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Calculu s Ch eat Sh eet

Calculu s Ch eat Sh eet

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 e > 0 there is a d > 0 such that whenever 0 < x - a < d then f ( x ) - L < e .

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

except we require x large and negative.

x ®-¥

x ®a

Right hand limit : lim+ f ( x) = L . This has

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

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 ®a

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

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

Left hand limit : lim- f ( x) = L. This has the

except we make f ( x) arbitrarily large and

x ®a

same definition as the limit except it requires negative. x 0 and sgn (a ) = -1 if a < 0 . 1. lim e x = ¥ & x®¥

2. lim ln ( x ) = ¥ x ®¥

x lim e = 0

x®- ¥

&

lim- ln ( x) = - ¥

x® 0

b r =0 x ®¥ x 4. If r > 0 and xr is real for negative x b then lim r = 0 x®-¥ x

3. If r > 0 then lim

Vis it h ttp://tutorial.math .lamar.edu for a complete set of Calculus notes.

(

)

x ®a

lim f ( g ( x )) = f lim g ( x ) = f (b ) x ®a

x ®a

Polynomials at Infinity p (x ) and q ( x) are polynomials. To compute

Factor and Cancel x 2 + 4x - 12 ( x -2 ) ( x +6 ) = 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 -1 9- x = lim 2 = lim x ®9 x - 81 3+ x x ®9 ( x+ 9 ) 3+

p (x ) and q ( x) and then compute limit.

= lim

x ®2

(

)(

)

(

p (x ) factor largest power of x out of both q (x )

lim

x ®-¥

(

)

x 2 3 - 42 3 - 24 3x 2 - 4 3 x = lim = lim 5 x = x ®-¥ 5x - 2x 2 x ®- ¥ x 2 x5 - 2 2 x -2

(

)

P iecewise F unction

)

x

-1 1 = =(18 )(6 ) 108 Combine Rational Expressions 1 1 1 1 æ x - ( x + h) ö - ö÷ = lim ç lim æç ÷ h ®0 h è x + h x ø h ®0 h çè x ( x + h ) ÷ø -1 1 æ -h ö 1 = lim çç =- 2 ÷ = lim ® h ® 0 h x (x + h ) ÷ x è ø h 0 x (x + h )

ì x 2 + 5 if x < - 2 lim g (x ) where g ( x) = í x ®- 2 î1 -3 x if x ³ -2 Compute two one sided limits, 2 lim - g ( x ) = lim - x +5 =9 x ®- 2

x ®-2

x ®- 2

x ®- 2

lim + g ( x ) = lim +1 -3 x =7

One sided limits are different so lim g ( x ) 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 F unctions 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 tan (x ) and sec ( x ) provided 8. division by zero. 3p p p 3p 3. n x (n odd) for all x. x ¹ L, - , - , , ,L 2 2 2 2 4. n x (n even) for all x ³ 0 . 9. cot (x ) and csc ( x ) provided 5. e x for all x. x ¹ L, -2 p, -p, 0, p , 2p ,L 6. ln x for x > 0 .

n

5.

Evaluation Techniques Continuous F unctions L’Hospital’s Rule f (x) ±¥ f (x) 0 If f ( x ) is continuous at a then lim f ( x ) = f (a ) x ®a = = or lim then, If lim x ®a g ( x ) x ®a g ( x ) 0 ±¥ Continuous F unctions and Composition f ( x) f ¢ ( x) = lim li®m a is a number, ¥ or -¥ f ( x) is continuous at b and lim g ( x ) = b then x a g (x ) x ®a g ¢ ( x )

5. n even : lim x n = ¥ x ®± ¥

6. n odd : lim x n = ¥ & lim x n = -¥ x ®¥

x ®- ¥

7. n even : lim a x n + L + 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 +L + 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 xn + L + c x + d = - sgn (a ) ¥ x ®-¥ © 2005 Paul Dawkins

Vis it h ttp://tutorial.math .lamar.edu for a complete set of Calculus notes.

© 2005 Paul Dawkins

Calculu s Ch eat Sh eet

Calculu s Ch eat Sh eet

Derivatives If y = f ( x

)

Definition and Notation f ( x + h) - 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) ) = Df ( x) f ¢( x) = y ¢ = dx dx dx

If y = f ( x ) all of the following are equivalent notations for derivative evaluated at x = a . df dy = = Df ( a ) f ¢( a ) = y ¢ x =a = dx x =a dx x =a

1. m = f ¢( a ) is the slope of the tangent

change of f ( x) at x = a .

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 ) . If f ( x) and g (x )

(

Interpretation of the Derivative 2. f ¢ (a ) is the instantaneous rate of

If y = f ( x ) then,

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 .

( c f )¢ = c f ¢( x )

2.

( f ± g ) ¢ = f ¢ ( x) ± g¢ ( x )

3.

( f g) ¢ =

d 5. ( c) = 0 dx d n n 1 x = n x - – Power Rule 6. dx d f ( g ( x )) = f ¢( g (x )) g ¢( x ) 7. dx This is the Chain Rule

( )

f ¢ g + f g ¢ – Product Rule

(

æ f ö¢ f ¢ g - f g¢ – Quotient Rule 4. ç ÷ = g2 è gø

)

(

)

(

(sin x ) = cos x (cos x ) = -sin x (tan x ) =sec

2

x

(sec x )= sec x tan x

d (csc x ) = - csc xcot x dx d (cot x ) = -csc 2 x dx d 1 1 sin - x = dx 1- x 2

(

)

(

)

(

)

d 1 1 cos - x = dx 1 - x2 d 1 tan -1 x = 2 dx 1 +x

7.

)

d sinëé f ( x)ûù = f¢ ( x) cosëé f ( x )ûù dx

8.

(

)

(

)

d cos éë f ( x )ùû = - f ¢ ( x )sin éë f (x )ùû dx d tan éë f ( x )ùû = f¢ ( x ) sec 2 éë f ( x )ùû dx d (sec[ f ( x )] ) = f ¢( x )sec [ f ( x) ] tan [ f ( x)] dx f¢( x ) d tan-1 ëé f (x )ûù = 2 dx 1 + ëé f ( x )ûù

(

)

Higher Order Derivatives The n th Derivative is denoted as The Second D erivative is denoted as n 2 d f d f (n ) (2 ) and is defined as f ¢¢ ( x ) = f ( x ) = 2 and is defined as f (x )= n dx dx ¢ ( ( ) f ¢¢ ( x ) = ( f ¢ ( x )) , i.e. the derivative of the f n ( x ) = f n-1) ( x ) ¢ , i.e. the derivative of ( - ) first derivative, f ¢( x ) . the (n-1)st derivative, f n 1 x .

(

)

Implicit Differentiation 2 9 3 2 Find y¢ if e x- y + x y = sin ( y ) + 11 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¢ . e 2x - 9y ( 2 -9 y ¢ ) +3 x 2y 2 +2 x 3y y¢ =cos ( y ) y¢ +11 2e 2 x-9 y - 9 y ¢e 2 x-9 y + 3x 2 y 2 + 2 x3 y y¢ = cos ( y ) y¢ + 11

(2 x y - 9e

Common Derivatives

( x )= 1

6.

)

3

d dx d dx d dx d dx d dx

5.

)

(

4.

Variants

( )

Basic Properties and F ormulas are differentiable functions (the derivative exists), c and n are any real numbers,

1.

Chain Rule The chain rule applied to some specific functions. n n -1 d é f ( x ) ûù = n ëé f (x ) ûù f ¢ (x ) 1. dx ë d f (x ) e = f ¢( x) e f ( x ) 2. dx f ¢( x ) d 3. ln ëé f ( x )ûù = dx f (x)

( )

d a x = a x ln ( a ) dx d x x e =e dx d ( ln (x ) ) = 1x , x > 0 dx d 1 ( ln x ) = x , x ¹ 0 dx d ( log a (x )) = x ln1 a , x > 0 dx

( )

2 x -9 y

)

- cos ( y ) y¢ = 11 - 2 e

2 x -9 y

-3x y 2

2

11 -2 e x- y -3 x y 3 2 9 2 x y - 9e x- y - cos ( y ) 2

Þ

y¢ =

9

2

2

Increasing/Decreasing – Concave Up/Concave Down Critical Points Concave Up/Concave D own = 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. 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. 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

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. Vis it h ttp://tutorial.math .lamar.edu for a complete set of Calculus notes.

© 2005 Paul Dawkins

Vis it h ttp://tutorial.math .lamar.edu for a complete set of Calculus notes.

© 2005 Paul Dawkins

Calculu s Ch eat Sh eet

Absolute Extrema 1. x = c is an absolute maximum of f ( x) if f ( c ) ³ f (x ) for all x in the domain.

Calculu s Ch eat Sh eet

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

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

1st Derivative Test If x = c is a critical point of f ( x) then x = c is

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

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 q changes at The bottom is initially 10 ft away and is being 1 0.01 rad/min. At what rate is the distance pushed towards the wall at 4 ft/sec. How fast between them changing when q = 0.5 rad? is the top moving after 12 sec?

1. a rel. max. of f ( x ) if f ¢ (x ) > 0 to the left

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

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

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

of x = c and f ¢( x) > 0 to the right of x = c .

[ 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 ].

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

Finding Absolute Extrema To find the absolute extrema of the continuous function f (x ) on the interval [ a, b] use the following process. 1. Find all critical points of f ( x ) in [ a ,b] .

f ¢ ( c ) = 0 then x = c 1. is a relative maximum of f (x) if f ¢¢ (c ) < 0 . 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 .

2. Evaluate f ( x) at all points found in Step 1. 3. Evaluate f (a) and f (b ) .

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

4. Identify the abs. max. (largest function value) and the abs. min.(smallest function value) from the evaluations in Steps 2 & 3.

x ¢ is negative because x is decreasing. Us ing Pythagorean Theorem and differentiating, x2 + y2 = 152 Þ 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 Þ 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 a nd one side of the closest to (0,2). field is a building. Determine dimensions that will maximize the enclosed area.

2. Use the 1 st derivative test or the 2 nd 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 M ethod If x n is the nth guess for the root/solution of f ( x) = 0 then (n+1)st guess is x n+1 = x n -

f ( xn ) f ¢ (x n )

provided f ¢ ( xn ) exists.

Vis it h ttp://tutorial.math .lamar.edu for a complete set of Calculus notes.

© 2005 Paul Dawkins

We have q ¢ = 0.01 rad/min. and want to find x ¢ . We can use various trig fcns but easiest is, x x¢ Þ sec q tan q q ¢ = sec q = 50 50 We knowq = 0.05 so plug in q ¢ and solve. x¢ sec ( 0.5) tan ( 0.5)( 0.01) = 50 x ¢ = 0.3112 ft/sec Remember to have calculator in radians!

2 Minimize f = d = ( 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 - 2 y ) x = 500 -2 y Þ = 500y - 2y 2 Differentiate and find critical point(s). A¢ = 500 -4 y Þ 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. Vis it h ttp://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 Þ f = x 2 +( y -2 )

2

= y - 1+ ( y - 2) = y 2 - 3 y + 3 Differentiate and find critical point(s). Þ f¢ = 2 y - 3 y = 32 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 Þ x = ± 12 2

The 2 points are then

(

1 2

,

3 2

) and ( -

1 2

)

, 32 .

© 2005 Paul Dawkins

Calculu s Ch eat Sh eet

Calculu s Ch eat Sh eet

Integ rals Definitions Definite Integral: Suppose f ( x) is continuous Anti-Derivative : An anti-deriva tive of f ( x )

Sta ndar d 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 b

g b

on [a ,b ] . Divide [a ,b ] into n subintervals of

is a function, F ( x) , such that F¢ ( x) = f ( x) .

u Substitutio n : The substitution u = g (x ) will convert

width D x and choose xi * from each interval.

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

du = g¢ ( x ) dx. For indefinite integrals drop the limits of integration.

¥

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

( )

Then ò f ( x )dx = lim å f x i* D x . a n b

®¥ i

=1

Ex.

2

ò1

( )

2

ò 1 5x

5x2 cos x3 dx

2

( )

b

then ò f ( x ) dx = F ( b ) - F ( a ) .

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

8

ò 1 53 cos (u )du 8 = 53sin (u ) 1 = 35 (sin (8 ) - sin (1 ))

cos x3 dx =

u = x 3 Þ d u = 3x 2d x Þ x 2d x = 13 du

Fundamental Theorem of Calculus Variants of P art I : Part I : If f ( x ) is continuous on [ a ,b] then d u (x ) x f (t )dt = u ¢ (x ) f ëéu (x )ûù g ( x ) = ò f (t ) dt is also continuous on [a ,b ] dx ò a a d b d x f ( t) dt = - v¢( x) f ëé v( x) ûù f ( t ) dt = f ( x ) . and g¢ ( x) = dx ò v( x) dx òa d u (x ) Part II : f ( x ) is continuous on[ a ,b] , F ( x ) is f (t ) dt = u ¢ (x ) f [u ( x ) ] -v ¢ (x ) f [v ( x ) ] dx ò v( x) an anti-derivative of f ( x) (i.e. F (x ) = ò f ( x )dx )

using

b

ò a u dv = uv

b a

b

- ò a v du . Choose u and dv from

integral and compute du by differentiating u and compute v using v = ò dv . Ex. ò xe - x dx u= x

ò xe

-x

Ex.

d v = e- x

d u = d x v = - e- x

Þ

dx = - xe - x + òe - x dx = - xe - x -e - x + c

a

5

ò 3 ln x dx

u = ln x 5

ò3

d v = d x Þ d u = 1x d x v = x

ln x dx = x ln x 3 - ò dx = (x ln ( x ) - x ) 5

5

5

3

3

= 5 ln (5 ) - 3 ln ( 3 ) - 2 Properties

ò f ( x) ± g( x) dx = ò f ( x) dx ± ò g( x) dx b b b òa f ( x) ± g ( x) dx = òa f ( x) dx ± òa g ( x) dx a òa f ( x) dx =0 b

ò cf ( x )dx =c ò f (x )dx , c is a constant b b ò a cf ( x) dx = c ò a f ( x) dx , c is a constant b b ò a f ( x ) dx = ò a f (t ) dt

a

òa f ( x) dx = - òb f ( x) dx

b

ò f ( x) dx £ ò a

If f ( x) ³ g ( x) ona £ x £ b then

b

a

f ( x) dx

ò f ( x ) dx ³ ò g ( x ) dx b

a

a

b

If f ( x ) ³ 0 on a £ x £ b then ò f ( x) dx ³ 0 b

a

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

P roducts and (some) Quotients of Trig F unctions Forò tann x secm x dx we have the following : For ò sinn x cosm x dx we have the following : 1. n odd. Strip 1 sine out and convert rest to 1. n odd. Strip 1 tangent and 1 secant out and convert the rest to secants using cosines using sin 2 x = 1- cos 2 x , then use tan2 x = sec2 x - 1 , then use the substitution the substitution u = cos x . u = sec x . 2. m odd. Strip 1 cosine out and convert rest 2. m even. Strip 2 secants out and convert rest to sines using cos 2 x = 1 - sin 2 x , then use to tangents using sec2 x = 1+ tan2 x , then the substitution u = sin x . 3. n and m both odd. Use either 1. or 2. use the substitutio...