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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

x

for every   0 there is a   0 such that whenever 0  x  a   then f  x   L   .

can make f  x  as close to L as we want by taking x large enough and positive.

“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

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

x  

xa

xa

the same definition as the limit except it requires x  a . Left hand limit : lim f  x   L . This has the xa

xa

There is a similar definition for lim f  x   x a

except we make f  x  arbitrarily large and negative.

same definition as the limit except it requires xa. Relationship between the limit and one-sided limits lim f  x   L  lim f  x  lim f  x  L lim f  x   lim f  x   L  lim f  x   L xa

xa

x a

xa

xa

x a

lim f  x  lim f  x  lim f  x  Does Not Exist

x  a

xa

x

a

x

xa

a

2. lim  f  x   g  x    lim f  x   lim g  x x a x a x a 3. lim  f  x  g  x    lim f  x  lim g  x  x a

x a

x a

f  x  f  x   lim provided lim g  x   0 4. lim    x a x a x a g  x  g  x   xlim a

Basic Limit Evaluations at  

x

2. lim ln  x    x 

lim e x  0

x  

&

lim ln x   

x 0

b 0 r x  x r 4. If r  0 and x is real for negative x b then lim r  0 x   x 3. If r  0 then lim



x a

x a

x a

Polynomials at Infinity p  x and q  x are polynomials. To compute

Factor and Cancel  x  2  x  6  x 2  4x 12  lim lim x 2 x 2 x2  2 x x  x  2

lim

x 

x 6 8  4 x 2 Rationalize Numerator/Denominator 3 x 3 x 3 x  lim 2 lim 2 x  9 x  81 x  9 x  81 3 x 9x 1  lim  lim x 9  x2  81 3  x x9 x  9  3  x

of both p  x  and q  x then compute limit.

 lim x 2









p  x factor largest power of x in q  x out q x 

lim

x  





x2 3  42 3  42 3 x2  4 3  lim 2 5 x  lim 5 x   2     x x 5x  2x 2 x x 2 x 2

Piecewise Function



1 1  18 6  108

1  h  1 1  lim   2   lim h 0 h  x  x  h  x   h  0 x  x  h

 x2  5 if x   2 g  x  where g  x   lim x  2 1  3 x if x  2 Compute two one sided limits, lim  g x   lim x 2 5  9 x  2

x  2

lim g x   lim 1 3x  7

x  2

x  2

g  x One sided limits are different so lim x  2

doesn’t exist. If the two one sided limits had been equal then lim g  x  would have existed x 2

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  x  and sin  x  for all x. 2. Rational function, except for x’s that give 8. tan  x  and sec  x provided division by zero. 3   3 3. n x (n odd) for all x. x  ,  ,  , , , 2 2 2 2 4. n x (n even) for all x  0 . 9. cot  x  and csc  x  provided x 5. e for all x. x   ,  2  ,  , 0,  , 2 , 6. ln x for x  0 .

n

n 5. lim  f  x    lim f  x   x a  x a 6. lim  n f  x    n lim f  x   x a  x a

Note : sgn a  1 if a  0 and sgn  a   1 if a  0 . 1. lim e x   &



lim f  g  x    f lim g  x   f  b

Combine Rational Expressions 1 1 1 1  x  x  h   lim     lim   h 0 h  x  h x  h 0 h  x  x  h  

Properties f  x  and lim g  x  both exist and c is any number then, Assume lim   1. lim  cf  x   clim f  x x a x a

Evaluation Techniques L’Hospital’s Rule Continuous Functions f  x 0 f  x   If f  x is continuous at a thenlim f  x  f  a  or lim xa If lim  then, x  a g  x x  a g  x 0  Continuous Functions and Composition f  x f   x  lim a is a number,  or  lim f x  is continuous at b and lim g  x  b then x  a g x  x  a g  x 

xn   5. n even : x lim   6. n odd : lim x n   & lim x n   x 

x 

7. n even : lim a x n   b x  c sgn  a 

Intermediate Value Theorem Suppose that f  x is continuous on [a, b] and let M be any number between f a  and f b  .

8. n odd : lim a x n    b x  c  sgn a  

Then there exists a number c such that a  c  b and f c   M .

x  

x 

9. n odd : lim a x n    c x  d   sgn  a  

Visit http://tutorial.math.lamar.edu for a complete set of Calculus notes.

x  

© 2005 Paul Dawkins

Visit http://tutorial.math.lamar.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     Df  a xa dx x  a dx x  a

Interpretation of the Derivative 2. f  a  is the instantaneous rate of

1. m  f   a  is the slope of the tangent line to y  f  x at x  a and the equation of the tangent line at x  a is given by y  f a   f  a  x  a  .

change of f  x at x  a . 3. If f  x  is the position of an object at time x then f   a  is the velocity of the object at x  a .

Basic Properties and Formulas If f  x  and g  x  are differentiable functions (the derivative exists), c and n are any real numbers, 1.

 c f   c f   x

2.

f

3.

 f g  

 g   f   x   g  x  f  g  f g – Product Rule

 f  f  g  f g  4.    – Quotient Rule 2 g g

d 5.  c  0 dx d n 6.  x   n xn 1 – Power Rule dx d 7. f  g  x   f   g x  g   x dx This is the Chain Rule



Chain Rule Variants The chain rule applied to some specific functions. n n 1 d d 1. 5.  f x    n  f  x   f   x  cos  f  x    f  x sin  f  x  dx dx  d d 2. 6. e f  x  f   x e f x tan  f  x   f  x sec 2  f  x  dx dx d f  x d 7.  sec  f ( x )   f  ( x ) sec  f ( x ) tan  f ( x)  ln  f  x    3. dx dx f x  f   x d d tan 1  f x    8. 2 sin  f  x    f   x  cos  f  x  4. dx 1   dx  f  x



 

d x  a   ax ln  a dx d x e   e x dx d 1  ln x   x , x  0 dx d 1  ln x   x , x  0 dx d 1  loga  x   x lna , x  0 dx



















Higher Order Derivatives The Second Derivative is denoted as The nth Derivative is denoted as 2 d f dn f 2  n  and is defined as f   x  f  x  2 and is defined as f  x  dx dx n   f   x   f   x  , i.e. the derivative of the f n   x   f  n 1  x , i.e. the derivative of n 1 st first derivative, f   x  . the (n-1) derivative, f    x .





 

Implicit Differentiation Find y if e2 x  9 y  x 3 y 2  sin  y   11x . 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 . e 2x  9y  2 9 y  3x 2y 2 2 x 3y y  cos y  y 11   2e 2 x 9 y  9 ye 2 x 9 y  3 x 2 y 2  2 x 3 y y  cos  y  y 11 3

d  csc x    csc x cot x dx d  cot x   csc2 x dx d  sin 1 x  1 2 dx 1 x d 1  cos 1 x    dx 1 x 2 d 1  tan 1 x  1 x 2 dx



 2 x y  9e x

Common Derivatives d x   1 dx d sin x   cos x dx d  cos x    sin x dx d  tanx   sec2 x dx d sec x   sec x tan x dx



2 9 y

 cos y  y  11  2e 2 x 9 y  3x 2 y 2



y 

11 2 e2 x 9 y 3x 2 y 2 2 x3 y  9e 2 x 9 y  cos  y 

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  c  0 or 2. f  c doesn’t exist.

 

 

Increasing/Decreasing 1. If f  x  0 for all x in an interval I then f  x is increasing on the interval I. 2. If f  x  0 for all x in an interval I then f  x  is decreasing on the interval I. 3. If f  x  0 for all x in an interval I then

f  x  is concave up on the interval I.

2. 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.math.lamar.edu for a complete set of Calculus notes.

© 2005 Paul Dawkins

Visit http://tutorial.math.lamar.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   f  x  for all x in the domain. 2. x  c is an absolute minimum of f  x if f  c  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  f  x for all x near c. 2. x  c is a relative (or local) minimum of f  x  if f  c  f  x for all x near c.

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

 a, b  then there exist numbers c and d so that, 1. a  c, d  b , 2. f c  is the abs. max. in  a, b  , 3. f d  is the abs. min. in  a, b  . Finding Absolute Extrema To find the absolute extrema of the continuous function f  x  on the interval  a , b use the

1st Derivative Test If x  c is a critical point of f  x then x  c is 1. a rel. max. of f  x if f  x   0 to the left of x  c and f  x  0 to the right of x  c . 2. a rel. min. of f  x if f   x  0 to the left of x  c and f  x  0to the right of x  c . 3. not a relative extrema of f x  if f   x  is the same sign on both sides of x  c . 2nd Derivative Test If x  c is a critical point of f  x such that f   c   0 then x  c 1. is a relative maximum of f x  if f   c   0 .

2. Evaluate f  x  at all points found in Step 1.

2. is a relative minimum of f x  if f  c  0. 3. may be a relative maximum, relative minimum, or neither if f   c   0 .

3. Evaluate f a  and f b  . 4. Identify the abs. max. (largest function value) and the abs. min.(smallest function value) from the evaluations in Steps 2 & 3.

Finding Relative Extrema and/or Classify Critical Points 1. Find all critical points of f  x  .

following process. 1. Find all critical points of f  x  in  a , b .

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. Two people are 50 ft apart when one Ex. A 15 foot ladder is resting against a wall. starts walking north. The angle changes at The bottom is initially 10 ft away and is being 0.01 rad/min. At what rate is the distance pushed towards the wall at 41 ft/sec. How fast between them changing when   0.5 rad? is the top moving after 12 sec?

x is negative because x is decreasing. Using Pythagorean Theorem and differentiating, x2  y2  152  2 x x  2 y y  0 After 12 sec we have x  10 12  14   7 and

so y  15 2  7 2  176 . Plug in and solve for y . 7 7   14   176 y  0  y  ft/sec 4 176

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 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). field is a building. Determine dimensions that will maximize the enclosed area.

2. 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  

f  b  f  a . b a

Newton’s Method If xn is the nth guess for the root/solution of f  x   0 then (n+1)st guess is xn  1  xn 

f  xn  f  xn 

provided f   xn  exists.

Visit http://tutorial.math.lamar.edu for a complete set of Calculus notes.

© 2005 Paul Dawkins

We have    0.01 rad/min. and want to find x . We can use various trig fcns but easiest is, x x sec   sec tan    50 50 We know  0.5 so plug in   and solve. x sec 0.5  tan  0.5 0.01  50 x  0.3112 ft/sec Remember to have calculator in radians!

Minimize f  d 2   x  0  y  2 and the 2

Maximize A  xy subject to constraint of x  2 y  500 . Solve constraint for x and plug into area. A  y  500 2y  x  500  2 y  2  500 y  2y Differentiate and find critical point(s). A  500  4 y  y  125 By 2nd deriv. test this is a rel. max. and so is the answer we’re after. Finally, find x. x  500  2 125   250 The dimensions are then 250 x 125. Visit http://tutorial.math.lamar.edu for a complete set of Calculus notes.

2

constraint is y  x 2  1 . Solve constraint for x 2 and plug into the function. x 2  y 1  f  x 2   y 2 

2

 y  1   y  2 2  y 2  3 y  3 Differentiate and find critical point(s).  y  32 f   2y 3 By the 2nd derivative test this is a rel. min. and so all we need to do is find x value(s). x2  32 1  12  x   12 The 2 points are then



1 2





, 32 and 

1 2



, 32 .

© 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 on  a , b  . Divide  a , b into n subintervals of

is a function, F x  , such that F  x   f  x  .

width  x and choose xi * from each interval.

Indefinite Integral :  f x dx  F x   c

Then

b

a

n

n 

i 1

Ex.

1 5 x cos  x  dx 2

2

2

x  1  u 13 1 :: x  2  u  23 8 Integration by Parts :  u dv  uv  v du and

v ( x ) 

then  f  x dx  F  b  F  a . Properties  cf  x dx  c f  x dx , c is a constant b

b

c dx  c b  a 

 a cf  x  dx  c  a f  x  dx , c is a constant

a

a f  x  dx  0



a

b

 f  x  dx  

a

a f  x  dx  b f  x  dx b

c

b

a

a

c

b

b

a

a

f  x  dx

 f  x  dx   f  x  dx   f  x dx for any value of c. If f  x   g  x  on a  x  b then

 f  x  dx   g  x dx

If f  x   0 on a  x  b then

f  x dx  0



b a

b

b

a

a

b

If m  f x   M on a  x  b then m  b  a    f  x dx  M  b  a a

Common Integrals

 k dx  k x  c n n 1  x dx  n1 x  c, n  1  x dx   x dx  ln x  c  a x b dx  a ln ax  b  c ln u du  u ln u   u  c  e du  e  c 1

1

1

1

u

1

u

cos u du  sin u  c sin u du  cos u  c sec u du  tan u  c sec u tan u du  sec u  c csc u cot udu   csc u  c csc u du   cot u  c 2

 tan u du  ln sec u  c  sec u du  ln sec u  tan u  c  a  u du  a tan  ua   c 1 u  a u du  sin  a  c 1

2

1

1

2

1

2

2

f u  du using

8

3

b

a u dv  uv

Ex.  x e x dx u x

 xe

x

Ex. x

dv  e

x



du  dx v  e

dx  xe x   e x dx  xe x e x c

a

b

g a

b a

b

  v du . Choose u and dv from a

integral and compute du by differentiating u and compute v using v   dv .

b

 f  x   g  x  dx   f  x  dx   g  x dx b b b a f  x   g  x  dx  a f  x  dx  a g  x dx

g b 

a

5 1 5 x cos  x  dx  1 3 cos  u  du 8  53 sin  u  35  sin 8  sin 1  1

3

u  x 3  du  3x 2dx  x 2dx  13 du

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 u  x  g  x   f  t dt is also continuous on  a , b  dx  a a d b d x f t dt   v  x  f  v  x  and g  x    f  t  dt  f  x . dx  v  x  dx a u x  d Part II : f x is continuous on  a , b  , F x  is f  t  dt  u  x  f u ( x )   v  x  f dx  v  x  an anti-...