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DIFFERENTIATION RULESDIFFERENTIATION RULES

3

Before starting this section,

you might need to review the

trigonometric functions.

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.

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.

DIFFERENTIATION RULES

DIFFERENTIATION RULES

3.6

Derivatives 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 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’.

DERIVATIVES OF TRIGONOMETRIC FUNCTIONS

Then, it looks as if the graph of f’ may

be the same as the cosine curve.

DERIVATIVES OF TRIGONOMETRIC FUNCTIONS

Let’s try to confirm

our guess that, if f(x) = sin x,

then f’(x) = cos x.

DERIVATIVES OF TRIGONOMETRIC FUNCTIONS

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 limcos lim

h h

h

h

h

h h h h

f x h f x x h xf x

h h

x h x x

h

x 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 limcos 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

We now use a geometric argument

to prove Equation 2.

!Assume first that _ lies between 0 and !/2.

DERIVS. OF TRIG. FUNCTIONS

The figure shows a sector of a circle with

center O, central angle _, and radius 1.

BC is drawn perpendicular to OA.

!By the definition of

radian measure, we have

arc AB = _.

!Also,

|BC| = |OB| sin _ = sin _.

DERIVS. OF TRIG. FUNCTIONS Proof

sinsin so 1

!! !

!< <

DERIVS. OF TRIG. FUNCTIONS

We see that

|BC| < |AB| < arc AB

Thus,

Proof

Let the tangent lines at A and B

intersect at E.

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.

So,

arc AB < |AE| + |EB|

DERIVS. OF TRIG. FUNCTIONS Proof

Thus,

_ = arc AB < |AE| + |EB|

< |AE| + |ED|

= |AD| = |OA| tan _

= tan _

DERIVS. OF TRIG. FUNCTIONS Proof

Therefore, we have:

So,

sin

cos

!!

!<

DERIVS. OF TRIG. FUNCTIONS

sincos 1

!!

!< <

Proof

We know that .

So, by the Squeeze Theorem,

we have:

0 0

lim1 1 and limcos 1! !

!" "

= =

0

sinlim 1!

!

!+

"

=

DERIVS. OF TRIG. FUNCTIONS Proof

However, the function (sin _)/_ is an even

function.

So, its right and left limits must be equal.

Hence, we have:

0

sinlim 1!

!

!"

=

DERIVS. OF TRIG. FUNCTIONS Proof

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:

0 0 0 0

cos 1 sin'( ) limsin lim limcos lim

(sin ) 0 (cos ) 1

cos

h h h h

h hf x x x

h h

x x

x

! ! ! !

"= # + #

= # + #

=

DERIVS. OF TRIG. FUNCTIONS

So, we have proved the formula for

the derivative of the sine function:

(sin ) cosd

x xdx

=

DERIV. OF SINE FUNCTION Formula 4

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

Using the same methods as in

the proof of Formula 4, we can prove:

(cos ) sind

x xdx

= !

Formula 5DERIV. OF COSINE 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.

DERIV. OF TANGENT FUNCTION

2

2

2 22

2 2

2

sin(tan )

cos

cos (sin ) sin (cos )

cos

cos cos sin ( sin )

cos

cos sin 1sec

cos cos

(tan ) sec

d d xx

dx dx x

d dx x x xdx dx

x

x x x x

x

x xx

x x

dx x

dx

! "= # $

% &

'=

( ' '=

+= = =

=

DERIV. OF TANGENT FUNCTION Formula 6

The derivatives of the remaining

trigonometric functions—csc, sec, and

cot—can also be found easily using the

Quotient Rule.

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.

2 2

(sin ) cos (csc ) csc cot

(cos ) sin (sec ) sec tan

(tan ) sec (cot ) csc

d dx x x x x

dx dx

d dx x x x x

dx dx

d dx x x x

dx 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 xdx dxf x

x

x x x x x

x

x x x x

x

x x

x

+ ! +

=+

+ ! "=

+

+ !=

+

!=

+

Example 2DERIVS. OF TRIG. FUNCTIONS

In simplifying the answer,

we have used the identity

tan2 x + 1 = sec2 x.

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 = n! + !/4,

where n is an integer.

Example 2DERIVS. OF TRIG. FUNCTIONS

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.

APPLICATIONS

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

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 3APPLICATIONS

The object oscillates from the lowest point

(s = 4 cm) to the highest point (s = -4 cm).

The period of the oscillation

is 2!, the period of cos t.

Example 3APPLICATIONS

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 3APPLICATIONS

The acceleration a = -4 cos t = 0 when s = 0.

It has greatest magnitude at the high and

low points.

Example 3APPLICATIONS

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 4DERIVS. 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.

DERIVS. OF TRIG. FUNCTIONS

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 _ = 7x, then _ _ 0 as x _ 0.

So, by Equation 2, we have:

0 0

0

sin 7 7 sin 7lim lim

4 4 7

7 sinlim4

7 71

4 4

x x

x x

x x

!

!!

" "

"

# $= % &

' (# $

= % &' (

= ) =

Example 5DERIVS. OF TRIG. FUNCTIONS

Calculate .

!We divide the numerator and denominator by x:

by the continuity of

cosine and Eqn. 2

0

lim cotx

x x!

Example 6DERIVS. OF TRIG. FUNCTIONS

0 0 0

0

0

cos coslim cot lim lim

sinsin

limcoscos0

sin 1lim

1

x x x

x

x

x x xx x

xx

x

x

x

x

! ! !

!

!

= =

= =

=

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