Chapter 7

7.5 Differentiation of Trigonometric Functions

Before you study the differentiation of trigonometric functions, we first evaluate an important limit, lim⁡x→0sin xx.\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 ∆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.
Since lim⁡x→01=1 and lim⁡x→0 cos x = 1,\small{ \text{Since}\, \lim_{x\to 0}1 = 1 \, \text{and} \, \lim_{x\to0}\, cos \, x\, = \, 1, } lim⁡x→0sin xx=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)
dydx=lim⁡δx→0δyδx\small{ \frac{dy}{dx} = \lim_{\delta x\to0} \frac{\delta y}{\delta x} } =lim⁡δx→02cos(x+δx2) ⋅ sinδx2δx\small{ = \lim_{\delta x\to0} \frac{2 \text{cos}(x + \frac{\delta x}{2})\, \sdot\, \text{sin} \frac{\delta x}{2}} {\delta x} } =lim⁡δx→0cos(x+δx2) ⋅ lim⁡δx→0sinδx2δx2\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 cos 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 ⋅ 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 x1⁄2
= - 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 4.
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

Solution
(a)   d⁄dx sin (2x + 3) = cos (2x + 3) d⁄dx (2x + 3)
= cos (2x + 3) ⋅ 2
= 2 cos (2x + 3)

(b)  d⁄dx cos 3⁄x = - sin 3⁄x d⁄dx 3x-1
= - sin 3⁄x ⋅ (-3) x-2
= sin 3⁄x ⋅ 3⁄x2

(c)  d⁄dx (x3 cos 2x) = x3 ⋅ d⁄dx cos 2x + cos 2x d⁄dx x3
= x3 (- sin 2x) d⁄dx 2x + cos 2x ⋅ 3x2
= x3 (- sin 2x) ⋅ 2 + 3x2 ⋅ cos 2x

(d)  d⁄dx (cos 7x + sin 3x) = d⁄dx cos 7x + d⁄dx sin 3x
= - sin 7x d⁄dx 7x + cos 3x d⁄dx 3x
= -7 sin 7x + 3 cos 3x

(e)   d⁄dx sin x.cos 2x = sin x d⁄dx cos 2x + cos 2x d⁄dx sin x
= sin x (-sin 2x) d⁄dx 2x + cos 2x ⋅ cos x
= sin x (-2 sin 2x) + cos 2x ⋅ cos x

(f)   d⁄dx cos2 (5x) = d⁄dx (cos 5x)2
= 2 (cos 5x) d⁄dx cos 5x
= 2 cos 5x ⋅ (-sin 5x) d⁄dx 5x
= - 2 cos 5x ⋅ sin 5x ⋅ 5
= -10 cos 5x ⋅ sin 5x

(g)  d⁄dx tan3 √ x = d⁄dx (tan √ x ) 3
= 3 (tan √ x )2 d⁄dx tan √ x
= 3 (tan √ x )2 sec2 √ x d⁄dx x 1⁄2
= 3 (tan √ x )2 ⋅ sec2 √ x ⋅ 1⁄2 x - 1⁄2
= 3 tan2 √ x ⋅ sec2 √ x ⋅ 1⁄2 √ x

(h)  d⁄dx sin (cos x) = cos (cos x) d⁄dx cos x
= cos (cos x) ⋅ (-sin x)

(i)  d⁄dx sin x⁄ 1 + tan x
= 1 + tan x ⋅ d⁄dx sin x - sin x d⁄dx (1 + tan x)⁄(1 + 10x)2
= 1 + tan x ⋅ cos x - sin x ⋅ sec2 x ⁄(1 + 10x)2

(j)  d⁄dx √ sin x + cos x = d⁄dx (sin x + cos x) 1⁄2
= 1⁄2 (sin x + cos x) - 1⁄2 ⋅ d⁄dx (sin x + cos x)
= 1⁄2 cos x - sin x⁄ √ 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.

Solution
(a) y = sin (1 - x2)
dy⁄dx = cos (1 - x2) ⋅ d⁄dx (1 - x2)
= cos (1 - x2) ⋅ (-2x)
= -2x ⋅ cos (1 - x2)

(b) y = 2 π x + 2 cos π x
dy⁄dx = d⁄dx 2πx + d⁄dx 2 cos π x
= (2π d⁄dx x + x ⋅ d⁄dx 2π) + (-2 sin πx ⋅ dy⁄dx πx)
= (2π + x⋅0) + (-2 sin πx ⋅ π)
= 2π - 2π ⋅ sin πx

(c) y = sin2 x . cos 3x
dy⁄dx = sin2 x ⋅ d⁄dx cos 3x + cos 3x ⋅ d⁄dx sin2 x
= sin2 x ⋅ (-sin 3x ⋅ d⁄dx 3x) + cos 3x ⋅ 2 sin x ⋅ dy⁄dx sin x
= sin2 x ⋅ (-sin 3x ⋅ 3) + cos 3x ⋅ 2 sin x ⋅ cos x

(d) y = x2 sin( 1⁄x)
dy⁄dx = x2 ⋅ d⁄dx sin (1⁄x) + sin 1⁄x ⋅ d⁄dx x2
= x2 ⋅ cos (1⁄x) ⋅ d⁄dx x-1 + sin 1⁄x ⋅ 2x
= x2 ⋅ cos (1⁄x) ⋅ (-x-2) + sin 1⁄x ⋅ 2x
= x2⋅ cos (1⁄x) ⋅ (- 1⁄x2) + 2x ⋅ sin (1⁄x)
= -cos (1⁄x) + 2x ⋅ sin (1⁄x)

(e) 3x2 + 2 sin y = y2
Differentiate both sides with respect to x
3 d⁄dx x2 + 2 d⁄dx sin y = d⁄dx y2
3 ⋅ 2x + 2 cos y dy⁄dx = 2y dy⁄dx
3 ⋅ 2x = 2y dy⁄dx - 2 cos y dy⁄dx
6x = (2y - 2 cos y) dy⁄dx
6x⁄ (2y - 2 cos y) = dy⁄dx
2 ⋅ 3x ⁄ 2 (y - cos y) = dy⁄dx
dy⁄dx = 3x⁄y - cos y


(f) sin x . cos y = 2y
Differentiate both sides with respect to x
sin x ⋅ d⁄dx cos y + cos y ⋅ d⁄dx sin x = 2 dy⁄dx
sin x ⋅ (-sin y) ⋅ dy⁄dx + cos y ⋅ cos x = 2 dy⁄dx
- sin x ⋅ sin y ⋅ dy⁄dx + cos y ⋅ cos x = 2 dy⁄dx
cos y ⋅ cos x = 2 dy⁄dx + sin x ⋅ sin y ⋅ dy⁄dx
cos x ⋅ cos y = (2 + sin x ⋅ sin y) ⋅ dy⁄dx
cos x ⋅ cos y⁄ 2 + sin x ⋅ sin y = dy⁄dx
dy⁄dx = cos x ⋅ cos y⁄2 + sin x ⋅ sin y

3. Given that y = cos2 x, prove that d2y⁄ dx2 + 4y = 2.
dy⁄dx = 2 cos x ⋅ d⁄dx cos x
= 2 cos x (-sin x)
= -2 sin x cos x
= - sin 2x

d2y⁄dx2 = d⁄dx (-sin 2x)
= -cos 2x ⋅ d⁄dx 2x
= -cos 2x ⋅ 2
= -2 cos 2x

d2y⁄dx2 + 4 y = -2 cos 2x + 4 cos2x
= -2 (2 cos2 x - 1) + 4 cos2x
= -4 cos2 x + 2 + 4 cos2 x
= 2

4. Given that y = 1⁄3 cos3x - cos x, prove that dy⁄dx = sin3x.
Solution
dy⁄dx = 1⁄3 d⁄dx cos3 x - d⁄dx cos x
= 1⁄3 (3 cos2 x) d⁄dx cos x - (-sin x)
= cos2 x (-sin x) + sin x
= sin x (-cos2 x + 1)
= sin x (1 - cos2 x)
= sin x (sin2 x)
= sin3 x