differentiation rules - university of alaska...

DIFFERENTIATION RULES

3

Before starting this section,

you might need to review the

trigonometric functions.


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.


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.



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.


Let’s try to confirm

our 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 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:


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


We now use a geometric argument

to prove Equation 2.

Assume first 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

sinsin so 1


We see that

|BC| < |AB| < arc AB

Thus,

Proof

Let the tangent lines at A and B

intersect at E.


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|


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 0lim1 1 and limcos 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:

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 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 x

dx 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.


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


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


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 = nπ + π/4,

where n is an integer.


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


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.


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

It has greatest magnitude at the high and

low points.


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


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.


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


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

limcos cos0

sin 1lim

1

x x x

x

x

x x xx x

xx

x

x

x

x

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