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Calculus_Cheat_Sheet_All (basically all the formulas for calc 1) Helpful

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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. (cosغf ( ) ß)- f ( ) sinº f ( )ß dx dx d f( ) f( ) d 2 2. dx ( ) = f¢( )e 6. dx (tanغf ( ) ß)= f¢( )sec غf ( ) 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
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