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Calculus Cheat Sheet

Calculus Cheat Sheet

Limits Definitions Precise Definition : We say lim f ( x ) = L'if Limit at Infinity : We say lim f ( x ) = L'if we x® a

for every e > 0 there is a d'> 0 such that whenever 0 < x

- L'< e

can make f ( x ) as close to L'as we want by

“Working” Definition : We say lim f ( x ) = L

There is a similar definition for lim f ( x ) = L

if we can make f ( x ) as close to L'as we want

except we require x large and negative

by taking x sufficiently close to a (on either side of a) without letting x = a

Infinite Limit : We say lim f ( x ) = ¥ if we

Right hand limit : lim+ f ( x ) = L'

This has x® a

the same definition as the limit except it requires x > a

Left hand limit : lim- f ( x ) = L'

This has the

taking x large enough and positive

can make f ( x ) arbitrarily large (and positive) by taking x sufficiently close to a (on either side of a) without letting x = a

There is a similar definition for lim f ( x ) =

-¥ x ®a

except we make f ( x ) arbitrarily large and x® a same definition as the limit except it requires negative

x®- ¥

If r > 0 and x r is real for negative x b then lim r = 0 x ®-¥ x 3

If r > 0 then lim

n even : lim x n = ¥ x ®± ¥

n odd : lim x n = ¥ & lim x n =

-¥ x ®¥

n even : lim a x n + L'+ b x + c'= sgn ( a ) ¥ x ®± ¥

n odd : lim a xn + L'+ b x + c'= sgn ( a ) ¥ x ®¥

Evaluation Techniques Continuous Functions L’Hospital’s Rule f ( x) ± ¥ f ( x) 0 If f ( x ) is continuous at a then lim f ( x ) = f ( a ) x ®a = then,

If lim = or lim x® a g ( x ) x® a g ( x ) ±¥ 0 Continuous Functions and Composition f ( x) f ¢ ( x) a is a number,

To compute Factor and Cancel p ( x) lim factor largest power of x out of both ( x

- 2x x ( x

- 42 2 x 3

- 42 3x2

If the two one sided limits had been equal then lim g ( x ) would have existed

-1 1 æ

h ®0 h x ( x + h ) ÷ h®0 x ( x + h ) x è ø

Some Continuous Functions Partial list of continuous functions and the values of x for which they are continuous

Polynomials for all x

cos ( x ) and sin ( x ) for all x

Rational function,

tan ( x ) and sec ( x ) provided division by zero

,L 2 2 2 2 4

cot ( x ) and csc ( x ) provided 5

2p ,L 6

Intermediate Value Theorem Suppose that f ( x ) is continuous on [a,

b] and let M be any number between f ( a ) and f ( b )

Then there exists a number c'such that a < c'< b and f ( c') = M

n odd : lim a xn + L'+ c'x + d'=

Visit http://tutorial

edu for a complete set of Calculus notes

© 2005 Paul Dawkins

Visit http://tutorial

edu for a complete set of Calculus notes

© 2005 Paul Dawkins

Calculus Cheat Sheet

Calculus Cheat Sheet

Derivatives Definition and Notation f ( x + h)

- f ( x)

If y = f ( x ) then the derivative is defined to be f ¢ ( x ) = lim h ®0 h If y = f ( x ) then all of the following are equivalent notations for the derivative

df dy d'f ¢ ( x ) = y¢ = = = ( f ( x) ) = Df ( x ) dx dx dx

If y = f ( x ) then,

If y = f ( x ) all of the following are equivalent notations for derivative evaluated at x = a

df dy f ¢ ( a ) = y ¢ x= a = = = Df ( a ) dx x =a dx x =a

Interpretation of the Derivative 2

f ¢ ( a ) is the instantaneous rate of

m = f ¢ ( a ) is the slope of the tangent line to y = f ( x ) at x = a and the

If f ( x ) is the position of an object at time x then f ¢ ( a ) is the velocity of

equation of the tangent line at x = a is given by y = f ( a ) + f ¢ ( a )( x

± g )¢ = f ¢ ( x ) ± g ¢ ( x ) f ¢ g + f g ¢ – Product Rule

æ f ö¢ f ¢ g

- f g ¢ 4

ç ÷ = – Quotient Rule g2 ègø

( x ) = n x n-1 – Power Rule dx d'f ( g ( x) ) = f ¢ ( g ( x) ) g¢ ( x) 7

Higher Order Derivatives The Second Derivative is denoted as The nth Derivative is denoted as 2 d'f dn f f ¢¢ ( x ) = f ( 2) ( x ) = 2 and is defined as f ( n) ( x ) = n and is defined as dx dx ¢ ¢ n n-1 ( ) f ¢¢ ( x ) = ( f ¢ ( x ) ) ,

the derivative of the f ( x ) = f ( ) ( x ) ,

the derivative of first derivative,

Implicit Differentiation Find y¢ if e2 x

Remember y = y ( x ) 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¢

Common Derivatives d'( x) = 1 dx d'( sin x ) = cos x dx d'( cos x ) =

Basic Properties and Formulas If f ( x ) and g ( x ) are differentiable functions (the derivative exists),

Chain Rule Variants The chain rule applied to some specific functions

é f ( x )ùû = n éë f ( x ) ùû f ¢ ( x ) 5

tan éë f ( x )ùû = f ¢ ( x ) sec2 éë f ( x ) ùû dx dx d'f ¢( x) d'7

( sec [ f ( x)]) = f ¢( x ) sec [ f ( x)] tan [ f ( x )] 3

ln ëé f ( x ) ûù = dx dx f ( x) f ¢( x ) d'd 8

d x ( a ) = a x ln ( a ) dx d'x ( e ) = ex dx d'( ln ( x ) ) = 1x ,

x ¹ 0 dx d'( log a ( x )) = x ln1 a ,

- 2e2 x

- 2e 2 x

- 9e2 x

Increasing/Decreasing – Concave Up/Concave Down Critical Points Concave Up/Concave Down x = c'is a critical point of f ( x ) provided either 1

If f ¢¢ ( x ) > 0 for all x in an interval I then 1

f ( x ) is concave up on the interval I

Increasing/Decreasing 1

If f ¢ ( x ) > 0 for all x in an interval I then f ( x ) is increasing on the interval I

If f ¢ ( x ) < 0 for all x in an interval I then f ( x ) is decreasing on the interval I

If f ¢ ( x ) = 0 for all x in an interval I then

If f ¢¢ ( x ) < 0 for all x in an interval I then f ( x ) is concave down on the interval I

Inflection Points x = c'is a inflection point of f ( x ) if the concavity changes at x = c'

f ( x ) is constant on the interval I

Visit http://tutorial

edu for a complete set of Calculus notes

© 2005 Paul Dawkins

Visit http://tutorial

edu for a complete set of Calculus notes

© 2005 Paul Dawkins

Calculus Cheat Sheet

Absolute Extrema 1

x = c'is an absolute maximum of f ( x ) if f ( c') ³ f ( x ) for all x in the domain

Calculus Cheat Sheet

Extrema Relative (local) Extrema 1

x = c'is a relative (or local) maximum of f ( x ) if f ( c') ³ f ( x ) for all x near c

x = c'is an absolute minimum of f ( x ) if f ( c') £ f ( x ) for all x in the domain

Fermat’s Theorem If f ( x ) has a relative (or local) extrema at x = c',

then x = c'is a critical point of f ( x )

Extreme Value Theorem If f ( x ) is continuous on the closed interval

b] then there exist numbers c'and d'so that,

Finding Absolute Extrema To find the absolute extrema of the continuous function f ( x ) on the interval [ a ,

Find all critical points of f ( x ) in [ a,

Evaluate f ( x ) at all points found in Step 1

Evaluate f ( a ) and f ( b )

Identify the abs

(largest function value) and the abs

(smallest function value) from the evaluations in Steps 2 & 3

x = c'is a relative (or local) minimum of f ( x ) if f ( c') £ f ( x ) for all x near c

Related Rates Sketch picture and identify known/unknown quantities

Write down equation relating quantities and differentiate with respect to t using implicit differentiation (i

add on a derivative every time you differentiate a function of t)

Plug in known quantities and solve for the unknown quantity

A 15 foot ladder is resting against a wall

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 0

01 rad/min

At what rate is the distance pushed towards the wall at 14 ft/sec

How fast between them changing when q = 0

? is the top moving after 12 sec

of f ( x ) if f ¢ ( x ) > 0 to the left of x = c'and f ¢ ( x ) < 0 to the right of x = c'

of f ( x ) if f ¢ ( x ) < 0 to the left of x = c'and f ¢ ( x ) > 0 to the right of x = c'

not a relative extrema of f ( x ) if f ¢ ( x ) is the same sign on both sides of x = c'

is a relative maximum of f ( x ) if f ¢¢ ( c') < 0

is a relative minimum of f ( x ) if f ¢¢ ( c') > 0

or neither if f ¢¢ ( c') = 0

Finding Relative Extrema and/or Classify Critical Points 1

Find all critical points of f ( x )

x¢ is negative because x is decreasing

Using Pythagorean Theorem and differentiating,

x 2 + y 2 = 15 2 Þ 2 x x¢ + 2 y y¢ = 0 After 12 sec we have x = 10

Plug in and solve for y¢

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

We’re enclosing a rectangular field with Ex

Determine point(s) on y = x 2 + 1 that are 500 ft of fence material and one side of the closest to (0,2)

Determine dimensions that will maximize the enclosed area

Use the 1st derivative test or the 2nd derivative test on each critical point

Mean Value Theorem If f ( x ) is continuous on the closed interval [ a ,

b ] and differentiable on the open interval ( a ,

b ) then there is a number a < c'< b such that f ¢ ( c') =

Newton’s Method If xn is the n guess for the root/solution of f ( x ) = 0 then (n+1)st guess is xn+1 = xn th

© 2005 Paul Dawkins

Minimize f = d'2 = ( x

Maximize A = xy subject to constraint of x + 2 y = 500

Solve constraint for x and plug into area

A = y ( 500

A¢ = 500

and so is the answer where after

Finally,

Visit http://tutorial

edu for a complete set of Calculus notes

We have q ¢ = 0

01 rad/min

We can use various trig fcns but easiest is,

x x¢ sec q = Þ sec q tan q q ¢ = 50 50 We know q = 0

Visit http://tutorial

edu for a complete set of Calculus notes

Solve constraint for x 2 and plug into the function

2 x2 = y

- 2 ) = y

- 1+ ( y

- 2) = y2

and so all we need to do is find x value(s)

The 2 points are then

© 2005 Paul Dawkins

Calculus Cheat Sheet

Calculus Cheat Sheet

Integrals Definitions Definite Integral: Suppose f ( x ) is continuous Anti-Derivative : An anti-derivative of f ( x )

Standard Integration Techniques Note that at many schools all but the Substitution Rule tend to be taught in a Calculus II class

( ) ò a f ( g ( x )) g ¢ ( x ) dx = ò g (a ) f ( u ) du

Divide [ a ,

F ( x ) ,

such that F ¢ ( x ) = f ( x )

u Substitution : The substitution u = g ( x ) will convert

width D'x and choose x from each interval

Indefinite Integral : ò f ( x ) dx = F ( x ) + c

For indefinite integrals drop the limits of integration

where F ( x ) is an anti-derivative of f ( x )

ò a f ( x ) dx = nlim å i b

ò 1 5x

Fundamental Theorem of Calculus Variants of Part I : Part I : If f ( x ) is continuous on [ a,

b ] then d'u(x) x f ( t ) dt = u ¢ ( x ) f éëu ( x ) ùû g ( x ) = ò f ( t ) dt is also continuous on [ a ,

b ] dx ò a a d'b d'x f ( t ) dt =

dx ò v( x) dx ò a d'u( x) Part II : f ( x ) is continuous on [ a ,

F ( x ) is f ( t ) dt = u ¢ ( x ) f [ u ( x) ]

F ( x ) = ò f ( x ) dx ) then ò f ( x ) dx = F ( b )

- F ( a )

ò 1 5x

Properties ò cf ( x) dx = c'ò f ( x) dx ,

ò a cf ( x ) dx = c'ò a f ( x ) dx ,

ò a f ( x ) dx = 0

ò a f ( x ) dx = ò a f ( t ) dt

ò a f ( x) dx =

ò f ( x ) dx £ ò b

If f ( x ) ³ g ( x ) on a £ x £ b then

ò f ( x) dx ³ ò g ( x ) dx

If f ( x ) ³ 0 on a £ x £ b then

If m £ f ( x ) £ M on a £ x £ b then m ( b

- a ) b

8 5 1 3

= 53 sin ( u ) 1 = 53 ( sin (8 )

x = 1 Þ u = 13 = 1 :: x = 2 Þ u = 2 3 = 8 Integration by Parts : ò u dv = uv

ò a u dv = uv

- ò v du

Choose u and dv from a

integral and compute du by differentiating u and compute v using v = ò dv

- e + c

ò3 ln x dx

ò3 ln x dx = x ln x 3

- x ) 3 5

Products and (some) Quotients of Trig Functions For ò tan n x sec m x dx we have the following : For ò sin n x cos m x dx we have the following : 1

Strip 1 sine out and convert rest to 1

then use the substitution u = cos x

Strip 1 cosine out and convert rest 2

then use the substitution u = sin x

Use either 1

Use double angle 3

and/or half angle formulas to reduce the 4

integral into a form that can be integrated

Trig Formulas : sin ( 2 x ) = 2sin ( x ) cos ( x ) ,

Common Integrals

ò k dx = k x + c'n n 1 ò x dx = n+1 x + c,

- 3ln ( 3)

ò f ( x) ± g ( x) dx = ò f ( x) dx ± ò g ( x) dx b b b ò a f ( x ) ± g ( x ) dx = ò a f ( x ) dx ± ò a g ( x ) dx

Þ du = 3 x dx Þ x dx = du 2

ò cos u du = sin u + c'ò sin u du =

ò tan u du = ln sec u + c'ò sec u du = ln sec u + tan u + c'u ò a + u du = a tan ( a ) + c'1 u ò a

Visit http://tutorial

edu for a complete set of Calculus notes

© 2005 Paul Dawkins

ò cos x dx (sin x) sin x sin x sin x sin x ò cos x dx = ò cos x dx = ò cos x dx (1- cos x ) sin x =ò dx ( u = cos x ) cos x (1-u ) 1 2 u + u du =

-ò du =

Strip 1 tangent and 1 secant out and convert the rest to secants using tan 2 x = sec 2 x

then use the substitution u = sec x

Strip 2 secants out and convert rest to tangents using sec 2 x = 1 + tan 2 x ,

then use the substitution u = tan x

Use either 1

Each integral will be dealt with differently

Visit http://tutorial

edu for a complete set of Calculus notes

Calculus Cheat Sheet

Calculus Cheat Sheet

Trig Substitutions : If the integral contains the following root use the given substitution and formula to convert into an integral involving trig functions

4- 9 x2

x = sin q Þ dx = cos q d'q 2 3

( 23 cos q ) dq = ò sin122 q dq

Use Right Triangle Trig to go back to x’s

From 3x x 2 = x

Because we have an indefinite substitution we have sin q = 2 so,

integral we’ll assume positive and drop absolute value bars

If we had a definite integral we’d need to compute q ’s and remove absolute value bars based on that and,

ì x if x ³ 0 x =í î- x if x < 0

From this we see that cot q =

In this case we have

-9 x2 3x

Net Area :

ò a f ( x ) dx represents the net area between f ( x ) and the

Area Between Curves : The general formulas for the two main cases for each are,

- 9x = 4

Applications of Integrals b

é ù ë upper function û

ò Q( x) dx where the degree of P ( x ) is smaller than the degree of

Integrate the partial fraction decomposition (P

For each factor in the denominator we get term(s) in the decomposition according to the following table

Term in P

D Factor in Q ( x )

A ax + b

Ax + B ax + bx + c

Term in P

D A1 A2 Ak + +L+ k ax + b ( ax + b )2 ( ax + b )

Ak x + Bk A1 x + B1 +L + k ax 2 + bx + c'( ax 2 + bx + c')

3x x2 +4

16 x2 +4

+C + Bx = x2 + 4

é right function ù ë û

Q ( x )

Factor denominator as completely as possible and find the partial fraction decomposition of

Factor in Q ( x )

If the curves intersect then the area of each portion must be found individually

Here are some sketches of a couple possible situations and formulas for a couple of possible cases

A = ò f ( x )

Partial Fractions : If integrating

A = ò f ( y )

A = ò f ( x )

Volumes of Revolution : The two main formulas are V = ò A ( x ) dx and V = ò A ( y ) dy

Here is some general information about each method of computing and some examples

Rings Cylinders A = p ( outer radius ) 2

Limits: x/y of right/bot ring to x/y of left/top ring Limits : x/y of inner cyl

Axis use f ( x ) ,

Axis use f ( y ) ,

Axis use f ( y ) ,

Axis use f ( x ) ,

A ( x ) and dx

A ( y ) and dy

A ( y ) and dy

A ( x ) and dx

Axis : y = a > 0

Axis : y = a £ 0

Axis : y = a > 0

Axis : y = a £ 0

- f ( x )

- g ( x )

- g ( y )

A( x 2 +4)+ ( Bx +C ) ( x

-1) ( x

Set numerators equal and collect like terms

A+ B = 7 C

An alternate method that sometimes works to find constants

Start with setting numerators equal in previous example : 7 x 2 + 13 x = A ( x 2 + 4 ) + ( Bx + C ) ( x

Chose nice values of x and plug in

For example if x = 1 we get 20 = 5A which gives A = 4

This won’t always work easily

Visit http://tutorial

edu for a complete set of Calculus notes

© 2005 Paul Dawkins

- g ( y )

These are only a few cases for horizontal axis of rotation

If axis of rotation is the x-axis use the y = a £ 0 case with a = 0

For vertical axis of rotation ( x = a > 0 and x = a £ 0 ) interchange x and y to get appropriate formulas

Visit http://tutorial

edu for a complete set of Calculus notes

© 2005 Paul Dawkins

Calculus Cheat Sheet

Work : If a force of F ( x ) moves an object

Average Function Value : The average value of f ( x ) on a £ x £ b is f avg =

the work done is W = ò F ( x ) dx b

ò f ( x ) dx

Arc Length Surface Area : Note that this is often a Calc II topic

The three basic formulas are,

L = ò ds

SA = ò 2p y ds (rotate about x-axis)

SA = ò 2p x ds (rotate about y-axis)

where ds is dependent upon the form of the function being worked with as follows

ds = r 2 + ( ddrq ) d'q if r = f (q ) ,

With surface area you may have to substitute in for the x or y depending on your choice of ds to match the differential in the ds

With parametric and polar you will always need to substitute

Improper Integral An improper integral is an integral with one or more infinite limits and/or discontinuous integrands

Integral is called convergent if the limit exists and has a finite value and divergent if the limit doesn’t exist or has infinite value

This is typically a Calc II topic

Infinite Limit ¥

f ( x ) dx = lim ò f ( x ) dx t

ò ¥ f ( x ) dx = ò ¥ f ( x ) dx + ò

ò ¥ f ( x ) dx = lim ò f ( x ) dx b

f ( x ) dx provided BOTH integrals are convergent

Discontinuous Integrand 1

Discont

at a: ò f ( x ) dx = lim+ ò f ( x ) dx b

Discont

at b : ò f ( x ) dx = lim- ò f ( x ) dx

ò f ( x ) dx = ò f ( x ) dx + ò f ( x ) dx provided both are convergent

Discontinuity at a < c'< b :

Comparison Test for Improper Integrals : If f ( x ) ³ g ( x ) ³ 0 on [ a,

¥ ) then,

Useful fact : If a > 0 then

dx converges if p > 1 and diverges for p £ 1

If ò g ( x ) dx divg

Approximating Definite Integrals For given integral

f ( x ) dx and a n (must be even for Simpson’s Rule) define Dx = b

-n a and

b ] into n subintervals [ x0 ,

xn ] with x0 = a and xn = b then,

Midpoint Rule : Trapezoid Rule : Simpson’s Rule :

ò f ( x ) dx » Dx éë f ( x ) + f ( x ) + L'+ f ( x ) ùû ,

Dx ò f ( x ) dx » 2 ëé f ( x ) + 2 f ( x ) + +2 f ( x ) + L'+ 2 f ( x ) + f ( x ) ûù b

Dx ò f ( x ) dx » 3 ëé f ( x ) + 4 f ( x ) + 2 f ( x ) +L + 2 f ( x ) + 4 f ( x ) + f ( x )ûù b

Visit http://tutorial

edu for a complete set of Calculus notes

© 2005 Paul Dawkins