos}(x + \frac{\delta x}{2})\, \sdot\, \text{sin} \frac{\delta x}{2}} {\delta x} }$$ $$\small{= \lim_{\delta x\to0} \text{cos}(x + \frac{\delta x}{2})\, \sdot\, \lim_{\delta x\to0} \frac{\text{sin} \frac{\delta x}{2}}{\frac{\delta x}{2}} }$$ = cos x × 1 (when δx→ 0, ^δx⁄₂ → 0)
^d⁄_dx sin x = cos x
In general,
^d⁄_dx sin u(x) = cos u(x) ⋅ ^d⁄_dx u(x).

Derivative of cos x
Since cos x = sin(^π⁄₂ + x)
^d⁄_dx cos x = ^d⁄_dx sin(^π⁄₂ + x)
= cos(^π⁄₂ + x) ⋅ ^d⁄_dx(^π⁄₂ + x)
= - sin x × 1 (∵ cos(^π⁄₂ + x) = - sin x)
^d⁄_dx cos x = - sin x
In general,
^d⁄_dx cos u(x) = - sin u(x) ⋅ ^d⁄_dx u(x).

Derivative of tan x
Since tan x = ^{sin x}⁄_{cos x}
^d⁄_dx tan x = ^d⁄_dx ^{sin x}⁄_{cos x}
= ^{cos x ⋅ ^d⁄_dx sin x - sin x
⋅ ^d⁄_dx cos x}⁄_{cos² x}
= ^{cos x ⋅ cos x - sin x ⋅ (- sin x)}⁄_cos²x (quotient formula)
^{cos² x + sin² x}⁄_{cos² x}
= ¹⁄_{cos² x} (cos² x + sin² x = 1)
= sec² x (sec x = ¹⁄_{cos x})
∴ ^d⁄_dx tan x = sec² x
In general,
^d⁄_dx tan u(x) = sec² u(x) ⋅ ^d⁄_dx u(x).
Similarly, we can easily find the formulas for the derivatives of cot x, sec x and csc x.

Formulas for derivatives of trigonometric functions

1 ^d⁄_dx sin (x) = cos x , ^d⁄_dx sin u(x) = cos u(x). ^d⁄_dx u(x).

2 ^d⁄_dx cos (x) = -sin x , ^d⁄_dx cos u(x) = -sin u(x). ^d⁄_dx u(x).

3 ^d⁄_dx tan (x) = sec² x , ^d⁄_dx tan u(x) = sec² u(x). ^d⁄_dx u(x).

4 ^d⁄_dx cot (x) = -cosec² x , ^d⁄_dx cot u(x) = -cosec² u(x). ^d⁄_dx u(x).

5 ^d⁄_dx sec (x) = sec x. tan x , ^d⁄_dx sec u(x) = sec u(x). tan u(x). ^d⁄_dx u(x).

6 ^d⁄_dx csc (x) = -csc x. cot x , ^d⁄_dx csc u(x) = -csc u(x). cot u(x). ^d⁄_dx u(x).

Example 9.
Differentiate the following functions with respect to x.
(a) sin 5x
(b) cos (7x² - 2)
(c) tan(6x + 7)
(d) 5 sec(3x + 1)
(e) ^{cot(1 - 2x)}⁄₃
(f) -2 csc 3x
Solution
(a) ^d⁄_dx sin 5x = cos 5x ⋅ ^d⁄_dx 5x
= cos 5x ⋅ 5 = 5 cos 5x

(b) ^d⁄_dx cos (7x² - 2) = - sin(7x² - 2) ⋅ ^d⁄_dx (7x² - 2)
= - sin (7x² - 2) ⋅ 14x

(c) ^d⁄_dx tan(6x + 7) = sec²(6x + 7) ⋅ ^d⁄_dx (6x + 7)
= sec² (6x + 7) ⋅ 6

(d) ^d⁄_dx 5 sec(3x + 1) = 5⋅ sec(3x + 1) ⋅ tan(3x + 1) ⋅ ^d⁄_dx (3x + 1)
= 5⋅ sec(3x + 1) ⋅ tan(3x + 1) ⋅ 3
= 15 ⋅ sec(3x + 1) ⋅ tan(3x + 1)

(e) ^d⁄_dx ^{cot(1 - 2x)}⁄₃ = ¹⁄₃ ⋅ ^d⁄_dx cot(1 - 2x)
= ¹⁄₃ ⋅ - csc²(1 - 2x) ⋅ ^d⁄_dx (1 - 2x)
= - ¹⁄₃ ⋅ csc²(1 - 2x) ⋅ (-2)
= ²⁄₃ ⋅ csc²(1 - 2x)

(f) ^d⁄_dx (-2 csc 3x) = -2 (-csc 3x ⋅ cot 3x) ⋅ ^d⁄_dx 3x
= 2 csc 3x ⋅ cot 3x ⋅ 3
= 6 csc 3x ⋅ cot 3x

Example 10.
Find ^dy⁄_dx.
(a) y = sin² x
(b) y = cos √ x
(c) y = tan² (x²)
(d) y = sin 2x - x cos x
(e) y = sin x ⋅ cos² x
(f) y = ^x ⁄_{tan x}
(g) y = √ x + 10x

Solution
(a) y = sin² x
^dy⁄_dx = ^d⁄_dx sin²x
= 2 sin x ^d⁄_dx sin x
= 2 sin x cos x

(b) y = cos √ x
^dy⁄_dx = ^d⁄_dx cos √ x
= - sin √ x ^d⁄_dx x^{^¹⁄₂}
= - sin √ x ⋅ ¹⁄₂ x^{^{(
¹⁄₂ - 1)}}
= - sin √ x ⋅ ¹⁄₂ x^{^{( -
¹⁄₂)}}

(c) y = tan² (x²)
^dy⁄_dx = 2 tan (x²) ^d⁄_dx tan (x²)
= 2 tan (x²).sec² (x²) ^dy⁄_dx x²
= 2 tan (x²).sec² (x²).2x

(d) y = sin 2x - x cos x
^dy⁄_dx = cos 2x. ^d⁄_dx 2x - [x.^d⁄_dx cos x + cos x ^dx⁄_dx ]
= 2 cos 2x - [x (-sin x) + cos x.1]
= 2 cos 2x + x sin x - cos x
(e) y = sin x.cos² x
^dy⁄_dx = sin x. ^d⁄_dx (cos² x) + cos² x. ^d⁄_dx (sin x)
= sin x.2 cos x. ^d⁄_dx (cos x) + cos² x.cos x
= sin x. 2 cos x.(-sin x) + cos² x. cos x
= -2 sin² x.cos x + cos³ x
(f) y = ^x⁄_{3 tan x}
^dy⁄_dx = ^{tan x.
^dx⁄_dx - x.
^d⁄_dx tan x} ⁄_{(tan x)²}
= ^{tan x.1 - x sec² x} ⁄_{tan² x}
= ^{tan x - x sec² x} ⁄_{tan² x}

(g) y = √ x + sin x = (x + sin x)^¹/₂
^dy⁄_dx = ¹⁄₂ (x + sin x)^{^-1/₂} . ^d⁄_dx (x + sin x)
= ¹⁄₂ (x + sin x)^{^-1/₂} . (1 + cos x)
= ^{1 + cos x}⁄ _{2 √

x + sin x}

Example 11.
Given that x + sin y = cos(xy) , find ^dy⁄_dx .
Solution
x + sin y = cos(xy) , find ^dy⁄_dx .
Differentiate with respect to x,
1 + cos y ⋅ ^dy⁄_dx = -sin xy ⋅ ^d⁄_dx (xy)
1 + cos y. ^dy⁄_dx = - sin (xy). [x. ^dy⁄_dx + y]
1 + cos y. ^dy⁄_dx = -x sin (xy). ^dy⁄_dx - y sin (xy)
x sin (xy). ^dy⁄_dx + cos y. ^dy⁄_dx = -y sin xy - 1
(x sin (xy) + cos y). ^dy⁄_dx = -y sin xy -1
^dy⁄_dx = ^{-y sin xy -1}⁄ _{x sin (xy) + cos y}
^dy⁄_dx = - ^{1 + y sin xy}⁄ _{cos y + x sin (xy)}

Example 12.
Given that y = x sin x, find ^d²y ⁄_dx².
Solution
y = x sin x
^dy⁄_dx = x ^d⁄_dx sin x + sin x.
^dx⁄_dx = x.cos x + sin x
^d²y⁄ _dx² = (x⋅^d⁄_dx cos x + cos x ⋅ ^dx⁄_dx ) + ^d⁄_dx sin x
= x (-sin x) + cos x + cos x
= 2 cos x - x sin x

Exercise 7.5
1. Differentiate the following function with respect to x.
(a) sin (2x + 3),
(b) cos ³⁄_x
(c) x³ cos 2x ,
(d) cos 7x + sin 3x
(e) sin x.cos 2x,
(f) cos² (5x)
(g) tan³ √ x ,
(h) sin (cos x)
(i) ^{sin x}⁄ _{1 + tan x} ,
(j) √ sin x + cos x

2. Find ^dy⁄_dx.
(a) y = sin (1 - x²) ,
(b) y = 2 π x + 2 cos π x.
(c) y = sin² x . cos 3x ,
(d) y = x² sin( ¹⁄_x)
(e) 3x² + 2 sin y = y² ,
(f) sin x . cos y = 2y.

3. Given that y = cos² x, prove that ^d²y⁄ _dx²+ 4y = 2.

4. Given that y = ¹⁄₃ cos³x - cos x, prove that ^dy⁄_dx = sin³x.

1	^d⁄_dx sin (x) = cos x ,	^d⁄_dx sin u(x) = cos u(x). ^d⁄_dx u(x).
2	^d⁄_dx cos (x) = -sin x ,	^d⁄_dx cos u(x) = -sin u(x). ^d⁄_dx u(x).
3	^d⁄_dx tan (x) = sec² x ,	^d⁄_dx tan u(x) = sec² u(x). ^d⁄_dx u(x).
4	^d⁄_dx cot (x) = -cosec² x ,	^d⁄_dx cot u(x) = -cosec² u(x). ^d⁄_dx u(x).
5	^d⁄_dx sec (x) = sec x. tan x ,	^d⁄_dx sec u(x) = sec u(x). tan u(x). ^d⁄_dx u(x).
6	^d⁄_dx csc (x) = -csc x. cot x ,	^d⁄_dx csc u(x) = -csc u(x). cot u(x). ^d⁄_dx u(x).