Calculus Cheat Sheet Derivatives ...

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Calculus Cheat Sheet
Calculus Cheat Sheet
Derivatives
Definition and Notation
Chain Rule Variants
The chain rule applied to some specific functions.
1.
d
d
fxhfx
!"
n
n
"
1
(
(
fx
$
nfx
fx
5.
cos
fx
$"
fx
sin
fx
8
If
yfx
$
then the derivative is defined to be
fx
$
lim
h
.
dx
dx
h
*
0
d
d
fx
fx
2.
$
fx
(
6.
tan
fx
$
fx
(
sec
2
fx
dx
dx
If
yf
$ then all of the following are
equivalent notations for the derivative.
If
yf
$ all of the following are equivalent
notations for derivative evaluated at
xa
(
d
dx
fx
d
&
'
&
'
&
'
.
7.
sec
$
(
sec
tan
$
3.
ln
fx
$
fx
()
fx
()
fx
()
fx
()
dx
fx
df
dyd
df
dy
$$$$
fxy
fx
$
Dfx
fay
$
$
$
$
Dfa
fx
(
d
dxdxdx
d
xa
$
dx
dx
8.
tan
"
1
fx
$
4.
sin
fx
$
fx
(
cos
fx
8
xa
$
xa
$
2
dx
1
!2
fx
dx
Interpretation of the Derivative
f
(
is the instantaneous rate of
change of
Higher Order Derivatives
If
yfx
$
then,
2.
The n
th
Derivative is denoted as
The Second Derivative is denoted as
1.
mfa
$
(
is the slope of the tangent
fx
at
xa
.
$
df
2
df
n
2
n
((
xf
$
x
dx
$
and is defined as
x
dx
$
and is defined as
line to
$ at
x
$ and the
equation of the tangent line at
xa
yfx
3.
If
fx
is the position of an object at
time
x
then
2
n
is
(
$
fa
is the velocity of
fx
(
(
((
$
,
i.e.
the derivative of the
n
n
"
1
$
,
i.e.
the derivative of
given by
yfafaxa
$
!
(
"
.
the object at
xa
.
$
(
first derivative,
fx
.
the (
n
-1)
st
derivative,
n
"
1
.
Basic Properties and Formulas
Implicit Differentiation
If
fx
and
gx
are differentiable functions (the derivative exists),
c
and
n
are any real numbers,
29
xy
"
32
Find
y
( if
yy
$ here, so products/quotients of
x
and
y
will use the product/quotient rule and derivatives of
y
will use the chain rule. The “trick” is to
differentiate as normal and every time you differentiate a
y
you tack on a
y
( (from the chain rule).
After differentiating solve for
y
(
.
!
xy
$
sin
!
11
. Remember
d
(
1.
0
cf
$
cfx
(
5.
c
$
dx
(
2.
+$
gfxgx
+
d
6.
n
$
nx
n
"
1
– Power Rule
dx
(
3.
fgfgfg
$!
– Product Rule
d
fgx
7.
$
fgxgx
xy
29
"
22
3
29
"!
3
xy
!
2
xyy
$
cos
yy
!
11
dx
/4
fgfg
"
xy
112
"
29
"
"
3
xy
22
4.
$
– Quotient Rule
This is the
Chain Rule
0
16
xy
xy
29
"
29
"
22
3
2
"
9
!
3
xy
!
2
xyy
$
cos
yy
!
11
.$
g
g
2
xy
2
xy
3
"
9
29
"
"
cos
xy
xy
2
xy
3
"
9
29
"
"
cos
yy
$"
112
29
"
"
3
xy
22
Common Derivatives
d
x
dx
d
d
1
Increasing/Decreasing – Concave Up/Concave Down
$
csc
$"
csccot
xx
aaa
$
ln
dx
dx
Critical Points
xc
d
d
d
dx
Concave Up/Concave Down
1.
If
is a critical point of
fx
provided either
sin
$
cos
cot
$"
csc
2
$
$
dx
dx
((
x
%
0
for all
x
in an interval
I
then
f
(
$ or
2.
(
1.
0
fc
doesn’t exist.
d
d
1
d
x
1
"
1
cos
$"
sin
sin
x
$
ln
$
%
0
fx
is concave up on the interval
I
.
2.
If
dx
dx
dx
1
"
x
2
Increasing/Decreasing
1.
If
((
x
#
0
for all
x
in an interval
I
then
d
d
1
d
1
2
tan
$
sec
ln
$
-
0
cos
"
1
x
$"
fx
% for all
x
in an interval
I
then
0
dx
dx
fx
is concave down on the interval
I
.
dx
2
1
"
x
d
d
1
fx
is increasing on the interval
I
.
2.
If
d
1
sec
$
sectan
xx
log
$
%
0
"
1
tan
x
$
a
dx
dx
xa
ln
Inflection Points
xc
fx
# for all
x
in an interval
I
then
0
dx
1
!
2
is a inflection point of
fx
if the
$
fx
is decreasing on the interval
I
.
3.
If
concavity changes at
xc
.
$
fx
$ for all
x
in an interval
I
then
0
fx
is constant on the interval
I
.
©
2005 Paul Dawkins
©
2005 Paul Dawkins
Visit
for a complete set of Calculus notes.
Visit
for a complete set of Calculus notes.
 Calculus Cheat Sheet
Calculus Cheat Sheet
Extrema
Related Rates
Sketch picture and identify known/unknown quantities. Write down equation relating quantities
and differentiate with respect to
t
using implicit differentiation (
i.e.
add on a derivative every time
you differentiate a function of
t
). Plug in know
n qua
ntities and solve for the unknown quantity.
Ex.
A 15 foot ladder is resting against a wall.
The bottom is initially 10 ft away and is being
pushed towards the wall at
4
ft/sec. How fast
is the top moving after 12 sec?
Absolute Extrema
1.
xc
Relative (local) Extrema
1.
xc
is a relative (or local) maximum of
is an absolute maximum of
fx
$
$
fx
if
fcfx
,
for all
x
near
c
.
if
fcfx
,
for all
x
in the domain.
Ex.
Two people are 50 ft apart when one
starts walking north. The angle
changes at
0.01 rad/min. At what rate is the distance
between them changing when
2.
xc
is a relative (or local) minimum of
$
2.
xc
is an absolute minimum of
fx
$
fx
if
fcfx
)
for all
x
near
c
.
if
fcfx
)
for all
x
in the domain.
$
0.5
rad?
1
st
Derivative Test
If
xc
Fermat’s Theorem
If
is a critical point of
fx
then
xc
is
$
$
fx
has a relative (or local) extrema at
xc
1.
a rel. max. of
fx
if
fx
%
0
to the left
, then
xc
is a critical point of
fx
.
$
$
of
xc
and
fx
# to the right of
xc
0
.
$
$
x
(
is negative because
x
is decreasing. Using
Pythagorean Theorem and differentiating,
2
We have
( $ rad/min. and want to find
x
( . We can use various trig fcns but easiest is,
sec
0.01
2.
a rel. min. of
fx
if
fx
# to the left
0
Extreme Value Theorem
If
fx
is continuous on the closed interval
of
xc
and
fx
%to the right of
xc
0
.
(
$
$
(
$
.
sectan
$
xy
!$
2
15
2
.!
2
xxyy
2
$
0
&
'
50
50
ab
then there exist numbers
c
and
d
so that,
1.
(
3.
not a relative extrema of
fx
if
fx
is
After 12 sec we have
$"
1012
1
4
$
7
and
so plug in
(
and solve.
We know
$
0.05
the same sign on both sides of
xc
.
acdb
)),
2.
,
fc
is the abs. max. in
$
x
(
2
2
so
y
$
157
"$
176
. Plug in and solve
sec0.5tan0.50.01
$
&
'
fd
is the abs. min. in
&
'
ab
,
3.
ab
.
50
0.3112 ft/sec
2
nd
Derivative Test
If
xc
for
y
( .
x
( $
is a critical point of
fx
such that
7
$
7
"!
176
$.$
0
ft/sec
Finding Absolute Extrema
To find the absolute extrema of the continuous
function
1
4
Remember to have calculator in radians!
f
(
$ then
x
$
1.
is a relative maximum of
4176
0
fx
on the interval
&
'
Optimization
Sketch picture if needed, write down equation to be optimized and constraint. Solve constraint for
one of the two variables and plug into first equation. Find critical points of equation in range of
variables and verify that they are min/max as ne
eded.
Ex.
We’re enclosing a rectangular field with
500 ft of fence material and one side of the
field is a building. Determine dimensions that
will maximize the enclosed area.
ab
use the
fx
if
fc
((
#
0
.
following process.
1.
Find all critical points of
2.
is a relative minimum of
fx
if
fc
((
%
0
.
fx
in
&
'
ab
.
3.
may be a relative maximum, relative
minimum, or neither if
2.
Evaluate
fx
at all points found in Step 1.
((
fc
$
0
.
Ex.
Determine point(s) on
y
$! that are
2
1
closest to (0,2).
3.
Evaluate
fb
.
4.
Identify the abs. max. (largest function
value) and the abs. min.(smallest function
value) from the evaluations in Steps 2 & 3.
fa
and
Finding Relative Extrema and/or
Classify Critical Points
1.
Find all critical points of
fx
.
2.
Use the 1
st
derivative test or the 2
nd
derivative test on each critical point.
2
2
2
Minimize
$$"!" and the
dx
0
2
Maximize
Axy
$
subject to constraint of
constraint is
y
$!. Solve constraint for
x
and plug into the function.
2
1
x
!$ . Solve constraint for
x
and plug
into area.
2500
Mean Value Theorem
fx
is continuous on the closed interval
&
ab
and differentiable on the open interval
'
If
ab
2
$".$!"
$"!"$"!
Differentiate and find critical point(s).
3
2
xy
2
1
2
2
Ay
$
5002
"
fbfa
"
$".
5002
2
then there is a number
acb
## such that
fc
(
$
.
1
2
2
33
2
$
5002
yy
"
ba
"
Differentiate and find critical point(s).
5004
(
$" .$
By the 2
nd
derivative test this is a rel. min. and
so all we need to do is find
x
value(s).
2
23
( $".$
By 2
nd
deriv. test this is a rel. max. and so is
the answer we’re after. Finally, find
x
.
A
125
Newton’s Method
fx
x
is the
n
th
guess for the root/solution of
fx
$ then (
n
+1)
st
guess is
n
If
0
$"
(
n
!
1
n
fx
n
$"$.$+
1
x
$" $
The dimensions are then 250 x 125.
5002125250
3
1
1
2
2
(
2
provided
fx
exists.
n
The 2 points are then
and
1
2
,
3
"
1
2
,
3
2
2
©
2005 Paul Dawkins
©
2005 Paul Dawkins
Visit
for a complete set of Calculus notes.
Visit
for a complete set of Calculus notes.
  [ Pobierz całość w formacie PDF ]