DERIVATIVESDERIVATIVES
3
DERIVATIVES
3.4Derivatives of
Trigonometric Functions
In this section, we will learn about:
Derivatives of trigonometric functions
and their applications.
Let’s sketch the graph of the function f(x) = sin
x, it looks as if the graph of f’ may be the same
as the cosine curve.
DERIVATIVES OF TRIGONOMETRIC FUNCTIONS
Figure 3.4.1, p. 149
From the definition of a derivative, we have:
0 0
0
0
0
0 0 0
( ) ( ) sin( ) sin'( ) lim lim
sin cos cos sin h sinlim
sin cos sin cos sinlim
cos 1 sinlim sin cos
cos 1limsin lim lim cos lim
h h
h
h
h
h h h h
f x h f x x h xf x
h hx h x x
hx h x x h
h h
h hx x
h h
hx x
h
0
sin h
h
DERIVS. OF TRIG. FUNCTIONS Equation 1
Two of these four limits are easy to
evaluate.
DERIVS. OF TRIG. FUNCTIONS
0 0 0 0
cos 1 sinlimsin lim lim cos limh h h h
h hx x
h h
Since we regard x as a constant
when computing a limit as h → 0,
we have:
DERIVS. OF TRIG. FUNCTIONS
limh 0
sin x sin x
limh 0
cos x cos x
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:
0
sinlim 1
DERIVS. OF TRIG. FUNCTIONS Equation 2
Assume that θ lies between 0 and π/2, 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.
DERIVS. OF TRIG. FUNCTIONS Proof of Eq.2
Figure 3.4.2a, p. 150
sinsin so 1
DERIVS. OF TRIG. FUNCTIONS
We see that
|BC| < |AB| < arc AB
Thus,
Proof of Eq.2
Figure 3.4.2a, p. 150
Let the tangent lines at A and B intersect at
E. Thus,
θ = arc AB < |AE| + |EB|
< |AE| + |ED|
= |AD| = |OA| tan θ
= tan θ
DERIVS. OF TRIG. FUNCTIONS Proof of Eq.2
Figure 3.4.2a, p. 150
Therefore, we have:
So,
We know that
So, by the Squeeze Theorem,
we have:
sin
cos
DERIVS. OF TRIG. FUNCTIONS
sincos 1
Proof of Eq.2
0 0lim1 1 and lim cos 1
0
sinlim 1
However, the function (sin θ)/θ is an even
function.
So, its right and left limits must be equal.
Hence, we have:
0
sinlim 1
DERIVS. OF TRIG. FUNCTIONS Proof of Eq.2
We can deduce the value of the remaining
limit in Equation 1 as follows.
0
0
2
0
cos 1lim
cos 1 cos 1lim
cos 1
cos 1lim
(cos 1)
DERIVS. OF TRIG. FUNCTIONS
2
0
0
0 0
0
sinlim
(cos 1)
sin sinlim
cos 1
sin sin 0lim lim 1 0
cos 1 1 1
cos 1lim 0
DERIVS. OF TRIG. FUNCTIONS Equation 3
If we put the limits (2) and (3) in (1),
we get:
So, we have proved the formula for sine,
0 0 0 0
cos 1 sin'( ) limsin lim lim cos lim
(sin ) 0 (cos ) 1
cos
h h h h
h hf x x x
h hx x
x
DERIVS. OF TRIG. FUNCTIONS Formula 4
(sin ) cosd
x xdx
Differentiate y = x2 sin x. Using the Product Rule and Formula 4,
we have:
2 2
2
(sin ) sin ( )
cos 2 sin
dy d dx x x x
dx dx dx
x x x x
Example 1DERIVS. OF TRIG. FUNCTIONS
Figure 3.4.3, p. 151
Using the same methods as in
the proof of Formula 4, we can prove:
(cos ) sind
x xdx
Formula 5DERIV. OF COSINE FUNCTION
2
2
2 22
2 2
2
sin(tan )
cos
cos (sin ) sin (cos )
coscos cos sin ( sin )
cos
cos sin 1sec
cos cos
(tan ) sec
d d xx
dx dx x
d dx x x x
dx dxx
x x x x
x
x xx
x xd
x xdx
DERIV. OF TANGENT FUNCTION Formula 6
We have collected all the differentiation
formulas for trigonometric functions here. Remember, they are valid only when x is measured
in radians.
2 2
(sin ) cos (csc ) csc cot
(cos ) sin (sec ) sec tan
(tan ) sec (cot ) csc
d dx x x x x
dx dxd d
x x x x xdx dxd d
x x x xdx dx
DERIVS. OF TRIG. FUNCTIONS
Differentiate
For what values of x does the graph of f
have a horizontal tangent?
sec( )
1 tan
xf x
x
Example 2DERIVS. OF TRIG. FUNCTIONS
The Quotient Rule gives:
2
2
2
2 2
2
2
(1 tan ) (sec ) sec (1 tan )'( )
(1 tan )
(1 tan )sec tan sec sec
(1 tan )
sec (tan tan sec )
(1 tan )
sec (tan 1)
(1 tan )
d dx x x x
dx dxf xx
x x x x x
x
x x x x
x
x x
x
Example 2Solution:
tan2 x + 1 = sec2 x
Since sec x is never 0, we see that
f’(x)=0 when tan x = 1. This occurs when x = nπ + π/4,
where n is an integer.
Example 2DERIVS. OF TRIG. FUNCTIONS
Figure 3.4.4, p. 152
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.
Example 3APPLICATIONS
Figure 3.4.5, p. 152
The velocity and acceleration are:
(4cos ) 4 (cos ) 4sin
( 4sin ) 4 (sin ) 4cos
ds d dv t t t
dt dt dt
dv d da t t t
dt dt dt
Example 3Solution:
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.
Example 3Solution:
Figure 3.4.5, p. 152
The speed is |v| = 4|sin t|, which is greatest
when |sin t| = 1, that is, when cos t = 0.
So, the object moves fastest as it passes through its equilibrium position (s = 0).
Its speed is 0 when sin t = 0, that is, at the high and low points.
Example 3Solution:
Figure 3.4.6, p. 153
The acceleration a = -4 cos t = 0 when s = 0.
It has greatest magnitude at the high and
low points.
Example 3Solution:
Figure 3.4.6, p. 153
Find the 27th derivative of cos x.
The first few derivatives of f(x) = cos x are as follows:
(4)
(5)
'( ) sin
''( ) cos
'''( ) sin
( ) cos
( ) sin
f x x
f x x
f x x
f x x
f x x
Example 4DERIVS. OF TRIG. FUNCTIONS
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
Example 4Solution:
Find
In order to apply Equation 2, we first rewrite the function by multiplying and dividing by 7:
0
sin 7lim
4x
x
x
sin 7 7 sin 7
4 4 7
x x
x x
Example 5DERIVS. OF TRIG. FUNCTIONS
If we let θ = 7x, then θ → 0 as x → 0.
So, by Equation 2, we have:
0 0
0
sin 7 7 sin 7lim lim
4 4 7
7 sinlim
4
7 71
4 4
x x
x x
x x
Example 5Solution:
Calculate .
We divide the numerator and denominator by x:
by the continuity of cosine and Eqn. 2
0lim cotx
x x
Example 6DERIVS. OF TRIG. FUNCTIONS
0 0 0
0
0
cos coslim cot lim lim
sinsin
lim cos cos0sin 1lim
1
x x x
x
x
x x xx x
xxx
x
x
x