derivatives 3. derivatives 3.4 derivatives of trigonometric functions in this section, we will learn...

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:


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


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


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


Figure 3.4.2a, p. 150

Therefore, we have:

So,

We know that

So, by the Squeeze Theorem,

we have:

sin

cos


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


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)


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


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


Differentiate

For what values of x does the graph of f

have a horizontal tangent?

sec( )

1 tan

xf x

x


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.


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 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


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


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


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

derivatives 3. derivatives 3.4 derivatives of trigonometric functions in this section, we will learn...

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