3.1 derivative of a function
DESCRIPTION
3.1 Derivative of a Function. We write:. There are many ways to write the derivative of. is called the derivative of at . “The derivative of f with respect to x is …”. 3.1 Derivative of a Function. “the derivative of f with respect to x”. “f prime x”. or. “y prime”. - PowerPoint PPT PresentationTRANSCRIPT
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0
limh
f a h f ah
is called the derivative of at .f a
We write: 0
limh
f a h f af x
h
“The derivative of f with respect to x is …”
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 respect to x”
y “y prime”
dydx “dee why dee ecks” or “the derivative of y with
respect to x”dfdx “dee eff dee ecks” or “the derivative of f with
respect 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 d times x !
dy does not mean d times y !
3.1 Derivative of a Function
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dydx does not mean !dy dx
(except when it is convenient to think of it as division.)
dfdx
does 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 !ddx
f x
3.1 Derivative of a Function
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The derivative is the slope of the original function.
The derivative is defined at the end points of a function on a closed interval.
y f x
y f x
3.1 Derivative of a Function
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2 3y x
2 2
0
3 3limh
x h xy
h
2y x
0lim 2h
y x h
3.1 Derivative of a Function
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A function is differentiable if it has a derivative everywhere in its domain. It must be continuous and smooth. Functions on closed intervals must have one-sided derivatives defined at the end points.
3.1 Derivative of a Function
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To be differentiable, a function must be continuous and smooth.Derivatives will fail to exist at:
corner
f x x
cusp
23f x x
vertical tangent 3f x x
discontinuity
1, 0 1, 0
xf 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 at x = a, then f is continuous at x = 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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12
f a
3f b
Intermediate Value Theorem for Derivatives
Between a and b, must take
on every value between and .
f 12 3
If a and b are any two points in an interval on which f is
differentiable, 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 a constant 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 , .2y x 2y x
This is part of a pattern.
1n nd x nxdx
examples:
4f x x
34f x x
8y x
78y x
power rule
3.3 Rules for Differentiation
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1n nd x nxdx
3.3 Rules for Differentiation
Proof:
hxhxx
dxd nn
h
n
)(lim0
hxhhnxxx
dxd nnnn
h
n
...lim1
0
hhhnxx
dxd nn
h
n
...lim1
0
1
0lim
n
h
n nxxdxd
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d ducu 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 ducu cdx dx
constant multiple rule:
sum and difference rules:
d du dvu vdx dx dx
d du dvu vdx dx dx
4 12y x x 34 12y x
4 22 2y x x
34 4dy x xdx
3.3 Rules for Differentiation
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Find the horizontal tangents of: 4 22 2y x x 34 4dy x x
dx
Horizontal tangents occur when slope = zero.34 4 0x x
3 0x x
2 1 0x x
1 1 0x x x
0, 1, 1x
Substituting the x values into the original equation, we get:
2, 1, 1y y y (The function is even, so we only get two horizontal tangents.)
3.3 Rules for Differentiation
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4 22 2y x x
2y
1y
3.3 Rules for Differentiation
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4 22 2y x x
First derivative (slope) is zero at:
0, 1, 1x
34 4dy x xdx
3.3 Rules for Differentiation
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product rule: d dv duuv u v
dx dx dx Notice that this is not just the
product of two derivatives.
This is sometimes memorized as: d uv u dv v du
2 33 2 5d x x xdx
5 3 32 5 6 15d x x x xdx
5 32 11 15d x x xdx
4 210 33 15x x
2 3x 26 5x 32 5x x 2x
4 2 2 4 26 5 18 15 4 10x x x x x
4 210 33 15x x
3.3 Rules for Differentiation
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product rule: d dv duuv u v
dx dx dx
3.3 Rules for Differentiation
Proof
hxvxuhxvhxuuv
dxd
h
)()()()(lim)(0
add and subtract u(x+h)v(x)in the denominator
hxvhxuxvhxuxvxuhxvhxuuv
dxd
h
)()()()()()()()(lim)(0
h
xuhxuxvxvhxvhxuuvdxd
h
)()()()()()(lim)(0
dxduv
dxdvuuv
dxd
)(
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quotient rule:
2
du dvv ud u dx dxdx v v
or 2
u v du u dvdv v
3
2
2 53
d x xdx 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:dyydx
is the first derivative of y with respect to x.
2
2
dy d dy d yydx dx dx dx
is the second derivative.(y double prime)
dyydx
is the third derivative.
4 dy ydx
is the fourth derivative.
We will learn later what these higher order derivatives are used for.
3.3 Rules for Differentiation
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3.3 Rules for Differentiation
Suppose u and v are functions that are differentiable atx = 3, and that u(3) = 5, u’(3) = -7, v(3) = 1, and v’(3)= 4.Find the following at x = 3 :
)(.1 uvdxd
'')( vuuvuvdxd
8)7)(1()3(5
vu
dxd.2
2
''v
uvvuvu
dxd
21)4)(5()7)(1( 27
uv
dxd.3
2
''u
vuuvuv
dxd
25)7)(1()4)(5(
2527
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3.3 Rules for Differentiation
hiho
dxd
))(()()()()(
hohohidhohodhi
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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 by taking:change in position
change in timest
t
sA
B
ave
f t t f tsVt t
The speedometer in your car does not measure average velocity, but instantaneous velocity.
0
limt
f t t f tdsV tdt 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:
2
ft32 sec
g
2
m9.8 sec
g
2
cm980 sec
g
21 32 2
s t
216 s t 32 dsV 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.
dvadt
2
2
d sdt
example: 32v t
32a If distance is in: feet
Velocity would be in:feetsec
Acceleration would be in:
ftsec sec
2
ftsec
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 x
h
Instantaneous rate of change = 0
limh
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 r2dA d r
dr 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 as r gets larger.
2 dA 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, and represents an approximation of the cost of producing one more 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 produce x 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
$195marginal 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 a great approximation – Don’t let that bother you.
Marginal cost is a linear approximation of a curved function. For large values it gives a good approximation of 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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20
2
Consider the function siny
We could make a graph of the slope: slope
1
0
1
0
1Now we connect the dots!The resulting curve is a cosine curve.
sin cosd x xdx
3.5 Derivatives of Trigonometric Functions
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3.5 Derivatives of Trigonometric Functions
hxhxx
dxd
h
sin)sin(limsin0
hxxhhxx
dxd
h
sincossincossinlimsin0
hxh
hhxx
dxd
hh
cossinlim)1(cossinlimsin00
hxhhxx
dxd
h
cossin)1(cossinlimsin0
Proof
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3.5 Derivatives of Trigonometric Functions
hxh
hhxx
dxd
hh
cossinlim)1(cossinlimsin00
= 0 = 1
sin cosd x xdx
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3.5 Derivatives of Trigonometric Functions
hxhxx
dxd
h
cos)cos(limcos0
hxxhhxx
dxd
h
cossinsincoscoslimcos0
hxh
hhxx
dxd
hh
sinsinlim)1(coscoslimcos00
hxhhxx
dxd
h
sinsin)1(coscoslimcos0
Find the derivative of cos x
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3.5 Derivatives of Trigonometric Functions
= 0 = 1
hxh
hhxx
dxd
hh
sinsinlim)1(coscoslimcos00
cos sind x xdx
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We can find the derivative of tangent x by using the quotient rule.
tand xdx
sincos
d xdx x
2
cos cos sin sincos
x x x xx
2 2
2
cos sincosx x
x
2
1cos x
2sec x
2tan secd x xdx
3.5 Derivatives of Trigonometric Functions
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Derivatives of the remaining trig functions can be determined the same way.
sin cosd x xdx
cos sind x xdx
2tan secd x xdx
2cot cscd x xdx
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 body’s positionat time t is s(t), the body’s jerk at time t is
3
3
2
2
)(dt
sddt
vddtdatj
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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
6dydx 2dy
du 3du
dx
dy dy dudx du dx
6 2 3
3.6 Chain Rule
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dy dy dudx du dx Chain Rule:
example: sinf x x 2 4g x x Find: at 2f g x
cosf 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
at at xu g xf g f g )('))((' xgxgf
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2sin 4f g x x
2sin 4y x
siny u 2 4u x
cosdy udu 2du x
dx
dy dy dudx du dx
cos 2dy u xdx
2cos 4 2dy x xdx
2cos 2 4 2 2dydx
cos 0 4dydx
4dydx
3.6 Chain Rule
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Here is a faster way to find the derivative:
2sin 4y x
2 2cos 4 4dy x xdx
2cos 4 2y 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
2cos 3d x
dx
2 cos 3 cos 3dx xdx
2cos 3 sin 3 3dx x xdx
2cos 3 sin 3 3x x
6cos 3 sin 3x x
The chain rule can be used more than once.
(That’s what makes the “chain” in the “chain rule”!)
3.6 Chain Rule
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Derivative formulas include the chain rule!
1n nd duu nudx dx
sin cosd duu udx dx
cos sind duu udx dx
2tan secd duu udx dx
etcetera…
3.6 Chain Rule
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3.6 Chain RuleFind
)3cos( 2 xxy )16)(3sin( 2 xxxdxdy
))sin(cos(xy
)24(cos 33 xxy
)sin)(cos(cos xxdxdy
)212))(24sin()(24(cos3 2332 xxxxxdxdy
))24sin()(24(cos)636( 3322 xxxxxdxdy
dxdy
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The chain rule enables us to find the slope of parametrically defined curves:
dy dy dxdt dx dt
dydydt
dx dxdt
The slope of a parametrized curve is given by:
dydy dt
dxdxdt
3.6 Chain Rule
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These are the equations for an ellipse.
Example:
3cosx t 2siny t
3sindx tdt 2cosdy t
dt
2cos3sin
dy tdx 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 be nice to be able to find the slope.
2 2 1d d dx ydx dx dx
Do the same thing to both sides.
2 2 0dyx ydx
Note use of chain rule.
2 2dyy xdx
22
dy xdx y
dy x
dx y
3.7 Implicit Differentiation
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22 siny x y
22 sind d dy x ydx dx dx
This can’t be solved for y.
2 2 cosdy dyx ydx dx
2 cos 2dy dyy xdx dx
22 cosdy xydx
22 cos
dy xdx y
This technique is called implicit differentiation.
1 Differentiate both sides w.r.t. x.
2 Solve for .dydx
3.7 Implicit Differentiation
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3.7 Implicit DifferentiationImplicit Differentiation Process
1. Differentiate both sides of the equation with respect to x.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 7x xy y ( 1, 2)2 2 7x xy y
2 2 0dydyx yx ydxdx
Note product rule.
2 2 0dy dyx x y ydx dx
22dy y xy xdx
22
dy y xdx y x
2 2 12 2 1
m
2 24 1
45
3.7 Implicit Differentiation
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Find the equations of the lines tangent and normal to the
curve at .2 2 7x xy y ( 1, 2)
45
mtangent:
42 15
y x
4 425 5
y x
4 145 5
y x
normal:
52 14
y x
5 524 4
y x
5 34 4
y x
3.7 Implicit Differentiation
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3.7 Implicit Differentiation
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Find if .2
2
d ydx
3 22 3 7x y
3 22 3 7x y
26 6 0x y y
26 6y y x 26
6xyy
2xyy
2
2
2y x x yyy
2
2
2x xy yy y
2 2
2
2x xyy
xyy
4
3
2x xyy y
Substitute back into the equation.
y
3.7 Implicit Differentiation
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3.7 Implicit DifferentiationRational Powers of Differentiable Functions
Power Rule for Rational Powers of x
If n is any rational number, then
1 nn nxx
dxd
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3.7 Implicit DifferentiationProof: Let p and q be integers with q > 0.
qp
xy
pq xy
Raise both sides to the q power
Differentiate with respect to x
11 pq pxdxdyqy Solve for dy/dx
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3.7 Implicit Differentiation
1
1
q
p
qypx
dxdy Substitute for y
1/
1
)(
qqp
p
xqpx
dxdy Remove parenthesis
qpp
p
qxpx
dxdy
/
1
Subtract exponents
qpx
dxdy qppp )/(1
1)/( qpx
qp
dxdy
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Because x and y are reversed to find the reciprocal function, the following pattern always holds:
2y x
y x
4m 2,4
4,2 14
m
Slopes are reciprocals.
Derivative Formula for Inverses:
dfdx df
dxx f a
x a
1 1
( )
evaluated at ( )f a
is equal to the reciprocal ofthe derivative of ( )f xevaluated at .a
The derivative of 1( )f x
3.8 Derivatives of Inverse Trigonometric Functions
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siny x
1siny xWe can use implicit differentiation to find:
1sind xdx
1siny x
sin y x sind dy xdx dx
cos 1dyydx
1cos
dydx y
3.8 Derivatives of Inverse Trigonometric Functions
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We can use implicit differentiation to find:
1sind xdx
1siny x
sin y x sind dy xdx dx
cos 1dyydx
1cos
dydx y
2 2sin cos 1y y
2 2cos 1 siny y 2cos 1 siny y
But 2 2y
so is positive.cos y
2cos 1 siny y
2
1
1 sin
dydx y
2
1
1
dydx x
3.8 Derivatives of Inverse Trigonometric Functions
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1siny x
1cos
dydx y
3.8 Derivatives of Inverse Trigonometric Functions
)cos(sin1
1 xdxdy
x1sin x
1
21 x211
xdxdy
xy sin
1cos dxdyy
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3.8 Derivatives of Inverse Trigonometric Functions
)(tansec1
12 xdxdy
x1tan x
1
21 x
211xdx
dy
xy tan
1sec2 dxdyy
Find xdxd 1tan
xy 1tan
ydxdy
2sec1
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3.8 Derivatives of Inverse Trigonometric Functions
)tan(sec)sec(sec1
11 xxdxdy
x1sec
x
1
12 x
1||
12
xxdx
dyxy sec
1tansec dxdyyy
Find xdxd 1sec
xy 1sec
yydxdy
tansec1
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1
2
1sin1
d duudx dxu
12
1tan1
d duudx u dx
1
2
1sec1
d duudx dxu u
1
2
1cos1
d duudx dxu
12
1cot1
d duudx u dx
1
2
1csc1
d duudx dxu u
1 1cos sin2
x x 1 1cot tan2
x x 1 1csc sec2
x x
3.8 Derivatives of Inverse Trigonometric Functions
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Your calculator contains all six inverse trig functions.However it is occasionallystill useful to know the following:
1 1 1sec cosxx
1 1cot tan2
x x
1 1 1csc sinxx
3.8 Derivatives of Inverse Trigonometric Functions
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3.8 Derivatives of Inverse Trigonometric Functions
Find
)3(cos 21 xy 422 916)6(
)3(1(1
xxx
xdxdy
xy 1cot 1
xxy 1sec
111
11
122
2
xxx
dxdy
)1)((sec1||
1 1
2x
xxx
dxdy
dxdy
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Look at the graph of xy e
The slope at x = 0 appears to be 1.
If we assume this to be true, then:
0 0
0lim 1
h
h
e eh
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
limx h x
x
h
d e eedx h
0lim
x h x
h
e e eh
0
1limh
x
h
eeh
0
1limh
x
h
eeh
1xe xe
This is the slope at x = 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 edx dx
We can now use this formula to find the derivative of xa
3.9 Derivatives of Exponential and Logarithmic Functions
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xd adx
ln xad edx
lnx ad edx
ln lnx a de x adx
Incorporating the chain rule:
lnu ud dua a adx dx
3.9 Derivatives of Exponential and Logarithmic Functions
aaadxd xx ln
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So far today we have:
u ud due edx dx
lnu ud dua a adx dx
Now it is relatively easy to find the derivative of .ln x
3.9 Derivatives of Exponential and Logarithmic Functions
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lny xye x
yd de xdx dx
1y dyedx
1y
dydx e
1lnd xdx x
1lnd duudx u dx
3.9 Derivatives of Exponential and Logarithmic Functions
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To find the derivative of a common log function, you could just use the change of base rule for logs:
logd xdx
lnln10
d xdx
1 ln
ln10d xdx
1 1
ln10 x
The formula for the derivative of a log of any base other than e is:
1loglna
d duudx u a dx
3.9 Derivatives of Exponential and Logarithmic Functions
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u ud due edx dx
lnu ud dua a adx dx
1loglna
d duudx u a dx
1lnd duu
dx u dx
3.9 Derivatives of Exponential and Logarithmic Functions
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3.9 Derivatives of Exponential and Logarithmic Functions
xey 22
3xy
3ln xy
)(sin 41 xey
Find y’
xey 22'
)2)(3ln(3'2
xy x
xx
xy 3)3(1' 2
3
)4)(()(1
1' 4
24
x
xe
ey
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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
xx
xdxdy
yln11
xydxdy ln1
xxdxdy x ln1