differentiation rules -...

DIFFERENTIATION RULESDIFFERENTIATION RULES

3

Before starting this section,you might need to review thetrigonometric functions.

DIFFERENTIATION RULES

In particular, it is important to remember that,when we talk about the function f defined forall real numbers x by f(x) = sin x, it isunderstood that sin x means the sine ofthe angle whose radian measure is x.


A similar convention holds forthe other trigonometric functionscos, tan, csc, sec, and cot.

Recall from Section 2.5 that all the trigonometricfunctions are continuous at every number in theirdomains.



3.6Derivatives of

Trigonometric Functions

In this section, we will learn about:Derivatives of trigonometric functions

and their applications.

Let’s sketch the graph of the functionf(x) = sin x and use the interpretation of f’(x)as the slope of the tangent to the sine curvein order to sketch the graph of f’.

DERIVATIVES OF TRIGONOMETRIC FUNCTIONS

Then, it looks as if the graph of f’ maybe the same as the cosine curve.


Let’s try to confirmour guess that, if f(x) = sin x,then f’(x) = cos x.


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 toevaluate.

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 constantwhen 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 madethe guess—on the basis of numerical andgraphical evidence—that:

0

sinlim 1!

!

!"= =


We now use a geometric argumentto prove Equation 2.

Assume first that θ lies between 0 and π/2.


The figure shows a sector of a circle withcenter O, central angle θ, and radius 1.BC is drawn perpendicular to OA.

By the definition ofradian measure, we havearc AB = θ.

Also,|BC| = |OB| sin θ = sin θ.

DERIVS. OF TRIG. FUNCTIONS Proof

sinsin so 1

!! !

!< <


We see that|BC| < |AB| < arc AB

Thus,

Proof

Let the tangent lines at A and Bintersect at E.


You can see from this figure thatthe circumference of a circle is smaller thanthe length of a circumscribed polygon.

So,arc AB < |AE| + |EB|


Thus,

θ = arc AB < |AE| + |EB|

< |AE| + |ED|

= |AD| = |OA| tan θ

= tan θ


Therefore, we have:

So,

sin

cos

!!

!<


sincos 1

!!

!< <

Proof

We know that .

So, by the Squeeze Theorem,we have:

0 0

lim1 1 and limcos 1! !

!" "

= =

0

sinlim 1!

!

!+"

=


However, the function (sin θ)/θ is an evenfunction.

So, its right and left limits must be equal.

Hence, we have:0

sinlim 1!

!

!"

=


We can deduce the value of the remaininglimit 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:

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

! ! ! !

"= # + #

= # + #

=


So, we have proved the formula forthe 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 inthe proof of Formula 4, we can prove:

(cos ) sind

x xdx

= !

Formula 5DERIV. OF COSINE FUNCTION

The tangent function can also bedifferentiated by using the definitionof a derivative.

However, it is easier to use the Quotient Ruletogether 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 remainingtrigonometric functions—csc, sec, andcot—can also be found easily using theQuotient Rule.


We have collected all the differentiationformulas for trigonometric functions here.

Remember, they are valid only when x is measuredin 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

= = !

= ! =

= = !


Differentiate

For what values of x does the graph of fhave 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 xdx dxf x

x

x x x x x

x

x x x x

x

x x

x

+ ! +

=+

+ ! "=

+

+ !=

+

!=

+


In simplifying the answer,we have used the identitytan2 x + 1 = sec2 x.

DERIVS. OF TRIG. FUNCTIONS Example 2

Since sec x is never 0, we see thatf’(x)=0 when tan x = 1.

This occurs when x = nπ + π/4,where n is an integer.


Trigonometric functions are often usedin modeling real-world phenomena.

In particular, vibrations, waves, elastic motions,and other quantities that vary in a periodic mannercan be described using trigonometric functions.

In the following example, we discuss an instanceof simple harmonic motion.

APPLICATIONS

An object at the end of a vertical springis stretched 4 cm beyond its rest positionand released at time t = 0.

In the figure, note that the downwarddirection is positive.

Its position at time t iss = f(t) = 4 cos t

Find the velocity and accelerationat time t and use them to analyzethe 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

= = = = !

= = ! = ! = !


The object oscillates from the lowest point(s = 4 cm) to the highest point (s = -4 cm).

The period of the oscillationis 2π, the period of cos t.


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

It has greatest magnitude at the high andlow points.


Find the 27th derivative of cos x.

The first few derivatives of f(x) = cos xare 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 occurin 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


Our main use for the limit in Equation 2has been to prove the differentiation formulafor the sine function.

However, this limit is also useful in findingcertain other trigonometric limits—as the followingtwo examples show.


Find

In order to apply Equation 2, we first rewritethe 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 sinlim4

7 71

4 4

x x

x x

x x

!

!

!

" "

"

# $= % &

' (

# $= % &

' (

= ) =


Calculate .

We divide the numerator and denominator by x:

by the continuity of cosine and Eqn. 2

0

lim cotx

x x!


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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