1. 3
DERIVATIVES

2. Trigonometric Functions
DERIVATIVES
3.4 Derivatives of
Trigonometric Functions
In this section, we will learn about:
Derivatives of trigonometric functions
and their applications.

3. DERIVATIVES OF TRIGONOMETRIC FUNCTIONS
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.
Figure 3.4.1, p. 149

4. From the definition of a derivative, we have:
DERIVS. OF TRIG. FUNCTIONS
Equation 1
From the definition of a derivative, we have:

5. Two of these four limits are easy to evaluate.
DERIVS. OF TRIG. FUNCTIONS
Two of these four limits are easy to evaluate.

6. DERIVS. OF TRIG. FUNCTIONS
Since we regard x as a constant when computing a limit as h → 0, we have:

7. The limit of (sin h)/h is not so obvious.
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:

8. DERIVS. OF TRIG. FUNCTIONS
Proof of Eq.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.
Figure 3.4.2a, p. 150

9. We see that |BC| < |AB| < arc AB
DERIVS. OF TRIG. FUNCTIONS
Proof of Eq.2
We see that |BC| < |AB| < arc AB
Thus,
Figure 3.4.2a, p. 150

10. Let the tangent lines at A and B intersect at E. Thus,
DERIVS. OF TRIG. FUNCTIONS
Proof of Eq.2
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
Figure 3.4.2a, p. 150

11. Therefore, we have: So, We know that
DERIVS. OF TRIG. FUNCTIONS
Proof of Eq.2
Therefore, we have:
So,
We know that
So, by the Squeeze Theorem, we have:

12. However, the function (sin θ)/θ is an even function.
DERIVS. OF TRIG. FUNCTIONS
Proof of Eq.2
However, the function (sin θ)/θ is an even function.
So, its right and left limits must be equal.
Hence, we have:

13. DERIVS. OF TRIG. FUNCTIONS
We can deduce the value of the remaining limit in Equation 1 as follows.

14. DERIVS. OF TRIG. FUNCTIONS
Equation 3

15. If we put the limits (2) and (3) in (1), we get:
DERIVS. OF TRIG. FUNCTIONS
Formula 4
If we put the limits (2) and (3) in (1), we get:
So, we have proved the formula for sine,

16. Differentiate y = x2 sin x.
DERIVS. OF TRIG. FUNCTIONS
Example 1
Differentiate y = x2 sin x.
Using the Product Rule and Formula 4, we have:
Figure 3.4.3, p. 151

17. Using the same methods as in the proof of Formula 4, we can prove:
DERIV. OF COSINE FUNCTION
Formula 5
Using the same methods as in the proof of Formula 4, we can prove:

18. DERIV. OF TANGENT FUNCTION
Formula 6

19. 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.

20. For what values of x does the graph of f have a horizontal tangent?
DERIVS. OF TRIG. FUNCTIONS
Example 2
Differentiate
For what values of x does the graph of f have a horizontal tangent?

21. The Quotient Rule gives:
Solution:
Example 2
The Quotient Rule gives:
tan2 x + 1 = sec2 x

22. Since sec x is never 0, we see that f’(x)=0 when tan x = 1.
DERIVS. OF TRIG. FUNCTIONS
Example 2
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.
Figure 3.4.4, p. 152

23. 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.
Figure 3.4.5, p. 152

24. The velocity and acceleration are:
Solution:
Example 3
The velocity and acceleration are:

25. The period of the oscillation is 2π, the period of cos t.
Solution:
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.
Figure 3.4.5, p. 152

26. Solution:
Example 3
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.
Figure 3.4.6, p. 153

27. The acceleration a = -4 cos t = 0 when s = 0.
Solution:
Example 3
The acceleration a = -4 cos t = 0 when s = 0.
It has greatest magnitude at the high and low points.
Figure 3.4.6, p. 153

28. Find the 27th derivative of cos x.
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:

29. Therefore, f (24)(x) = cos x
Solution:
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

30. DERIVS. OF TRIG. FUNCTIONS
Example 5
Find
In order to apply Equation 2, we first rewrite the function by multiplying and dividing by 7:

31. If we let θ = 7x, then θ → 0 as x → 0. So, by Equation 2, we have:
Solution:
Example 5
If we let θ = 7x, then θ → 0 as x → 0. So, by Equation 2, we have:

32. Calculate . Example 6 DERIVS. OF TRIG. FUNCTIONS
We divide the numerator and denominator by x: by the continuity of cosine and Eqn. 2