kalkulus 03-derivat
TRANSCRIPT
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0
limh
f a h f a
h
is called the derivative of at .f a
We write:
0limh
f a h f af x
h
The derivative off with respect toxis
There are many ways to write the derivative of y f x
3.1 Derivative of a Function
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f x f prime x or the derivative of f with respectto x
y
y primedy
dxdee why dee ecks or the derivative of y with
respect to x
df
dx dee eff dee ecks or the derivative of f withrespect to x
d f xdx dee dee ecks uv eff uv ecks or the derivative
of f of x( of of )d dx f x
3.1 Derivative of a Function
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dx does not mean dtimesx !
dy does not mean dtimesy !
3.1 Derivative of a Function
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dy
dxdoes not mean !dy dx
(except when it is convenient to think of it as division.)
df
dxdoes not mean !df dx
(except when it is convenient to think of it as division.)
3.1 Derivative of a Function
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(except when it is convenient to treat it that way.)
d
f xdx does not mean times !
d
dx f x
3.1 Derivative of a Function
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The derivative is
the slope of theoriginal function.
The derivative is definedat the end points of afunction on a closedinterval.
0
1
2
3
4
1 2 3 4 5 6 7 8 9
y f x
-2
-1
0
1
2
3
1 2 3 4 5 6 7 8 9
y f x
3.1 Derivative of a Function
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-3
-2
-1
0
1
2
3
4
5
6
- 3 - 2 - 1 1 2 3x
2 3y x
2 2
0
3 3limh
x h xy
h
2y x
-6
-5-4
-3
-2
-10
1
23
4
5
6
-3 -2 -1 1 2 3x
0lim2h
y x h
3.1 Derivative of a Function
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A function is differentiable if it has aderivative everywhere in its domain. Itmust be continuous and smooth.Functions on closed intervals must haveone-sided derivatives defined at the endpoints.
3.1 Derivative of a Function
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To be differentiable, a function must be continuousand smooth.Derivatives will fail to exist at:
corner
f x x
cusp
2
3 f x x
vertical tangent
3 f x x
discontinuity
1, 0
1, 0
x f x
x
3.2 Differentiability
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Most of the functions we study in calculus will be differentiable.
3.2 Differentiability
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There are two theorems on page 110:
If f has a derivative atx = a, thenf is continuous atx = a.
Since a function must be continuous to have a derivative,if it has a derivative then it is continuous.
3.2 Differentiability
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1
2f a
3f b
Intermediate Value Theorem for Derivatives
Between a and b, must take
on every value between and .
f1
23
If a and b are any two points in an interval on which f isdifferentiable, then takes on every value between
and .
f f a
f b
3.2 Differentiability
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If the derivative of a function is its slope, then for aconstant function, the derivative must be zero.
0d
cdx
example: 3y
0y
The derivative of a constant is zero.
3.3 Rules for Differentiation
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We saw that if , .2
y x 2y x
This is part of a pattern.
1n nd
x nxdx
examples:
4 f x x
34 f x x
8y x
78y x
power rule
3.3 Rules for Differentiation
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1n nd
x nx
dx
3.3 Rules for DifferentiationProof:
h
xhx
xdx
dnn
h
n
)(
lim0
h
xhhnxxx
dx
dnnnn
h
n
...lim
1
0
h
hhnxx
dx
d nn
h
n
...lim
1
0
1
0
lim
n
h
n nxx
dx
d
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d du
cu cdx dx
examples:
1n nd cx cnxdx
constant multiple rule:
5 4 47 7 5 35d
x x xdx
3.3 Rules for Differentiation
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(Each term is treated separately)
d du
cu cdx dx
constant multiple rule:
sum and difference rules:
d du dv
u vdx dx dx
d du dv
u vdx dx dx
4 12 y x x
3
4 12y x
4 22 2 y x x
3
4 4
dy
x xdx
3.3 Rules for Differentiation
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Find the horizontal tangents of:4 22 2 y x x 34 4
dyx x
dx
Horizontal tangents occur when slope = zero.34 4 0x x
3 0x x
2 1 0x x 1 1 0 x x x
0, 1, 1x
Substituting the x values into theoriginal equation, we get:
2, 1, 1 y y y (The function is even, so weonly get two horizontaltangents.)
3.3 Rules for Differentiation
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-2
-1
0
1
2
3
4
-2 -1 1 2
4 22 2 y x x
2y
1y
3.3 Rules for Differentiation
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-2
-1
0
1
2
3
4
-2 -1 1 2
4 22 2 y x x
First derivative(slope) is zero at:
0, 1, 1x
34 4
dyx x
dx
3.3 Rules for Differentiation
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product rule:
d dv duuv u vdx dx dx
Notice that this is not just theproduct of two derivatives.
This is sometimes memorized as: d uv u dv v du 2 33 2 5
d x x x
dx
5 3 3
2 5 6 15
d
x x x xdx
5 32 11 15d
x x xdx
4 2
10 33 15x x
2 3x 26 5x 32 5x x 2x
4 2 2 4 26 5 18 15 4 10 x x x x x
4 2
10 33 15x x
3.3 Rules for Differentiation
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product rule:
d dv duuv u vdx dx dx
3.3 Rules for Differentiation
Proof
h
xvxuhxvhxuuv
dx
d
h
)()()()(lim)(
0
add and subtract u(x+h)v(x)in the denominator
h
xvhxuxvhxuxvxuhxvhxuuv
dx
d
h
)()()()()()()()(lim)(
0
h
xuhxuxvxvhxvhxuuv
dx
d
h
)()()()()()(lim)(
0
dx
duv
dx
dvuuv
dx
d)(
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quotient rule:
2
du dvv u
d u dx dx
dx v v
or 2u v du u dv
dv v
3
2
2 5
3
d x x
dx x
2 2 3
22
3 6 5 2 5 2
3
x x x x x
x
3.3 Rules for Differentiation
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Higher Order Derivatives:dy
y
dx
is the first derivative of y with respect to x.
2
2
dy d dy d yy
dx dx dx dx
is the second derivative.
(y double prime)
dy
y dx
is the third derivative.
4 dy ydx
is the fourth derivative.
We will learnlater what thesehigher orderderivatives are
used for.
3.3 Rules for Differentiation
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3.3 Rules for DifferentiationSuppose u and v are functions that are differentiable at
x = 3, and that u(3) = 5, u(3) = -7, v(3) = 1, and v(3)= 4.
Find the following atx = 3 :
)(.1 uvdx
d'')( vuuvuv
dx
d 8)7)(1()3(5
vu
dxd.2
2''
vuvvu
vu
dxd
21
)4)(5()7)(1( 27
u
v
dx
d.3
2
''
u
vuuv
u
v
dx
d
2
5
)7)(1()4)(5(
25
27
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3.3 Rules for Differentiation
hi
ho
dx
d
))((
)()()()(
hoho
hidhohodhi
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3.3 Rules for Differentiation
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Consider a graph of displacement (distance traveled) vs. time.
time (hours)
distance(miles)
Average velocity can be found bytaking:
change in positionchange in time
s
t
t
sA
B
ave
f t t f t sV
t t
The speedometer in your car does not measure averagevelocity, but instantaneous velocity.
0
limt
f t t f t dsV t
dt t
(The velocity at one
moment in time.)
3.4 Velocity and other Rates
of Change
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3.4 Velocity and other Rates
of Change
Velocity is the first derivative of position.
Acceleration is the second derivative
of position.
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Example: Free Fall Equation
21 2
s g t
GravitationalConstants:
2ft32
secg
2
m9.8
sec
g
2
cm980
secg
21 322
s t
216s t 32ds
V tdt
Speed is the absolute value of velocity.
3.4 Velocity and other Rates
of Change
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Acceleration is the derivative of velocity.
dv
a dt
2
2
d s
dt example:32v t
32a If distance is in: feet
Velocity would be in: feetsec
Acceleration would be in:
ft
sec
sec
2
ft
sec
3.4 Velocity and other Rates
of Change
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time
distance
acc posvel pos &increasing
acc zerovel pos &constant
acc negvel pos &decreasing
velocityzero
acc negvel neg &decreasing acc zero
vel neg &constant
acc posvel neg &increasing
acc zero,velocity zero
3.4 Velocity and other Rates
of Change
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Rates of Change:
Average rate of change = f x h f xh
Instantaneous rate of change =
0limh
f x h f xf x
h
These definitions are true for any function.
( x does not have to represent time. )
3.4 Velocity and other Rates
of Change
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For a circle: 2A r
2dA d rdr dr
2dA
rdr
Instantaneous rate of change of the area withrespect to the radius.
For tree ring growth, if the change in area is constant then dr
must get smaller asr
gets larger.
2dA r dr
3.4 Velocity and other Rates
of Change
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from Economics:
Marginal cost is the first derivative of the cost function, andrepresents an approximation of the cost of producing onemore unit.
3.4 Velocity and other Rates
of Change
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Example 13:Suppose it costs: 3 26 15c x x x x to producex stoves. 23 12 15c x x x If you are currently producing 10 stoves,the 11th stove will cost approximately:
210 3 10 12 10 15c 300 120 15
$195
marginal cost
The actual cost is: 11 10C C
3 2 3 211 6 11 15 11 10 6 10 15 10
770 550 $220 actual cost
3.4 Velocity and other Rates
of Change
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Note that this is not agreat approximationDont let that bother you.
Marginal cost is a linear approximation of a curvedfunction. For large values it gives a good approximationof the cost of producing the next item.
3.4 Velocity and other Rates
of Change
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3.4 Velocity and other Rates
of Change
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2
0
2
Consider the function siny We could make a graph of the slope:
slope
1
0
1
0
1
Now we connect the dots!The resulting curve is a cosine curve.
sin cosd
x x
dx
3.5 Derivatives of
Trigonometric Functions
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3.5 Derivatives of
Trigonometric Functions
hxhxx
dxd
hsin)sin(limsin
0
h
xxhhxx
dx
d
h
sincossincossinlimsin
0
h
xh
h
hx
xdx
d
hh
cossin
lim
)1(cossin
limsin 00
h
xhhxx
dx
d
h
cossin)1(cossinlimsin
0
Proof
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3.5 Derivatives of
Trigonometric Functions
h
xh
h
hxx
dx
d
hh
cossinlim
)1(cossinlimsin
00
= 0 = 1
sin cosd
x xdx
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3.5 Derivatives of
Trigonometric Functions
hxhxx
dxd
hcos)cos(limcos
0
h
xxhhxx
dx
d
h
cossinsincoscoslimcos
0
h
xh
h
hx
xdx
d
hh
sinsin
lim
)1(coscos
limcos 00
h
xhhxx
dx
d
h
sinsin)1(coscoslimcos
0
Find the derivative of cos x
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3.5 Derivatives of
Trigonometric Functions= 0 = 1
h
xh
h
hxx
dx
d
hh
sinsinlim
)1(coscoslimcos
00
cos sin
d
x xdx
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We can find the derivative of tangentx by using thequotient rule.
tand xdx
sin
cos
d x
dx x
2
cos cos sin sin
cos
x x x x
x
2 2
2cos sin
cosx x
x
2
1
cos
x2
sec x
2tan secd x xdx
3.5 Derivatives of
Trigonometric Functions
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Derivatives of the remaining trig functionscan be determined the same way.
sin cosd x xdx
cos sind
x xdx
2tan sec
dx x
dx
2cot cscd x x
dx
sec sec tand
x x xdx
csc csc cotd
x x xdx
3.5 Derivatives of
Trigonometric Functions
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3.5 Derivatives of
Trigonometric Functions
Jerk A sudden change in acceleration
Definition JerkJerk is the derivative of acceleration. If a bodys positionat time tis s(t), the bodys jerk at time t is
3
3
2
2
)(dt
sd
dt
vd
dt
datj
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3.5 Derivatives of
Trigonometric Functions
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Consider a simple composite function:6 10y x
2 3 5y x If 3 5u x
then 2y u
6 10y x 2y u 3 5u x
6dy
dx 2
dy
du 3
du
dx
dy dy du
dx du dx
6 2 3
3.6 Chain Rule
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dy dy du
dx du dx Chain Rule:
example: sin f x x 2 4g x x Find: at 2 f g x
cos f x x 2g x x 2 4 4 0g
0 2f g cos 0 2 2 1 4 4
3.6 Chain Rule
If is the composite of and ,then:
f g y f u u g x
atat xu g x f g f g )('))((' xgxgf
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2sin 4 f g x x
2sin 4y x siny u 2 4u x
cosdy
udu
2du
xdx
dy dy du
dx du dx
cos 2
dy
u xdx
2cos 4 2dy
x xdx
2cos 2 4 2 2dy
dx
cos 0 4dydx
4dy
dx
3.6 Chain Rule
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Here is a faster way to find the derivative:
2sin 4y x
2 2cos 4 4d
y x xdx
2cos 4 2 y x x
Differentiate the outside function...
then the inside function
At 2, 4x y
3.6 Chain Rule
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2cos 3d
xdx
2
cos 3d
xdx
2 cos 3 cos 3dx xdx
2cos 3 sin 3 3d
x x x
dx
2cos 3 sin 3 3x x
6cos 3 sin 3x x
The chain rule can be usedmore than once.
(Thats what makes the
chain in the chain rule!)
3.6 Chain Rule
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Derivative formulas include the chain rule!
1n n
d duu nu
dx dx sin cos
d duu u
dx dx
cos sind du
u udx dx
2tan secd du
u udx dx
etcetera
3.6 Chain Rule
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3.6 Chain RuleFind
)3cos( 2 xxy )16)(3sin( 2 xxxdx
dy
))sin(cos(xy
)24(cos 33 xxy
)sin)(cos(cos xx
dx
dy
)212))(24sin()(24(cos3 2332 xxxxxdx
dy
))24sin()(24(cos)636(3322
xxxxxdx
dy
dx
dy
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The chain rule enables us to find the slope ofparametrically defined curves:
dy dy dx
dt dx dt
dydydt
dx dx
dt
The slope of a parametrizedcurve is given by:
dy
dy dtdxdx
dt
3.6 Chain Rule
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These are the equations for
an ellipse.
Example:
3cosx t 2siny t
3sindx
tdt
2cosdy
tdt
2cos
3sin
dy t
dx t
2
cot3
t
3.6 Chain Rule
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2 2 1x y This is not a function,but it would still benice to be able to find
the slope.2 2 1d d d
x ydx dx dx
Do the same thing to both sides.
2 2 0
dy
x y dx
Note use of chain rule.
2 2dy
y xdx
22
dy xdx y
dy xdx y
3.7 Implicit Differentiation
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22 sin y x y
22 sind d d
y x ydx dx dx
This cant be solved fory.
2 2 cosdy dy
x ydx dx
2 cos 2dy dyy xdx dx
22 cosdy
xy
dx
2
2 cos
dy x
dx y
This technique is calledimplicit differentiation.
1 Differentiate both sides w.r.t.x.
2 Solve for .dy
dx
3.7 Implicit Differentiation
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3.7 Implicit Differentiation
Implicit Differentiation Process
1. Differentiate both sides of the equation with respect tox.2. Collect the terms with dy/dx on one side of the equation.3. Factor out dy/dx .4. Solve for dy/dx .
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Find the equations of the lines tangent and normal to the
curve at .2 2 7 x xy y ( 1,2)
2 2 7 x xy y
2 2 0
dydy
x yx y dxdx
Note product rule.
2 2 0dy dy
x x y ydx dx
22dy
y xy x dx
2
2
dy y x
dx y x
2 2 1
2 2 1m
2 2
4 1
4
5
3.7 Implicit Differentiation
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Find the equations of the lines tangent and normal to the
curve at .2 2 7 x xy y ( 1,2)
4
5m
tangent:
4
2 15
y x
4 42 5 5y x
4 14
5 5
y x
normal:
5
2 14
y x
5 52 4 4y x
5 3
4 4y x
3.7 Implicit Differentiation
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3.7 Implicit Differentiation
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Find if .
2
2
d y
dx3 22 3 7x y
3 2
2 3 7x y 26 6 0 x y y
26 6 y y x
26
6
xy
y
2xy
y
2
2
2y x x yy
y
2
2
2x xy y
y y
2 2
2
2x xyy
x
yy
4
3
2x xy
y y
Substituteback into theequation.
y
3.7 Implicit Differentiation
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3.7 Implicit Differentiation
Rational Powers of Differentiable Functions
Power Rule for Rational Powers ofx
If n is any rational number, then
1 nn nxxdxd
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3.7 Implicit Differentiation
Proof: Letp and q be integers with q > 0.
q
p
xy pq
xy
Raise both sides to the q power
Differentiate with respect tox
11 pq pxdx
dyqy Solve for dy/dx
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3.7 Implicit Differentiation
1
1
q
p
qy
px
dx
dy Substitute for y
1/
1
)(
qqp
p
xq
px
dx
dyRemove parenthesis
qpp
p
qx
px
dx
dy/
1
Subtract exponents
q
px
dx
dyqppp )/(1
1)/( qpx
q
p
dx
dy
3 8 D i ti f I
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Becausex andy are
reversed to find the
reciprocal function, the
following pattern always
holds:86420
8
6
4
2
0
x
y
x
y
2y x
y x
4m 2,4
4,21
4m
Slopes are
reciprocals.
Derivative Formula for Inverses:
df
dx df
dx x f a
x a
1 1
( )
evaluated at ( )f a
is equal to the reciprocal of
the derivative of ( )f x
evaluated at .a
The derivative of 1( )f x
3.8 Derivatives of InverseTrigonometric Functions
3 8 D i ti f I
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-1.5
-1
-0.5
0
0.5
1
1.5
-1.5 -1 -0.5 0.5 1 1.5
siny x
1siny xWe can use implicit
differentiation to find:1sin
dx
dx
1siny x
siny x
sind d
y xdx dx
cos 1dy
ydx
1
cos
dy
dx y
3.8 Derivatives of InverseTrigonometric Functions
3 8 D i ti f I
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We can use implicitdifferentiation to find:
1sind
xdx
1siny x
sin y x sind dy xdx dx
cos 1dy
y
dx
1
cos
dy
dx y
2 2sin cos 1y y
2 2
cos 1 siny y 2cos 1 siny y
But
2 2
y
so is positive.cos y
2cos 1 siny y
2
1
1 sin
dy
dx y
2
1
1
dy
dx x
3.8 Derivatives of InverseTrigonometric Functions
3 8 D i ti f I
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1siny x
1
cos
dy
dx y
3.8 Derivatives of InverseTrigonometric Functions
)cos(sin
11xdx
dy
x
1
sin
x
1
21 x21
1
xdx
dy
xy sin
1cos dxdyy
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3.8 Derivatives of InverseTrigonometric Functions
)(tansec
112xdx
dy
x
1
tan
x
1
21 x
211xdx
dy
xy tan
1sec2 dxdyy
Find xdx
d 1tan
xy 1tan
ydx
dy2
sec
1
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3.8 Derivatives of InverseTrigonometric Functions
)tan(sec)sec(sec1
11xxdx
dy
x1sec
x
1
12 x
1||
1
2
xxdx
dyxy sec
1tansec dxdyyy
Find xdx
d 1sec
xy 1sec
yydx
dy
tansec
1
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1
2
1sin
1
d duu
dx dxu
121tan 1
d duudx u dx
1
2
1sec
1
d duu
dx dxu u
1
2
1cos
1
d duu
dx dxu
1
21cot
1d duu
dx u dx
1
2
1csc
1
d duu
dx dxu u
1 1cos sin2
x x 1 1cot tan
2x x
1 1csc sec2
x x
3.8 Derivatives of InverseTrigonometric Functions
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Your calculator contains allsix inverse trig functions.However it is occasionally
still useful to know thefollowing:
1 1 1sec cosxx
1 1cot tan2
x x
1 1 1csc sinxx
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3.8 Derivatives of InverseTrigonometric Functions
Find
)3(cos 21 xy
422 916)6(
)3(1(1
xxx
xdxdy
x
y1
cot1
xxy 1sec
1
11
11
122
2
xx
xdx
dy
)1)((sec1||
1 12
xxx
xdx
dy
dx
dy
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-1
0
1
2
3
-3 -2 -1 1 2 3x
Look at the graph ofx
y e
The slope atx = 0appears to be 1.
If we assume this to betrue, then:
0 0
0lim 1
h
h
e e
h
definition of derivative
3.9 Derivatives of Exponential and
Logarithmic Functions
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Now we attempt to find a general formula for the
derivative of using the definition.xy e
0
lim x h x
x
h
d e ee
dx h
0lim
x h x
h
e e e
h
0
1lim
hx
h
ee
h
0
1lim
hx
h
ee
h
1x
e
xe
This is the slope atx = 0, which
we have assumed to be 1.
3.9 Derivatives of Exponential
and Logarithmic Functions
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x xd
e edx
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xe is its own derivative!
If we incorporate the chain rule: u ud due e
dx dx
We can now use this formula to find the derivative ofxa
3.9 Derivatives of Exponentialand Logarithmic Functions
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xd
adx
lnx
ad edx
lnx ad edx
ln lnx ad
e x adx
Incorporating the chain rule:
lnu ud du
a a adx dx
3.9 Derivatives of Exponentialand Logarithmic Functions
aaa
dx
d xx ln
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So far today we have:
u ud due e
dx dx lnu u
d dua a a
dx dx
Now it is relatively easy to find the derivative of .ln x
3.9 Derivatives of Exponentialand Logarithmic Functions
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lny xye x
yd d
e x
dx dx
1ydy
e
dx
1y
dy
dx e
1ln
dx
dx x
1ln
d duu
dx u dx
3.9 Derivatives of Exponentialand Logarithmic Functions
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To find the derivative of a common log function, youcould just use the change of base rule for logs:
logd xdx
lnln10
d xdx
1 lnln10
d xdx
1 1ln10 x
The formula for the derivative of a log of any base
other than e is:
1log
ln
a
d duu
dx u a dx
3.9 Derivatives of Exponentialand Logarithmic Functions
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u ud due e
dx dx
lnu ud du
a a adx dx
1logln
ad duudx u a dx
1lnd duudx u dx
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3.9 Derivatives of Exponential
and Logarithmic Functions
x
ey
2
2
3xy
3
ln xy
)(sin 41 xey
Findy
x
ey
2
2')2)(3ln(3'
2
xy x
xxxy3
)3(1
'2
3
)4)((
)(1
1' 4
24
x
xe
e
y
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3.9 Derivatives of Exponential
and Logarithmic Functions
Logarithmic differentiation
Used when the variable is in the base and the exponent
y = xx
ln y = ln xx
ln y = x ln x
xxd
dyln
11
xydx
dyln1
xxdxdy x ln1