false

Class Notes
(839,092)

Canada
(511,185)

University of Ottawa
(33,539)

Mathematics
(1,115)

MAT1320
(117)

Don't Know
(1)

Lecture

Description

Calculus Cheat Sheet
Limits
Definitions
Precise Definition : We say lixﬁa x( )L if Limit at Infinity : We say xﬁ¥ f ( ) L if we
for every e > 0 there is ad > 0such that can make f x( ) close to L as we want by
whenever 0 < x-a a.
There is a similar definition for lim f x = -¥
xﬁa ( )
Left hand limit : lim - x( )L. This has the
xﬁa except we make f ( )rbitrarily large and
same definition as the limit except it requires negative.
x < a.
Relationship between the limit and one-sided limits
lim f ( ) L Þ lim f x + l( )f x = - ( ) lim+f ( ) lim f-x =( )Þ lim f x = L ( )
xﬁ a xﬁa xﬁa xﬁa xﬁa xﬁa
lim f ( ) lim f x ( )im f x Does( )t Exist
xﬁa+ xﬁa- xﬁa
Properties
Assume lim f( ) and limg ( ) both exist and c is any number then,
xﬁa xﬁa
1. limØcf ( ) = clim f x( ) Ø f x ø lim f ( )
xﬁa º ß xﬁa 4. lim Œ ( ) œ= ﬁ a provided lim g( ) „ 0
ﬁ a g( ) limg x( ) xﬁa
º ß ﬁ a
2. limØ º x( )g x ( )ßlim f x ( )mg x ( ) n n
xﬁa xﬁa xﬁa 5. lixﬁaº x( )ß limºxﬁa ( )ß
3. limØ f x g x ø = lim f x limg x 6. lim Øn f ( )ø = lim f x( )
xﬁa º ( ) ( ) ß xﬁa ( ) xﬁa ( ) xﬁaº ß xﬁa
Basic Limit Evaluations at – ¥
Note : sgn a =1 if a > 0 and sgn a = -1 if a < 0 .
( ) ( )
1. lim e = ¥ & lim e = 0 5. n even : lim x = ¥
xﬁ¥ xﬁ- ¥ xﬁ–¥
n n
2. lxﬁ¥ln ( ) ¥ & xﬁ0m+ln( )= -¥ 6. n odd : lxﬁ¥x = ¥ & limxﬁ- ¥-¥
n
3. If r > 0then lim b = 0 7. n even : lim a x +L+bx+c = sgn a ¥ ( )
xﬁ¥ x r xﬁ–¥
r 8. n odd : lim ax +L+bx+c = sgn a ¥ ( )
4. If r > 0 and x is real for negative x xﬁ¥
b 9. n odd : lim ax +L+cx+d = -sgn a ¥ ( )
thenxﬁ-¥ r = 0 xﬁ-¥
x
Visitttp://tutorial.math.lamar.edu for a complete set of Calculus notes. © 2005 Paul Dawkins Calculus Cheat Sheet
Evaluation Techniques
Continuous Functions L’Hospital’s Rule
If f( )is continuous at a thxﬁlam ( )= f ( ) If lim f( ) = 0 or lim f ( )= –¥ then,
xﬁa g( ) 0 xﬁa g ( ) –¥
Continuous Functions and Composition f( ) f¢( )
f( )is continuous at b and lim ( ) = b then xiag x = xﬁa g x a is a number, ¥ or -¥
xﬁa ( ) ( )
lim f g x = f limg x = f b Polynomials at Infinity
xﬁa ( ( )) (xﬁa ( )) ( ) p( )and q x( )e polynomials. To compute
Factor and Cancel
2 p ( )
lim x +4x-12 = lim (x-2 )(6 ) xﬁ–¥ factor largest power of x i( ) x out
xﬁ2 x -2x xﬁ2 x (-2 ) q ( )
of both p ( )nd q x( )en compute limit.
x+6 8
= xﬁ2 x = 2 = 4 2 x2 3- 4 3 - 4
lim 3x -4 = lim ( x2 )= lim x = - 3
Rationalize Numerator/Denominator x-¥ 5 -2 x2 xﬁ-¥ x2( 5-2 ) xﬁ-¥ 5-2 2
3- x 3- x 3+ x x x
lxﬁ9x -81 = xﬁ9x -81 Piecewise Function
3+ x ìx +5 if x < -2
9- x -1 lim g( )where g x( ) í
= xﬁ9 2 = xﬁ9 xﬁ-2 î1-3x if x ‡ -2
(x -81 )( x ) (x+9 )( x ) Compute two one sided limits,
-1 1 2
= = - xﬁ-2g ( ) lxﬁ-2-+5 = 9
( )( ) 108
Combine Rational Expressions xﬁ-2g ( ) lxﬁ-2+3x = 7
1æ 1 1ö 1 æx- (+h )ö One sided limits are different xﬁ-2i( )
hﬁ0 h x + - x÷ =lhﬁ0h ç x x + ÷
Ł ł Ł ( ) ł doesn’t exist. If the two one sided limits had
1 æ -h ö -1 1 been equal then lim g( )would have existed
= lim ç ÷ = lim = - 2 xﬁ-2
hﬁ0h Ł x(h )ł hﬁ0x (+h ) x and had the same value.
Some Continuous Functions
Partial list of continuous functions and the values of x for which they are continuous.
1. Polynomials for all x. 7. cos ( )nd sin x ( ) all x.
2. Rational function, except for x’s that give
division by zero. 8. tan ( )nd sec x ( )vided
3. nx (n odd) for all x. 3p p p 3p
x „L,- ,- , , ,L
4. nx (n even) for alx ‡ 0. 2 2 2 2
5. e for all x. 9. cot ( )nd csc x ( )vided
x „L,-2p,-p,0,p,2p,L
6. ln xfor x > 0.
Intermediate Value Theorem
Suppose that f ( )s continuous on [a, b] and let M be any number between f( )and f b( )
Then there exists a number c such tha < c < band f ( ) M .
Visittp://tutorial.math.lamar.edu for a complete set of Calculus notes. © 2005 Paul Dawkins Calculus Cheat Sheet
Derivatives
Definition and Notation
f(x +h ) f x( )
If y = f( )then the derivative is defined to be ( )= lim .
hﬁ0 h
If y = f( )then all of the following are If y = f( )all of the following are equivalent
equivalent notations for the derivative. notations for derivative evaluated at x = a.
f¢( )= y = df = dy = d (f ( ))= Df x( ) f ( ) = y¢ = df = dy = Df ( )
dx dx dx x a dx x a dx x a
Interpretation of the Derivative
If y = f( )then, 2. f ( )s the instantaneous rate of
1. m = f a is the slope of the tangent change of f x at x = a.
( ) ( )
line toy = f ( )t x = aand the 3. If f ( )s the position of an object at
equation of the tangent line at= a is
time x then f ( )s the velocity of
given by y = f ( ) f a ( )(. ) the object at x = a.
Basic Properties and Formulas
If f( )and g x ( ) differentiable functions (the derivative exists), c and n are any real numbers,
¢ ¢ d
1. (c f) = c f ( ) 5. dx ( )= 0
¢ ¢ ¢ d
2. (f – g)= f ( ) – g ( ) 6. ( ) = nxn-1– Power Rule
3. f g = f g + f g – Product Rule dx
( ) d ¢ ¢
¢ 7. dx (f ( x( )))= f (g ( ))( )
4. ç f÷ = f g - f g¢ – Quotient Rule This is the Chain Rule
Łg ł g 2
Common Derivatives
d d d
( )=1 (cscx ) -cscxcot x ( ) = a ln ( )
dx dx dx
d d 2 d x x
dx(sin x)= cosx dx(cot x)= -csc x dx ( ) = e
d d 1 d 1
(cosx ) -sin x (sin x =) (ln( ))= , x > 0
dx dx 1- x 2 dx x
d tan x = sec x d -1 1 d ln x = 1 , x „ 0
dx( ) dx(cos x =)- 2 dx ( ) x
d 1- x d 1
(secx ) secxtan x d tan x = 1 (loga( ))= , x > 0
dx dx( ) 1+ x2 dx xlna
Visittp://tutorial.math.lamar.edu for a complete set of Calculus notes. © 2005 Paul Dawkins Calculus Cheat Sheet
Chain Rule Variants
The chain rule applied to some specific functions.
d n n-1 d
1. (º f ( )ß ) = nº f( )øß f ( ) 5. (cosØºf ( ) ß)- f ( ) sinº f ( )ß
dx dx
d f( ) f( ) d 2
2. dx ( ) = f¢( )e 6. dx (tanØºf ( ) ß)= f¢( )sec Øºf ( ) ß
3. d lnØ f ( ) = f¢( ) 7. d (sec [f (x])= f (xsec [f (x] [n f (x]
dx ( º ß ) f( ) dx
d f¢( )
4. d (sinØ f ( ) =)f ( ) cosØ f ( ) 8. (tan Ø º x( )ß ) 2
dx º ß º ß dx 1+ º f( )øß
Higher Order Derivativesth
The Second Derivative is denoted as The n Derivative is denoted as
( ) d f ( ) d f
f¢( )= f ( ) = 2 and is defined as f ( )= n and is defined as
dx dx
f¢( ) = ( f¢( ))¢, i.e. the derivative of the ( ) (n-) ¢
f ( )= f( ( ) ) , i.e. the derivative of
first derivative, ¢( ). the (n-1) derivative, f (-1) x .
( )
Implicit Differentiation
Find y if e 2x-9y + x y = sin ( )11x . Remember y = y x her( )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 .
2x-9y 2 2 3
e (2-9 y ) +3 x y +2 x y y¢= cos ( )y ¢+11 2x-9y 2 2
2e 2x-y -9y e 2x-9y+3x y +2x y y = cos y y ( ) ¢ Þ y = 11-2e -3x y
2 x y -9e 2x-9y-cos ( )
(2 x y -9e 2x-y -cos ( ) )y ¢=11-2e 2x-9y -3 x y2
Increasing/Decreasing – Concave Up/Concave Down
Critical Points
x = c is a critical point of f x provided either Concave Up/Concave Down
( )
1. f c( )0 or 2. f c d( )n’t exist. 1. If f ¢¢( )> 0 for all x in an interval I then
f( )is concave up on the interval I.
Increasing/Decreasing 2. If f ¢¢( )< 0 for all x in an interval I then
1. If f ¢( )> 0 for all x in an interval I then
f( )is concave down on the interval I.
f( )is increasing on the interval I.
2. If f¢( )< 0 for all x in an interval I then Inflection Points
f( )is decreasing on the interval I. x = c is a inflection point of ( ) if the
concavity changes at x = c.
3. If f ¢( )= 0 for all x in an interval I then
f( )is constant on the interval I.
Visitttp://tutorial.math.lamar.edu for a complete set of Calculus notes. © 2005 Paul Dawkins Calculus Cheat Sheet
Extrema
Absolute Extrema Relative (local) Extrema
1. x = cis an absolute maximum of f x ( ) 1. x = c is a relative (or local) maximum of
f( )if f ( ) f x ( ) all x near c.
if f( )‡ f ( )or all x in the domain.
2. x = c is an absolute minimum of f x( ) 2. x = c is a relative (or local) minimum of
f( )if f ( ) f x ( ) all x near c.
if f( )£ f ( )or all x in the domain.
st
1 Derivative Test
Fermat’s Theorem If x = c is a critical point o( ) x then x = c is
If f ( )as a relative (or local) extrema at 1. a rel. max. of f x if f ¢ x > 0 to the left
x = c, then x = c is a critical point of( )x . ( ) ( )
of x = c and f¢( )< 0 to the right of x = c.
Extreme Value Theorem 2. a rel. min. of f( )if f¢( )< 0 to the left
If f ( )s continuous on the closed interval ofx = cand f ¢( )> 0to the right of x = c.
[ ]b then there exist numbers c and d so that, 3. not a relative extrema of f( )if f ¢( )is
1. a £ c,d £ b , 2. ( ) is the abs. max. in the same sign on both sides of x = c.
[ ]b , 3. f ( )s the abs. min. in [ ] . 2 Derivative Test
Finding Absolute Extrema If x = c is a critical point o( ) x such that
f ( ) 0 then x = c
To find the absolute extrema of the continuous
function f ( )n the interval a[ ]use the 1. is a relative maximum of f x( ) f ¢( )< 0.
following process. ¢¢
2. is a relative minimum of f( )if f ( )> 0.
1. Find all critical points of ( ) in [ ] . 3. may be a relative maximum, relative
2. Evaluate f x( ) all points found in Step 1. minimum, or neither if f ¢( )= 0.
3. Evaluate f( )and f b ( )
4. Identify the abs. max. (largest function Finding Relative Extrema and/or
value) and the abs. min.(smallest function Classify Critical Points
value) from the evaluations in Steps 2 & 3. 1. Find all critical points of ( )
st nd
2. Use the 1 derivative test or the 2
derivative test on each critical point.
Mean Value Theorem
If f ( )s continuous on the closed interval [ ] and differentiable on the open interval a(b )
then there is a number a < c < bsuch that f ( ) f( )- f ( ) .
b-a
Newton’s Method
f x
If xnis the n guess for the root/solution of ( ) = 0 then (n+1) guess is x n+1= xn- ( )n
f¢( n
provided f ¢( ) exists.
n
Visitttp://tutorial.math.lamar.edu for a complete set of Calculus notes. © 2005 Paul Dawkins Calculus Cheat Sheet
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 known quantities and solve for the unknown quantity.
Ex. A 15 foot ladder is resting against a wall. Ex. Two people are 50 ft apart when one
The bottom is initially 10 ft away and is being starts walking north. The angle q changes at
pushed towards the wall at 1ft/sec. How fast 0.01 rad/min. At what rate is the distance
4
is the top moving after 12 sec? between them changing when q = 0.5 rad?
x¢is negative because x is decreasing. Using We have q = 0.01 rad/min. and want to find
x¢. We can use various trig fcns but easiest is,
Pythagorean Theorem and differentiating, x x¢
x + y =15 2 Þ 2xx +2y y = 0 secq = Þ secq tanq q ¢=
1 50 50
After 12 sec we have x =10-12 (4) = 7and We know q = 0.05 so plug in q¢ and solve.
2 2 x ¢
so y = 15 -7 = 176 . Plug in and solve sec ( ) tan 0(5 0)(1 = )
for y . 50
x = 0.3112 ft/sec
1 ¢ ¢ 7
7 ( )4 + 176 y = 0 Þ y = 4 176 ft/sec Remember to have calculator in radians!
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 needed.
Ex. We’re enclosing a rectangular field with 2
Ex. Determine point(s) on y = x +1 that are
500 f

More
Less
Unlock Document

Related notes for MAT1320

Only pages 1,2 and half of page 3 are available for preview. Some parts have been intentionally blurred.

Unlock DocumentJoin OneClass

Access over 10 million pages of study

documents for 1.3 million courses.

Sign up

Join to view

Continue

Continue
OR

By registering, I agree to the
Terms
and
Privacy Policies

Already have an account?
Log in

Just a few more details

So we can recommend you notes for your school.

Reset Password

Please enter below the email address you registered with and we will send you a link to reset your password.