Derivatives of Trigonometric Functions

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Sharp Tutor Tue, Jan 18, 2022 08:43 PM UTC

In this section, we will learn about: Derivatives of trigonometric functions and their applications.

1. 3
DIFFERENTIATION RULES

2. DIFFERENTIATION RULES
Before starting this section,
you might need to review the
trigonometric functions.

3. DIFFERENTIATION RULES
In particular, it is important to remember that,
when we talk about the function f defined for
all real numbers x by f(x) = sin x, it is
understood that sin x means the sine of
the angle whose radian measure is x.

4. DIFFERENTIATION RULES
A similar convention holds for
the other trigonometric functions
cos, tan, csc, sec, and cot.
 Recall from Section 2.5 that all the trigonometric
functions are continuous at every number in their
domains.

5. DIFFERENTIATION RULES
3.6
Derivatives of
Trigonometric Functions
In this section, we will learn about:
Derivatives of trigonometric functions
and their applications.

6. DERIVATIVES OF TRIGONOMETRIC FUNCTIONS
Let’s sketch the graph of the function
f(x) = sin x and use the interpretation of f’(x)
as the slope of the tangent to the sine curve
in order to sketch the graph of f’.

7. DERIVATIVES OF TRIGONOMETRIC FUNCTIONS
Then, it looks as if the graph of f’ may
be the same as the cosine curve.

8. DERIVATIVES OF TRIGONOMETRIC FUNCTIONS
Let’s try to confirm
our guess that, if f(x) = sin x,
then f’(x) = cos x.

9. DERIVS. OF TRIG. FUNCTIONS Equation 1
From the definition of a derivative, we have:
f ( x  h)  f ( x ) sin( x  h)  sin x
f '( x) lim lim
h 0 h h 0 h
sin x cos h  cos x sin h  sin x
lim
h 0 h
 sin x cos h  sin x cos x sin h 
lim   
h 0
 h h
  cos h  1   sin h  
lim  sin x    cos x  
h 0
  h   h 
cos h  1 sin h
lim sin x lim  lim cos x lim
h 0 h 0 h h 0 h 0 h

10. DERIVS. OF TRIG. FUNCTIONS
cos h  1 sin h
lim sin x lim  lim cos x lim
h 0 h 0 h h 0 h 0 h
Two of these four limits are easy to

11. DERIVS. OF TRIG. FUNCTIONS
Since we regard x as a constant
when computing a limit as h → 0,
we have:
lim sin x sin x
h 0
lim cos x cos x
h 0

12. DERIVS. OF TRIG. FUNCTIONS Equation 2
The limit of (sin h)/h is not so obvious.
In Example 3 in Section 2.2, we made
the guess—on the basis of numerical and
graphical evidence—that:
sin 
lim  1
0 

13. DERIVS. OF TRIG. FUNCTIONS
We now use a geometric argument
to prove Equation 2.
 Assume first that θ lies between 0 and π/2.

14. DERIVS. OF TRIG. FUNCTIONS Proof
The figure shows a sector of a circle with
center O, central angle θ, and radius 1.
BC is drawn perpendicular to OA.
 By the definition of
radian measure, we have
arc AB = θ.
 Also,
|BC| = |OB| sin θ = sin θ.

15. DERIVS. OF TRIG. FUNCTIONS Proof
We see that
|BC| < |AB| < arc AB
sin 
sin    so 1


16. DERIVS. OF TRIG. FUNCTIONS Proof
Let the tangent lines at A and B
intersect at E.

17. DERIVS. OF TRIG. FUNCTIONS Proof
You can see from this figure that
the circumference of a circle is smaller than
the length of a circumscribed polygon.
arc AB < |AE| + |EB|

18. DERIVS. OF TRIG. FUNCTIONS Proof
θ = arc AB < |AE| + |EB|
< |AE| + |ED|
= |AD| = |OA| tan θ
= tan θ

19. DERIVS. OF TRIG. FUNCTIONS Proof
Therefore, we have: sin 

cos 
sin 
So, cos   1


20. DERIVS. OF TRIG. FUNCTIONS Proof
We know that lim1 1 and lim cos  1.
0 0
So, by the Squeeze Theorem,
we have: sin 
lim 1
0 

21. DERIVS. OF TRIG. FUNCTIONS Proof
However, the function (sin θ)/θ is an even
So, its right and left limits must be equal.
Hence, we have: sin 
lim 1
0 

22. DERIVS. OF TRIG. FUNCTIONS
We can deduce the value of the remaining
limit in Equation 1 as follows.
cos   1
0 
 cos   1 cos   1 
lim   
0
  cos   1 
2
cos   1
  0  (cos   1)

23. DERIVS. OF TRIG. FUNCTIONS Equation 3
2
 sin 
  0  (cos   1)
 sin  sin  
 lim   
0
  cos   1 
sin  sin   0 
 lim lim  1   0
  0 cos   1
0 
 1 1 
cos   1
lim 0
0 

24. DERIVS. OF TRIG. FUNCTIONS
If we put the limits (2) and (3) in (1),
we get:
cos h  1 sin h
f '( x) lim sin x lim  lim cos x lim
h 0 h 0 h h 0 h 0 h
(sin x) 0  (cos x) 1
cos x

25. DERIV. OF SINE FUNCTION Formula 4
So, we have proved the formula for
the derivative of the sine function:
d
(sin x) cos x
dx

26. DERIVS. OF TRIG. FUNCTIONS Example 1
Differentiate y = x2 sin x.
 Using the Product Rule and Formula 4,
we have: dy d d
 x2 (sin x)  sin x ( x2 )
dx dx dx
2
 x cos x  2 x sin x

27. DERIV. OF COSINE FUNCTION Formula 5
Using the same methods as in
the proof of Formula 4, we can prove:
d
(cos x)  sin x
dx

28. DERIV. OF TANGENT FUNCTION
The tangent function can also be
differentiated by using the definition
of a derivative.
However, it is easier to use the Quotient Rule
together with Formulas 4 and 5—as follows.

29. DERIV. OF TANGENT FUNCTION Formula 6
d d  sin x 
(tan x)   
dx dx  cos x 
d d
cos x (sin x)  sin x (cos x)
 dx dx
cos 2 x
cos x cos x  sin x( sin x)
cos 2 x
cos 2 x  sin 2 x 1 2
 2
 2 sec x
cos x cos x
d
(tan x) sec 2 x
dx

30. DERIVS. OF TRIG. FUNCTIONS
The derivatives of the remaining
trigonometric functions—csc, sec, and cot—
can also be found easily using the Quotient

31. DERIVS. OF TRIG. FUNCTIONS
We have collected all the differentiation
formulas for trigonometric functions here.
 Remember, they are valid only when x is measured
in radians.
d d
(sin x) cos x (csc x)  csc x cot x
dx dx
d d
(cos x)  sin x (sec x) sec x tan x
dx dx
d 2 d 2
(tan x) sec x (cot x)  csc x
dx dx

32. DERIVS. OF TRIG. FUNCTIONS Example 2
sec x
Differentiate f ( x ) 
1  tan x
For what values of x does the graph of f
have a horizontal tangent?

33. DERIVS. OF TRIG. FUNCTIONS Example 2
The Quotient Rule gives:
d d
(1  tan x) (sec x)  sec x (1  tan x)
f '( x)  dx dx
(1  tan x) 2
2
(1  tan x) sec x tan x  sec x sec x
 2
(1  tan x)
2 2
sec x(tan x  tan x  sec x)
 2
(1  tan x)
sec x(tan x  1)

(1  tan x) 2

34. DERIVS. OF TRIG. FUNCTIONS Example 2
In simplifying the answer,
we have used the identity
tan2 x + 1 = sec2 x.

35. DERIVS. OF TRIG. FUNCTIONS Example 2
Since sec x is never 0, we see that f’(x)
when tan x = 1.
 This occurs when x = nπ + π/4,
where n is an integer.

36. Trigonometric functions are often used
in modeling real-world phenomena.
 In particular, vibrations, waves, elastic motions,
and other quantities that vary in a periodic manner
can be described using trigonometric functions.
 In the following example, we discuss an instance
of simple harmonic motion.

37. APPLICATIONS Example 3
An object at the end of a vertical spring
is stretched 4 cm beyond its rest position
and released at time t = 0.
 In the figure, note that the downward
direction is positive.
 Its position at time t is
s = f(t) = 4 cos t
 Find the velocity and acceleration
at time t and use them to analyze
the motion of the object.

38. APPLICATIONS Example 3
The velocity and acceleration are:
ds d d
v   (4 cos t ) 4 (cos t )  4sin t
dt dt dt
dv d d
a   ( 4sin t )  4 (sin t )  4 cos t
dt dt dt

39. APPLICATIONS Example 3
The object oscillates from the lowest point
(s = 4 cm) to the highest point (s = -4 cm).
The period of the oscillation
is 2π, the period of cos t.

41. APPLICATIONS Example 3
The acceleration a = -4 cos t = 0 when s = 0.
It has greatest magnitude at the high and
low points.

42. DERIVS. OF TRIG. FUNCTIONS Example 4
Find the 27th derivative of cos x.
 The first few derivatives of f(x) = cos x
are as follows:
f '( x)  sin x
f ''( x)  cos x
f '''( x) sin x
(4)
f ( x) cos x
(5)
f ( x)  sin x

43. DERIVS. OF TRIG. FUNCTIONS Example 4
 We see that the successive derivatives occur
in a cycle of length 4 and, in particular,
f (n)(x) = cos x whenever n is a multiple of 4.
 Therefore, f (24)(x) = cos x
 Differentiating three more times,
we have:
f (27)(x) = sin x

44. DERIVS. OF TRIG. FUNCTIONS
Our main use for the limit in Equation 2
has been to prove the differentiation formula
for the sine function.
 However, this limit is also useful in finding
certain other trigonometric limits—as the following
two examples show.

45. DERIVS. OF TRIG. FUNCTIONS Example 5
Find lim sin 7 x
x 0 4x
 In order to apply Equation 2, we first rewrite
the function by multiplying and dividing by 7:
sin 7 x 7  sin 7 x 
  
4x 4  7x 

46. DERIVS. OF TRIG. FUNCTIONS Example 5
If we let θ = 7x, then θ → 0 as x → 0.
So, by Equation 2, we have:
sin 7 x 7  sin 7 x 
lim  lim  
x 0 4x 4 x  0
 7x 
7  sin  
 lim  
4   0
  
7 7
 1 
4 4

47. DERIVS. OF TRIG. FUNCTIONS Example 6
Calculate lim x cot x .
x 0
 We divide the numerator and denominator by x:
x cos x cos x
lim x cot x lim lim
x 0 x  0 sin x x  0 sin x
x
lim cos x cos 0
 x 0  by the continuity
of
sin x 1 cosine
lim
and Eqn. 2 x  0 x
1

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