Before you 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.
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.
$$\small{ \text{Since}\, \lim_{x\to 0}1 = 1 \, \text{and} \, \lim_{x\to0}\, cos \, x\, = \, 1, } $$
$$\small{ \lim_{x\to0} \frac{sin \, x}{x} = 1 } $$
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⁄dxu(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⁄dxu(x).
Derivative of tan x
Since tan x = sin x⁄cos x d⁄dx tan x =
d⁄dxsin x⁄cos x
= cos x ⋅ d⁄dx sin x - sin x
⋅ d⁄dx cos x⁄cos2x
= cos x ⋅ cos x - sin x ⋅ (- sin x)⁄cos2x
(quotient formula) cos2x + sin2x⁄cos2x
= 1 ⁄cos2x
(cos2x + sin2x = 1)
= sec2x (sec x = 1 ⁄cos x)
∴ d⁄dx tan x = sec2x
In general, d⁄dx tan u(x) = sec2u(x) ⋅
d⁄dxu(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
Example 10.
Find dy⁄dx.
(a) y = sin2x
(b) y = cos
√x
(c) y = tan2 (x2)
(d) y = sin 2x - x cos x
(e) y = sin x ⋅ cos2x
(f) y = x⁄tan x
(g) y =
√x + 10x
Solution
(a) y = sin2x dy⁄dx =
d⁄dx sin2x
= 2 sin xd⁄dx sin x
= 2 sin x cos x
(b) y = cos
√x dy⁄dx =
d⁄dx cos
√x
= - sin √xd⁄dxx1⁄2
= - sin
√x ⋅
1⁄2x (
1⁄2 - 1)
= - sin
√x ⋅
1⁄2x( -
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 xdy⁄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)1/2dy⁄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⁄xd⁄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 2xd⁄dxx3
= 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 7xd⁄dx 7x + cos 3x
d⁄dx 3x
= -7 sin 7x + 3 cos 3x
(e) d⁄dx sin x.cos 2x =
sin xd⁄dx cos 2x + cos 2xd⁄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
)2d⁄dx
tan √
x
= 3 (tan √
x
)2
sec2
√
x
d⁄dxx1⁄2
= 3 (tan √
x
)2
⋅
sec2
√
x
⋅
1⁄2x -
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⁄dxsin x⁄1 + tan x
=
1 + tan x ⋅ d⁄dx
sin x - sin xd⁄dx
(1 + tan x)⁄(1 + 10x)2
= 1 + tan x ⋅ cos x - sin x ⋅ sec2x⁄(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)
= 1⁄2cos 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⁄dxx + 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 = sin2x ⋅
d⁄dx cos 3x + cos 3x ⋅
d⁄dx sin2x
= sin2x ⋅ (-sin 3x ⋅ d⁄dx
3x) + cos 3x ⋅ 2 sin x ⋅
dy⁄dx sin x
= sin2x ⋅ (-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⁄dxx2
= x2 ⋅ cos
(1⁄x) ⋅
d⁄dxx-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⁄dxx2 + 2
d⁄dx sin y =
d⁄dxy2
3 ⋅ 2x + 2 cos ydy⁄dx = 2ydy⁄dx
3 ⋅ 2x = 2ydy⁄dx - 2 cos ydy⁄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
4. Given that y =
1⁄3
cos3x - cos x, prove that
dy⁄dx =
sin3x. Solution dy⁄dx = 1⁄3d⁄dx cos3x -
d⁄dx cos x
= 1⁄3 (3 cos2x)
d⁄dx cos x - (-sin x)
= cos2x (-sin x) + sin x
= sin x (-cos2x + 1)
= sin x (1 - cos2x)
= sin x (sin2x)
= sin3x
