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Calculus I Chapter Three 1
Calculus I Chapter Three 2
Calculus Timeline: Descartes 1596-1650Cavalieri 1598-1647Fermat 1601-1665Wallis 1616-1703Barrow 1630-1677Gregory 1638-1675Newton 1642-1727Leibniz 1646-1716
Calculus I Chapter Three 3
The Tangent Problem:
were called infinitesimals, very small numbers, but not zero!
This idea was problematic because infinitesimals were called ghosts of
departed quantities by Berkley!
dx and dy
GeoGebra
Slope of the tangent line:
dyslope
dx
Calculus I Chapter Three 4
0
Slope of the tangent line:
: What is the slope of the tangent line at ?
dyslope
dxExample x
3
3 3 2 2 30 0 0 0
3 2 2 30 0 0 0 0
2 3
3 20 0 0 0
3 2 30 0 0 0 0 0 0
( ) 2 .
: ( ) 3 3
( ) 3 3 2 2
, :
( ) 3 2 2
( ) ( ) 3 2 2 2
Let y f x x x
Note x dx x x dx x dx dx
f x dx x x dx x dx dx x dx
Since dx is very small dx and dx go to zero
f x dx x x dx x dx
dy f x dx f x x x dx x dx x x
dy
20
2 20 0 0 0
3 2
(3 2 ) 3 2
x dx dx
dydy x x dx x x
dx
5Calculus I Chapter Three
Explain how slopes of secant lines approach the slopes of the tangent line at a point.(Definition(1))
Definition(1) Derivative at a point GeoGebra file
Slope of the secant line:
( ) ( )f x f a
x a
Slope of the tangent line:
( ) ( )limx a
f x f a
x a
Calculus I Chapter Three 6
Explain how slopes of secant lines approach the slopes of the tangent line at a point. (Definition(2))
Slope of the secant line:
( ) ( )f x h f x
h
0
Slope of the tangent line:
( ) ( )limh
f x h f x
h
Definition(2) Derivative at a point GeoGebra file
Calculus I Chapter Three 7
The Derivative Function
Calculus I Chapter Three 8
3( )f x xAverage Rates
The pattern of the average rates looks quadratic!
2( ) 3f x x
1 2
2 3
3 44 55 6
6 7
(8 1)7
(2 1)
x Average Rate
1 7
2 19
3 37
4 61
5 91
6 127
3 2
2
The derivative of
( ) is 3 .
( ) 3 is a function called
the derivative function.
f x x x
f x x
Calculus I Chapter Three 9
The derivative can be interpreted in two ways:• the slope of the tangent line to the graph of the relation at the given x value.• the rate of change of the function with respect to x at the given x value.
Calculus I Chapter Three 10
Slope of the tangent line:
( ) ( )limx a
f x f a
x a
Calculus I Chapter Three 11
0
Slope of the tangent line:
( ) ( )limh
f x h f x
h
Calculus I Chapter Three 12
0
Slope of the tangent line:
( ) ( )limh
f x h f x
h
Calculus I Chapter Three 13
Calculus I Chapter Three 14
a-D b-C c-B d-A
Calculus I Chapter Three 15
Calculus I Chapter Three 16
Differentiable Implies Continuous
( ) is not defined at f x a( ) is not defined at f x a
( ) is not defined at f x a
( ) is not defined at f x a
Calculus I Chapter Three 17
Calculus I Chapter Three 18
Calculus I Chapter Three 19
Calculus I Chapter Three 20
Calculus I Chapter Three 21
0 0 0
( )
( ) ( )( ) lim lim lim 0 0
h h h
f x c
f x h f x c cf x
h h
Calculus I Chapter Three 22
Calculus I Chapter Three 23
Calculus I Chapter Three 24
1 11
0
What is the derivative of the sum of differentiable functions?
Let ( ) ( ) ( ), does ( ) exist?
( ) ( )( ) lim
x
h x f x g x h x
h x x h xh x
x
1 1 1
1 1 1
1 1
1 1 1 1
1 1 1 1
1 1 1 1
( ) ( ) ( )
( ) ( ) ( )
( ) ( )
( ) ( ) ( ( ) ( ))
( ) ( ) ( ) ( )
( ) ( ) ( ) ( )
h x x f x x g x x
h x f x g x
h x x h x
xf x x g x x f x g x
xf x x f x g x x g x
xf x x f x g x x g x
x x
Calculus I Chapter Three 25
1 11 0
1 1 1 1
0
1 1 1 1
0 0
1 1
( ) ( )( ) lim
( ) ( ) ( ) ( )lim
( ) ( ) ( ) ( )lim lim
( ) ( )
x
x
x x
h x x h xh x
xf x x f x g x x g x
x xf x x f x g x x g x
x xf x g x
What is the derivative of the sum of differentiable functions?
Let ( ) ( ) ( ), does ( ) exist? Yes!h x f x g x h x
Calculus I Chapter Three 26
Calculus I Chapter Three 27
Calculus I Chapter Three 28
2
Is the derivative of a product the product of the derivatives?
( ( ) ( )) ( ) ( )
( ) 1 ( ) 1
( ) 1 ( ) 1
( ) ( ) 1
But:
( ) ( ) 1
( ( ) ( )) 2
Is the deriv
f x g x f x g x
Example
f x x g x x
f x g x
f x g x
f x g x x
f x g x x
ative of a product the product
of the derivative? !NO
Calculus I Chapter Three 29
1 11
0
What is the derivative of the product of
differentiable functions?
Let ( ) ( ) ( ), does ( ) exist?
We use the definition of derivative:
( ) ( )( ) lim
Using the definition we'll p
x
h x f x g x h x
h x x h xh x
x
1 1 1 1 1
rove that:
If ( ) ( ) ( ), then
( ) ( ) ( ) ( ) ( )
h x f x g x
h x g x f x g x f x
Calculus I Chapter Three 30
1 1 1
1 1 1
1 1 1
1 1 1 1
1 1 1
1 1 1 1
1 1 1 11 1
( ) ( ) ( )
( ) ( ) ( )
( ) ( ) (
( ) ( )
) (
( )
) ( ( ) ( ))
( ) ( ) ( ( ) ( ))
(
( )
( )) ( )( ( ))(( ( ) ( ))
h x x f x x g x x
h x f x g x
h x x h x f x x g x x f x g xx x
f x x g x x f x g xx
g x x f x
f x g x x f x g x x
f x f x g xx xx
x g
1 11
1
1 11
1 110
1 1 1 1
11
1 1
11
1
1
11
(( ( ) ( ))( )
(( ( ) ( ))lim ( )
( ) ( ) ( ) ( )
( )) ( ( )( )
( )) ( ( )
(( ( )(
( )
( ) ( )
( ))
( ( )
)
( ))( )
x
f x g x xf x
f x g x x
f x x g xg x x
x xf x x g x
g x xx x
g x f x g x f x
f x
f x
f x f x
f x
g
xf x
xxx x g
g xx
1 1
Note that we add and subtract the blue expressions.Note that when 0, ( ) ( ).
Note that g (x) exists, and we can say that g(x) is continuous.
x g x x g x
Calculus I Chapter Three 31
( )uv u v uv
Calculus I Chapter Three 32
11 0
What is the derivative of the quotient of two differentiable functions?
( )Let ( ) , does ( ) exist?
( )
( )Note that ( ) !
( )
We use the definition of derivative:
( )( ) lim
x
f xh x h x
g x
f xh x
g x
h x x hh x
1
1 1 1 11 2
1
( )
Using the definition we can prove that:
( )If ( ) , then
( )
( ) ( ) ( ) ( )( )
( ( ))
The proof is similar to the proof of the product rule.
x
x
f xh x
g x
g x f x g x f xh x
g x
Calculus I Chapter Three 33
2
1 11 2 1
2 2
Extending the power rule to negative integers
If ( ) ( ) ?
1Let ( ) and use the qoutient rule:
(0)( ) ( )( )
( )
This shows that we
n
n
n n nn n n
n n
f x x f x
f xx
u u v uv
v v
x nx nxf x nx nx
x x
0
can extend the power rule to negative integers!
Remember that when 0 we use the constant function rule (x ) 0.
So far we have shown that we can use the power rule for all integers!
Later we'll show t
n
hat it can be used when n is a fraction,
and when n is a real number.
Calculus I Chapter Three 34
2
2
2
1( )
1
wh w
w
u u v uv
v v
Calculus I Chapter Three 35
Calculus I Chapter Three 36
0 0
0
0
0 0
Derivative of ( ) sin :
( ) ( ) sin( ) sin( )( ) lim lim
: sin( ) sin cosh cos sinh
sin cosh cos sinh sinlim
sin (cosh 1) cos sinhlim
sin (cosh 1) cos sinhlim lim
h h
h
h
h h
f x x
f x h f x x h xf x
h hNote x h x x
x x x
hx x
hx x
h
0 0
(cosh 1) inhsin lim cos lim
(sin )(0) (cos )(1)
cos
h h
hs
x xh h
x x
x
0
2 2
2 2
0 0
0
0
cosh 1 cosh 1: lim
cosh 1
Given: sin cos 1
cos h 1 sinlim lim
(cosh 1) (cosh 1)
sin sinlim 0
(cosh 1)
sinhlim 1 , So, (sin )
h
h h
h
h
Noteh
x x
h
h h
h h
h
and
dx
h dx
cos x
Calculus I Chapter Three 37
cos
sin .
Derivative of x is proved in a way similar to the way we proved
derivative of x Derivative of other trigonometric functions are found
by the Quotient Rule because we can write them as ratios of sin cos .x and x
Calculus I Chapter Three 38
2
2 22
2 2
(tan ) ?
sintan
cos
sin cos cos sin sin(tan )
cos cos
cos sin 1 = sec
cos cos
dx
dxx
xx
x x x x xx
x x
x xx
x x
Calculus I Chapter Three 39
2
2 2
2
(cot ) ?
coscot
sin1
Note: you could use cot to prove the result.tan
cos sin sin cos cos(cot )
sin sin
sin cos =
sin
dx
dxx
xx
xx
x x x x xx
x x
x x
x
2 22
2 2
(cos sin ) 1 csc
sin sin
x xx
x x
Calculus I Chapter Three 40
2
2
(csc ) ?
1csc
sin
1 (0) in (1)cos(csc )
sin sin
cos =
sin1 cos
csc cotsin sin
dx
dx
xx
s x xx
x x
x
xx
x xx x
Calculus I Chapter Three 41
2 2 2
(sec ) ?
1sec
cos
1(sec )
cos
(0)cos (1)( sin ) (0)cos (1)sin sin= =
cos cos cos1 sin
sec tancos cos
dx
dx
xx
xx
x x x x x
x x xx
x xx x
Calculus I Chapter Three 42
Calculus I Chapter Three 43
0
2 2
2 2
0 0
0
0 0
cosh 1 cosh 1: lim
cosh 1
Given: sin cos 1
cos h 1 sinlim lim
(cosh 1) (cosh 1)
sin sinlim 0
(cosh 1)
sinh coshlim 1 , So lim
h
h h
h
h h
Noteh
x x
h
h h
h h
h
and
h
10
h
Calculus I Chapter Three 44
2
( 4sin )( ) (4cos )(1)cos
x x xy x
x
Calculus I Chapter Three 45
Calculus I Chapter Three 46
Calculus I Chapter Three 47
Approximating using the derivative
ytany
x 1
1
tan
1
tan
tan
slope of the tangent line at
x x
x x
yx
x
y dy
x dx
dyy x
dx
tan
1
tan 1 1
1 1
1 1 1
1
1
t
by . This means thatwe can approximate )We ca
( ) by
( ) ( ). Note that is the tangent line
( ) ( )( - ).
We'll
n approxi
have:) ( ) ( )
mate (
(
y yy x f x
y y x x
f x
f x
y x y
y f x f x x x
x f x y f x y
1
an
tan1 1
tan1 11 1
1
1
) ( ) ( )
) ( ) ( ) ( )( - ).
(
(x x
dyx f x y f x xdx
f x
f x x f x y f x f x x x
Calculus I Chapter Three 48
3
3
2 2
tan1
3
1
(1.01) if ( ) .
0.01
( )
(1) 3 3(1) 3
(0.01) (3)(.01) .03
(1.01) (1) (1)(0.01) 1 0.03 1.03
Compare to (1.01) 1.030301 b
:
1
y calculato
r
x
Approximate f f x x
x
f x x
f x
Examp
d
le
yy
f
x
dx
f f
1
1
3 31 1
3 2 2 3 31 1 1 1
2 2 31 1
2 31
( ) ( )
3 3 ( ) ) ( )
3 3 ( ) ) ( )
3 ( ) ) (
:
)x x
x x
y x x x
y x x x x x x x
y x x x x x
dyy x x x x
dx
y dyk
x d
No e
x
t
Calculus I Chapter Three 49
Alternative proof of the product rule using linearization
Calculus I Chapter Three 50
1 1
0 :
Note: is the error whe
H
n we let
Note that is a variable and depends on .
ere we'll see that the err
Take the limit when 0 :
l
o
i
r goes to zero w
m
hen
x x x x
x
x
y dy y dyk k
x dx x dx
k x
x
1
1 1
1 1
1 1
0 0 0
0
0
0
0
lim lim
Note: limit of constant is constant: lim
lim
lim 0
lim 0
x xx x
xx x x x
xx x x x
xx x x x
x
y dyk
x dx
dy dy
dx dx
dy dyk
dx dx
dy dyk
dx dx
k
Calculus I Chapter Three 51
1
1
0
tan
tan
Given lim 0 and the equation below
we multiply both sides by :
So, we can approximate by
because 0 when 0
which makes (
x
x x
x x
k
x
y dyk
x dx
y dyx x k x
x dx
y y k x
y y
k x
k
) 0
But the reason approximation works even when is not
close to 0 is that ( )(the error) goes to zero faster than goes to 0.
See the figure above.
x
x
k x x
Calculus I Chapter Three 52
0
0
0
We know that
Given that lim 0
( ) ( ) ( ) Given that lim 0
( ) ( ) ( ) Given that lim 0
We can use this to approximate a function
xx a
x
x
dyy x k x k
dx
f a x f a f a x k x k
f a x f a x f a k x k
at
if we know ( ) and ( ). a x f a f a
Linear Approximation to ( ) at ( is a number close to )f x a x a
Calculus I Chapter Three 53
tan
3 2
Another result from this section is that
can be thought of as two quantities and where
(the y according to the tangent line)
So given ( ) , 3 can be written as 3
dydy dx
dxdy y
dx x
dyf x x x dy
dx
2 .
But, more importantly, we'll use this result
to discuss the Chain Rule next.
x dx
ytany
x
Calculus I Chapter Three 54
Calculus I Chapter Three 55
1
1 1
1
If is a function of and is a function of , then what is ?
t tx x
t t
t t
dxdydtdx
dy
dt
dyy x x t
dt
y x t
1 1 1
?
?
Is this always true?
( )( )
If ( ) and ( ), then
( ( ( )) ( ( )) ( )
t t x x t t
dy dy dx
dt dx dt
or
y f x x g t
f g t f g t g t
Calculus I Chapter Three 56
2
2
2 2
2
Example
( ) and ( ) 3 1
?
One way to find this is to find ( ) :
( ) (3 1) 9 6 1
18 6 18(2) 6 30
Can we do this without finding y(t)?
2 3(2) 1 5
2
t
t
y x x x t t
dy
dt
y t
y t t t t
dy dyt
dt dt
t x
dy
dx
5
2
2 5 2
2(5) 10
3 3
(10)(3) 30
The same answer! But does this formula always work?
x
t
t x t
dyx
dx
dx dx
dt dt
dy dy dx
dt dx dt
2( )y x x
( ) 3 1x t t
2( ) (3 1)y t t
Calculus I Chapter Three 57
1 1
?
If y is a function of x, and x is a function of t,
then y is a function of t.
But if ( ) exits and ( ) exists, then
does ( ) exist? If yes, how do we find it?
t t x x
y x x t
y t
dy dy dx
dt dx dt
1
is this true?t t
Differentiation of composite functions (the Chain Rule)
Calculus I Chapter Three 58
1
1
1
0
0
We start by the result we proved:
( lim 0)
We want , so we divide by :
( lim 0)
Take the limit of
x x
xx x
xx x
y dyk
x dx
dyy x k x kdx
dy tdt
y dy x xk kt dx t t
1
11 1
0 0 0 0 0
0
both sides when 0
lim lim lim lim lim
0 , ( lim 0)
t t t t tx x
xt tt t x x
t
y dy x xkt dx t t
dy dy dx dx kdt dx dt dt
dy dy dxdt dx dt
the desired result.
Calculus I Chapter Three 59
2 2
2 2 1 2
2 2 4 2
3 2
: ( 1) ?
If we use the power rule, we'll have:
2( 1) 2 2
But ( 1) 2 1
4 4 2 2
There is something wrong, and we are wrong
by a correction factor of 2 .
dyExample y x
dx
dyx x
dx
y x x x
dyx x x
dx
x
2
2
2
2 3
If we think of as a function of and as a
function of :
By the Chain Rule:1
(2 )(2 ) (2( 1))(2 )
(2 2))(2 ) 4 4
y u u
x
y u
u x
dy dy duu x x x
dx du dx
x x x x
The short cut is this ,
but we should not cancel the ' because
wh
:
en 0 , could equal zero.
y y x
t x t
x s
t x
Note
Calculus I Chapter Three 60
If , then ( ( ( )) ( ( )) ( )
If , then
f g x f g x f g x g x
dy dy duy u x
dx du dxExample
( ( ( )) ( ( )) ( )f g x f g x g x
Calculus I Chapter Three 61
Calculus I Chapter Three 62
( ( ( )) ( ( )) ( )f g z f g z g z
cos(4cos )( 4sin )
4sin cos(4cos )
z z
z z
Calculus I Chapter Three 63
Calculus I Chapter Three 64
0
0 0
0 0
0 0
Derivative of exponential function is:
( ) lim( )
( ) ( 1)lim( ) lim( )
( 1) (1 1)lim( ) lim( )
lim( ) lim(1)
x h xx
h
x h x x h
h h
hx x
h h
x x x
h h
e ee
h
e e e e e
h h
e he e
h hh
e e eh
1
0
1
1
Given that:
1lim(1 )
1
lim(1 )
(1 )
(1 ) )
(1 ) for small h
n
n
h
h
h
h hh
h
en
nh
h e
h e
h e
h e
Calculus I Chapter Three 66
(ln ) ?
ln
( ) ( )
( ) 1
1 1
y
y
y
y
x
y x
e x
e x
y e
ye x
Derivative of lnx:
Calculus I Chapter Three 68
( ) ?
ln ln
ln ln
1( ) ln
ln
ln
x
x
x
x
a
y a
y a
y x a
y ay
y y a
y a a
Derivative of : xa
Calculus I Chapter Three 70
log
(log ) ?
log
ln 1
1
ln1
ln
a
a
a
xy
y
y
y
x
y x
a a
a x
y a a
ya a
yx a
Derivative of : loga x
2
2
(log ) ?
1(ln ) (0)(ln )
ln
ln (ln )
1(ln )
ln
ln (ln )
ln 1
ln ln
1(log )
ln
a
a
x
a xx xa a
ax xa a
x
a x a
xx a
OR
Calculus I Chapter Three 72
Calculus I Chapter Three 73
Calculus I Chapter Three 74
0
0 0
1 1
1
1 11 0
00
0
00 0
0
If ( )
: ( ) ( ( ))
Thus: ( )
( ) ( )( ) ( ) lim
1 1lim = lim =
( ) ( )( ) ( ) ( )
y y
x x y y
y f x
Then f y f f x
f y x
f y f yf y
y y
x xf x f xf x f x f x
x x
0
10 0 0 0
0
0 0
Derivative of the inverse function
: x where is differentiable:
y ( ) and ( )
, since is continuous at :
, we have
Note At f
f x f y x
Also f x
as y y x x
Calculus I Chapter Three 75
3
3 3 3
1 21
3 32 233
Find the derivative of the inverse:
( ) 3
3 3 ( ) 3
This is the equation of the inverse without interchanging and .
1 1 1 1( ) ( 3) ( 3)
3 3 3 ( 3)3( 3)
Interchanging an
f x x
y x x y x y y
x y
x y y yyy
x
23
1 1
23
1 3
3
1d we get: ( ) ( )
3 ( 3)
1Letting ( ) ( ) ( ) ( ) .
3 ( 3)
But, using the formula for the derivative of the inverse:
1( ) ( ) Given: 3
( )
Note: ( 3) But, interchanging and
y y xx
y x f x f xx
f y y xf x
x y x
3
1 12 2 23 23
we get: y= ( 3)
1 1 1 1( ) ( ) ( ) ( )
3 3 3( 3) 3 ( 3)
y x
f y f xx y x x
Calculus I Chapter Three 76
1
3
( ) ( )-1
1
Find the derivative of the inverse at (4,1) on the graph of ( ) :
( ) 3
Using the formula for the derivative of inverse:
( 4 , 1 ) , (1,4) :
1( ) ( )
(
y coordOnf x x coordOnf x
f x
f x x
is on f so is on f
f yf x
2
12
1
) 3
1 1 1( ) (4)
(1) 3(1) 3
x
ff
1
:
The slope of the tangent line at
1(4,1) to f is .
3The slope of the tangent line at
(1,4) to f is 3.
Note
( )f x
1( )f x
1
1
1
1
1
2
(sin ( )) ?
( ) sin( )
we use the inverse
derivative formu
1( ) ( )
( )
la:
1
cos1
cos
1(sin (
(
)
) ( )
( )
)1
1
(
1
)
f yf x
f y
x
f x x
x
yf x
xx
x
Derivative of sine inverse function:2
1
1
Note: If this is easy to figure out.
But if sin ?
We set up the triangle below
to see the relation of :
Let sin an angle whose sine is x.
Note: this y is the inverse, ( )
y x x y
y x what is x
x and y
y x
f x
, not ( ).f x
y
x
21 x
cosadjacent
yhypotenuse
Calculus I Chapter Three 78
1
2
sin ( )
1( )
1 ( )
y something
y somethingsomething
1 3sin ( 3 )y x
Derivative of cosine inverse function:
1
1
2
1
1
1
1( ) ( )
( )
( ) ( )
(
(cos ( )) ?
( ) cos( )
1
sin1
sin
1(cos ( ))
11 1
) ( )
f yf x
f y
f
x
f x x
x
y
xx
x
x
y
x
21 xsin
oppositey
hypotenuse
1
In the triangle:
cos an angle whose cos is xy x
Calculus I Chapter Three 80
1
2
cos ( )
1( )
1 ( )
y something
y somethingsomething
1 5cos ( 3 )y x
y
Derivative of tangent inverse function:
x
21 x
1
1
1
1
1
2
2
2
(tan ( )) ?
( ) tan
1
(sec )
1
(sec )
1( ) ( )
( )
( ) ( )
1
1
( ) ( )
( ) ( )
( , )
x
f
f yf x
f y
x x
x
y
xx
f x
f x
2
2 2
1
2
tan an angle whose tan is x
1cos
1
sec 1
sec 1
yx
y x
y x
y x
2
2
1cos
1
sec 1
yx
y x
Calculus I Chapter Three 82
1
2
tan ( )
1( )
( ) 1
y something
y somethingsomething
1 3tan ( 4 )y x
Derivative of cotangent inverse function:
1
1
2
2
1
1
2
1 1
(cot ( )) ?
( ) cot
1
( csc )
1
( csc
( ) (
)
1
1( , )
)( )
( ) ( )
( ) ( )
( ) ( )
x
f x x
x
y
xx
f yf x
f y
f x
f x
2
2 2
1
2
cot an angle whose cot is x
1sin
1
csc 1 csc 1
yx
y x
y x
y x
y
1
x
2 1x
2
2
1sin
1
csc 1
yx
y x
Calculus I Chapter Three 84
1
2
cot ( )
1( )
( ) 1
y something
y somethingsomething
1 3cot ( 2 )y x
1
1
1
1
2
1
2
1
(sec ( )) ?
( ) sec
1
(sec tan )
1
(sec tan
1( ) ( )
( )
)
1
1
1(
( ) ( )
( ) ( )
( ) ( )
sec ( ))1
1
f yf x
f y
f x
f
x
f x x
x x
y y
x x
xx
x
x
x
Derivative of secant inverse function:1
2
sec an angle whose sec
1cos sec
tan 1
is x
y y xx
y x
y x
y
x
1
2 1x
2
1cos sec
tan 1
y y xx
y x
Calculus I Chapter Three 86
1
2
sec ( )
1( )
( ) 1
y something
y somethingsomething something
1 3sec ( 3 )y x
1
1
1
2
1
2
1
1
1( ) ( )
( )
( ) ( )
(
(csc ( )) ?
( ) csc
1
( csc cot )
1
( csc cot )
1
1
1(csc ( ))
1
1
) ( )
( ) ( )
x
f x x
x x
y
f yf x
y
x x
xx x
x
f y
f x
f x
Derivative of cosecant inverse function:1
2
csc an angle whose csc is x
1sin csc
cot 1
x
y y
y
y
xx
x
y
x
1
2 1x
2
1sin csc
cot 1
y y xx
y x
Calculus I Chapter Three 88
1
2
csc ( )
1( )
( ) 1
y something
y somethingsomething something
1
55
4
5
( ) ( )
1( ) ( )
5
y something something
y something something
1 25 csc ( )y x
Calculus I Chapter Three 89
Derivative of ( )
ln ln
1 1(1) ln ( )
ln 1
(ln 1)
(ln 1)
x
x
x
f x x
y x
y x x
y x xy x
yx
y
y y x
y x x
Calculus I Chapter Three 90
Calculus I Chapter Three 91
8 6 4 6
Implicit Differentiation
Given ( ) we can always find ?. The assumption we
make is that we can explicitly solve for .
What if we are given the following, and ask to find ?
x 3
dyy f x
dxy
dy
dx
x y y
Calculus I Chapter Three 92
8 6 4 6
2
2
x 3
Note: There is a difference between an equation
and an identity: 0 is an equation that is
true for ( , ) such that ( , );for example (3,9) works.
But an identity is true for all val
x y y
x y
x y x x
2 2 2
2
ues of the
variables and it is represented by instead of =.
( ) 2 is true for all ( , ).
Or, 1 ( 1)( 1) is true for all values of x.
When we take the derivative of both sides of an id
x y x xy y x y
x x x
2
enty
we get an equality.
( 1) [( 1)( 1)] 2 2x x x x x
Calculus I Chapter Three 93
Assume ( ) such that when we substitute ( ) into:
1 for we get an identity.
1Let , then 1
for all values of except 0.
Since we are assuming that y is a function of x we can use
the pro
y f x f x
xy y
y xyx
x x
2
2
duct rule to take the derivative of both sides of
( ) (1)
( ( )) ( ) 0
11
0
1 1But if , then we can compute directly:
d dxy
dx dxd d
x y y xdx dx
dy dy dy y xy x x ydx dx dx x x x
dyy
x dx x
Calculus I Chapter Three 94
2 2
2 2
: Assume that ( ) such that 25
2( ) (25) 2 2 0
2
If we now differentiate directly we get the same answer:
Example y y x x y
dy dy x xx y x y
dx dx y y
dy x
dx y
2 2
2
2 21 2
121 2
12 12
122 2
1 12 2 22 2
Afetr solving for in 25 :
25 which means we have two functions:
y 25 , y 25
1 2(25 ) ( 2 )
22(25 )
1 2(25 ) ( 2 )
22(25 ) (25 )
y x y
y x
x x
dy x xx x
dx yx
dy x x xx x
dx yx x
Calculus I Chapter Three 95
8 6 4 6
8 6 4 6
8 6
Find the equation of the tangent line to
3 at the point (1,1).
We assume that ( ) is a function of x such that
if substituted for y it would make the equation an identity:
3
(
x x y y
y f x
x x y y
x x
4 6
7 5 4 6 3 5
7 5 4
6 3 5
(1,1)
) (3)
8 6 4 6 0
(8 6 )
4 6
14 7
10 5
71 ( 1) 7 5 12
5
y y
dy dyx x y x y y
dx dx
dy x x y
dx x y y
dy
dx
y x x y
Calculus I Chapter Three 96
6(0, 3)
6(0, 3)
8( 3,0)8( 3,0)
(1,1)( 1,1)
( 1, 1) (1, 1)
8 6 4 6
Note that given:
3
If we were told to find the tangent line to the graph at 1, there
would be difficult to figure out what the value for 1 is!
We would have to solve a 6th degree polynomi
x x y y
x
y x
8 6 4 6
al!
1 1 3y y
Calculus I Chapter Three 97
Note: The previous problem was "cooked" so that we could
graph it, and we were given points with x and y coordinates
known. In general, it is very difficult to graph these
equations. For example, we can tell from the graph that
there are two tangent lines at 1, and two tangent lines
at 1.
x
y
(1,1)( 1,1)
( 1, 1) (1, 1)
This is why we need to know where
the equation has vertical tangent lines
because around these points the graph
becomes a multi-valued curve. We can
also assume that both and are
functions of another
x y
variable, namely, t.
Calculus I Chapter Three 98
Use implicit differentiation to find in terms of and .dy
x ydx
3 23 2x y 2
2
9 2 0
9
2
dyx y
dx
dy x
dx y
Calculus I Chapter Three 99
Calculus I Chapter Three 100
2 2
3, 4
2 2
3, 4
Relate Rates
A particle moves along the circular path 25.
( and in feet). At (3,4), 8 sec
Find ?
( ) (25)
2 2 0
2
2
3(8) 6 s4
x y
x y
x y
dx ftx ydt
dy
dt
d dx y
dt dtdx dy
x ydt dt
dxxdy x dxdt
dt y y dt
dy ftdt
ec
dx
dt
dy
dt
Calculus I Chapter Three 101
A hypothetical square grows so that the length of its sides
are increasing at a rate of 9 m/min. How fast is the area
of the square increasing when the sides are 14 m each?
2
2
14
2 2(14)(9) 252mins
A s
dA ds dA ms
dt dt dt
Calculus I Chapter Three 102
A 17 ft ladder is leaning against a wall and sliding toward
the floor. The top of the ladder is sliding down the wall
at a rate of 6 ft/sec. How fast is the base of the ladder
sliding away from the wall when the base of the ladder is
15 ft from the wall?
dy
dtdx
dt
2 2 2 2 2 2
2 2 2
: 17 15 17 15 64, 8
( ) 17 2 2 0
Note x y x y y
d d dx dyx y x y
dt dt dt dt
Calculus I Chapter Three 103
dr
dtdV
dt
3 2
2 210
44
3
15 30.012
4 4 10 80 min minr
dV drV r r
dt dtdV
dr dr in indtdt r dt
Calculus I Chapter Three 104
250 Given: 30sec
50 2
10 50(30) 2(10)(30) 900sec
40 50(30) 2(40)(30) 900sec
dx fty x x
dtdy dx dx
xdt dt dt
dy ftx
dtdy ft
xdt
Calculus I Chapter Three 105
Higher-order Derivatives
1 1 1( 1) ( )2 2 2
1 3( 1)2 2
52
52
5
Find the third derivative of ( ) .
1 1( ) ( )2 2
1 1 1( ) ( )( )2 2 4
1 3 3 1( ) ( )( ) ( ) ( )4 2 8
3 1( ) ( )8
f x x
f x x f x x x
f x x x
f x x f xx
f xx
Geogebra’s answer
Calculus I Chapter Three 106
(1)
2(2)
2
3(3)
3
4(4)
4
( )
Notation for higher derivatives:
( ), , y , ( )
( ), , y , ( )
( ), , y , ( )
.........., ,....., ( )
.........., ,....., ( )n
nn
dyf x f x
dx
d yf x f x
dx
d yf x f x
dx
d yf x
dx
d yf x
dx
(0): ( ) ( )Note f x f x
Calculus I Chapter Three 107
Tangent lines and Normal linesThe normal line to a curve C at a point A is the line through A that is perpendicular to the tangent line at A.
The concept of the normal line to a curve has applications in the study of optics where one needs to consider the angle between a light ray and the normal line to a lens.
Calculus I Chapter Three 108
2
Find equations for the tangent and normal lines to the graph
of ( ) at (2,4).
( )
f x
f x x
( ) 2 (2) 4
: (2) (2)( 2) 4 4( 2)
1 1: (2) ( 2) 4 ( 2)
(2) 4
f x x f
Tangent y f f x y x
Normal y f x y xf