Week 1 : Revision
4.5 . Substitution Rule
/ fcgcx ) ) gllxldx =
/flu) du
if 4=41×1 is a differentiable function
set u to
your inner function
•
•
when e is involved set u = the power of e
,
•
differentiate U
◦
manipulate so that function cancels out
" "
•
remember + C
6. 1 . Inverse functions
only one lone output for each input )
•
one -
to -
•
horizontal line -
test
f- ( x ) reflection
'
f- ( y ) = x
y
= in ↳= x
if is with domain A and B then f-
'
has domain B and
f l l
range range A
• -
,
cancellation
equations
"
f- ( fix ) ) = x V ✗ C- A ( Domain of inner function )
f- ( f-
'
( x)) = X U KE B
Find the inverse :
① Let
y
= f- ( x )
② Find domain and range of fcx )
③ Solve for have find the
equation x i. t.co .
y ( sometimes
you to
square )
④ Swap x and
y to find
'
f- ( x )
, Derivative of an inverse function
* If f is a 1-1 continuous function, then f-
'
is also continuous
slope of inverse
f at a =
IF
f is an odd function :
9
'
(
f 1)
-
I
(a) = = I
f- (b)
'
f ' ( fila ) ) fcx ) - DX = 0
I
f-
'
f- (b) = a (a) = b
6. G. Inverse trig functions
•
trig functions are not 1- I
•
we must restrict their domains to make them 1- I
Arcsine
sin
_ '
✗ =
y siny=x and
-
≤
y≤ ¥
sin
_ '
Csinx ) = ≥ for
-
É ≤ ✗ ≤ ¥
'
for ≤ 1
sin ( sin
_
-1 ≤ ✗
x ) = x
Input domain :
-
I ≤ ✗ ≤ i
sink =
y
±z
I
'
off
I
( sin
-
x ) =
,
1- ✗ 2 -
-
I
Arccos
'
IT for
-
◦ ≤ x ≤ cos -1 ≤
cosx =
y ,
y = x
y ≤ 1
'
( COSI ) for IT
-
COS = x 0 ≤ ✗ ≤
COS ( COS
'
) for
_
x = x
-
I ≤ × ≤ I
( cos
- '
x ) = -
1 -
I < x e I
;
I 2
-
✗
,Arctan
tan
- '
✗ =
y any
+ =x and
-
¥ <
y < E
( tan
_
'
X) =
I
, 1-1×2
method
triangle
I
' '
Prove sin cos =
_ -
e. ✗ x
g. + 2
¥ Iz
-
' ≤ a ≤
Let
_
a =
sin x
;
✗ Sina
b
=
,
×
b=
'
Let b
-
cos X O ≤ ≤ it
✗ =
costs
a
1- ✗ 2
at b + ¥ =
IT
i. at b =
¥
"
b = b
"
/ nb
Integration
≥
a
:-/
a
"
DX =
Inca )
( É)
/
I
xz+az
=
ta - arctan
/{ du
= 81h ( IU ) )
/¥ dx = In (1×1)
, week 2
6.7 .
Hyperbolic functions
I
" "
sinhx = e -
e- cosechx =
sinhx
2
I
' "
coshx = e' + e- sechx =
cosh >c
2
COSHX
tanhx = sinhx [ ◦ thx =
sinhx
coshx
Hyperbolic identities
cosh >
sin C -
x > = -
sinhx cos he -
x) = coshx sinhcxty ) = sinh >
ccoshy
+
csinhy
coshzx sinhzx cosh >
cushy
= I 1- tanhzx = sech 2x coshlxty ) = +
sinhxsinhy
-
sinhx
"
coshx + =
e Sinha>c) = Zsinhsccosh >c
of
Properties infinity
as ± K = as
Derivatives of Hyperbolic functions
+ A = A
ddxlcosechx)
d
d-✗ ( sinh >c) =
coshx =
-
cosechxcothx
d d
d-
× ( cosh >c) = sin hide fsechx ) =
-
sechxtanhx A. ( IK ) = ± as if k≠O
d
dI ( tanh >c) SECHZX A A
( coth >c) cosechzx
=
= -
= -
= 0 if k≠o
6.8 .
Indeterminate forms and L' Hopital 's Rule
¥ = ± as if k≠O
if k≠o
8- = indeterminate form
ago =
indeterminate form
= as
% =
indeterminate form I = indeterminate form } y=1n . . .
¥ = 0
9- = A
as
form
=
indeterminate
}
as as :
write
quotient
-
as a
A. 0 = indeterminate form
4.5 . Substitution Rule
/ fcgcx ) ) gllxldx =
/flu) du
if 4=41×1 is a differentiable function
set u to
your inner function
•
•
when e is involved set u = the power of e
,
•
differentiate U
◦
manipulate so that function cancels out
" "
•
remember + C
6. 1 . Inverse functions
only one lone output for each input )
•
one -
to -
•
horizontal line -
test
f- ( x ) reflection
'
f- ( y ) = x
y
= in ↳= x
if is with domain A and B then f-
'
has domain B and
f l l
range range A
• -
,
cancellation
equations
"
f- ( fix ) ) = x V ✗ C- A ( Domain of inner function )
f- ( f-
'
( x)) = X U KE B
Find the inverse :
① Let
y
= f- ( x )
② Find domain and range of fcx )
③ Solve for have find the
equation x i. t.co .
y ( sometimes
you to
square )
④ Swap x and
y to find
'
f- ( x )
, Derivative of an inverse function
* If f is a 1-1 continuous function, then f-
'
is also continuous
slope of inverse
f at a =
IF
f is an odd function :
9
'
(
f 1)
-
I
(a) = = I
f- (b)
'
f ' ( fila ) ) fcx ) - DX = 0
I
f-
'
f- (b) = a (a) = b
6. G. Inverse trig functions
•
trig functions are not 1- I
•
we must restrict their domains to make them 1- I
Arcsine
sin
_ '
✗ =
y siny=x and
-
≤
y≤ ¥
sin
_ '
Csinx ) = ≥ for
-
É ≤ ✗ ≤ ¥
'
for ≤ 1
sin ( sin
_
-1 ≤ ✗
x ) = x
Input domain :
-
I ≤ ✗ ≤ i
sink =
y
±z
I
'
off
I
( sin
-
x ) =
,
1- ✗ 2 -
-
I
Arccos
'
IT for
-
◦ ≤ x ≤ cos -1 ≤
cosx =
y ,
y = x
y ≤ 1
'
( COSI ) for IT
-
COS = x 0 ≤ ✗ ≤
COS ( COS
'
) for
_
x = x
-
I ≤ × ≤ I
( cos
- '
x ) = -
1 -
I < x e I
;
I 2
-
✗
,Arctan
tan
- '
✗ =
y any
+ =x and
-
¥ <
y < E
( tan
_
'
X) =
I
, 1-1×2
method
triangle
I
' '
Prove sin cos =
_ -
e. ✗ x
g. + 2
¥ Iz
-
' ≤ a ≤
Let
_
a =
sin x
;
✗ Sina
b
=
,
×
b=
'
Let b
-
cos X O ≤ ≤ it
✗ =
costs
a
1- ✗ 2
at b + ¥ =
IT
i. at b =
¥
"
b = b
"
/ nb
Integration
≥
a
:-/
a
"
DX =
Inca )
( É)
/
I
xz+az
=
ta - arctan
/{ du
= 81h ( IU ) )
/¥ dx = In (1×1)
, week 2
6.7 .
Hyperbolic functions
I
" "
sinhx = e -
e- cosechx =
sinhx
2
I
' "
coshx = e' + e- sechx =
cosh >c
2
COSHX
tanhx = sinhx [ ◦ thx =
sinhx
coshx
Hyperbolic identities
cosh >
sin C -
x > = -
sinhx cos he -
x) = coshx sinhcxty ) = sinh >
ccoshy
+
csinhy
coshzx sinhzx cosh >
cushy
= I 1- tanhzx = sech 2x coshlxty ) = +
sinhxsinhy
-
sinhx
"
coshx + =
e Sinha>c) = Zsinhsccosh >c
of
Properties infinity
as ± K = as
Derivatives of Hyperbolic functions
+ A = A
ddxlcosechx)
d
d-✗ ( sinh >c) =
coshx =
-
cosechxcothx
d d
d-
× ( cosh >c) = sin hide fsechx ) =
-
sechxtanhx A. ( IK ) = ± as if k≠O
d
dI ( tanh >c) SECHZX A A
( coth >c) cosechzx
=
= -
= -
= 0 if k≠o
6.8 .
Indeterminate forms and L' Hopital 's Rule
¥ = ± as if k≠O
if k≠o
8- = indeterminate form
ago =
indeterminate form
= as
% =
indeterminate form I = indeterminate form } y=1n . . .
¥ = 0
9- = A
as
form
=
indeterminate
}
as as :
write
quotient
-
as a
A. 0 = indeterminate form
