differentiation rules - facultyfaculty.essex.edu/~bannon/2007.fall.121/lectures/chap3_sec3.pdf ·...
TRANSCRIPT
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DIFFERENTIATION RULESDIFFERENTIATION RULES
3
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Before starting this section,
you might need to review the
trigonometric functions.
DIFFERENTIATION RULES
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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
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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
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DIFFERENTIATION RULES
3.6
Derivatives of
Trigonometric Functions
In this section, we will learn about:
Derivatives of trigonometric functions
and their applications.
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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
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Then, it looks as if the graph of f’ may
be the same as the cosine curve.
DERIVATIVES OF TRIGONOMETRIC FUNCTIONS
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Let’s try to confirm
our guess that, if f(x) = sin x,
then f’(x) = cos x.
DERIVATIVES OF TRIGONOMETRIC FUNCTIONS
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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
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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! ! ! !
"# + #
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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
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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
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We now use a geometric argument
to prove Equation 2.
!Assume first that _ lies between 0 and !/2.
DERIVS. OF TRIG. FUNCTIONS
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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
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sinsin so 1
!! !
!< <
DERIVS. OF TRIG. FUNCTIONS
We see that
|BC| < |AB| < arc AB
Thus,
Proof
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Let the tangent lines at A and B
intersect at E.
DERIVS. OF TRIG. FUNCTIONS Proof
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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
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Thus,
_ = arc AB < |AE| + |EB|
< |AE| + |ED|
= |AD| = |OA| tan _
= tan _
DERIVS. OF TRIG. FUNCTIONS Proof
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Therefore, we have:
So,
sin
cos
!!
!<
DERIVS. OF TRIG. FUNCTIONS
sincos 1
!!
!< <
Proof
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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
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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
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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
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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
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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
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So, we have proved the formula for
the derivative of the sine function:
(sin ) cosd
x xdx
=
DERIV. OF SINE FUNCTION Formula 4
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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
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Using the same methods as in
the proof of Formula 4, we can prove:
(cos ) sind
x xdx
= !
Formula 5DERIV. OF COSINE FUNCTION
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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
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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
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The derivatives of the remaining
trigonometric functions—csc, sec, and
cot—can also be found easily using the
Quotient Rule.
DERIVS. OF TRIG. FUNCTIONS
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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
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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
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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
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In simplifying the answer,
we have used the identity
tan2 x + 1 = sec2 x.
DERIVS. OF TRIG. FUNCTIONS Example 2
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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
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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
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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
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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
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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
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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
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The acceleration a = -4 cos t = 0 when s = 0.
It has greatest magnitude at the high and
low points.
Example 3APPLICATIONS
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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
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!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
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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
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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
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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
![Page 47: DIFFERENTIATION RULES - Facultyfaculty.essex.edu/~bannon/2007.fall.121/lectures/Chap3_Sec3.pdf · DIFFERENTIATION RULES. ... We have collected all the differentiation formulas for](https://reader035.vdocuments.site/reader035/viewer/2022062402/5af102007f8b9aa17b8fb774/html5/thumbnails/47.jpg)
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
! ! !
!
!
= =
= =
=