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Calculus Cheat Sheet - Pauls Online Math Notes

Calculus Cheat Sheet Visit http://tutorial.math.lamar.edu for a complete set of Calculus notes. © 2005 Paul Dawkins Derivatives Definition and Notation

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Calculus Cheat Sheet Visit for a complete set of Calculus notes. 2005 Paul Dawkins Limits Definitions Precise Definition : We say ()limxafxL = if for every 0e> there is a 0d>such that whenever 0xad<-< then ()fxLe-<. Working Definition : We say ()limxafxL = if we can make ()fx as close to L as we want by taking x sufficiently close to a (on either side of a) without letting xa=. Right hand limit : ()limxafxL+ =. This has the same definition as the limit except it requires xa>. Left hand limit : ()limxafxL- =. This has the same definition as the limit except it requires xa<. Limit at Infinity : We say()limxfxL = if we can make ()fx as close to L as we want by taking x large enough and positive. There is a similar definition for ()limxfxL - = except we require x large and negative. Infinite Limit : We say ()limxafx = if we can make ()fx arbitrarily large (and positive) by taking x sufficiently close to a (on either side of a) without letting xa=. There is a similar definition for ()limxafx =- except we make ()fx arbitrarily large and between the limit and one-sided limits ()limxafxL = fi ()()limlimxaxafxfxL+- == ()()limlimxaxafxfxL+- == fi ()limxafxL = ()()limlimxaxafxfx+- fi ()limxafx Does Not Exist Properties Assume ()limxafx and ()limxagx both exist and c is any number then, 1. ()()limlimxaxacfxcfx = 2. ()()()()limlimlimxaxaxafxgxfxgx = 3. ()()()()limlimlimxaxaxafxgxfxgx = 4. ()()()()limlimlimxaxaxafxfxgxgx = provided ()lim0xagx 5. ()()limlimnnxaxafxfx = 6. ()()limlimnnxaxafxfx = Basic Limit Evaluations at Note : ()sgn1a= if 0a> and ()sgn1a=- if 0a<.1. limxx = e & lim0xx - =e 2. ()limlnxx = & ()0limlnxx+ =- 3. If 0r>thenlim0rxbx = 4. If 0r> and rxis real for negative x thenlim0rxbx - = 5. n even : limnxx = 6. n odd : limnxx = & limnxx - =- 7. n even : ()limsgnnxaxbxca +++= L 8. n odd : ()limsgnnxaxbxca +++= L 9. n odd : ()limsgnnxaxcxda - +++=- L Calculus Cheat Sheet Visit for a complete set of Calculus notes. 2005 Paul Dawkins Evaluation Techniques Continuous Functions If()fxis continuous at a then()()limxafxfa = Continuous Functions and Composition ()fx is continuous at b and ()limxagxb = then ()()()()()limlimxaxafgxfgxfb == Factor and Cancel ()()()2222226412limlim2268lim42xxxxxxxxx xxxx -++-=--+=== Rationalize Numerator/Denominator ()()()()()()2299299333limlim8181391limli m8139311186108xxxxxxxxxxxxxxx --+=--+--==-+++-==- Combine Rational Expressions ()()()()002001111limlim111limlimhhhhxxhh xhxhxxhhhxxhxxhx -+ -= ++ --===- ++ L Hospital s Rule If ()()0lim0xafxgx = or ()()limxafxgx = then, ()()()()limlimxaxafxfxgxgx = a is a number, or - Polynomials at Infinity ()px and ()qx are polynomials. To compute ()()limxpxqx factor largest power of x in ()qxout of both ()px and ()qx then compute limit. ()()222222445533343limlimlim52222xxxxxxx xxxxx - - - ---===----Piecewise Function ()2limxgx - where ()25if 213if 2xxgxxx +<-= - - Compute two one sided limits, ()222limlim59xxgxx-- - -=+= ()22limlim137xxgxx++ - -=-= One sided limits are different so ()2limxgx - doesn t exist. If the two one sided limits had been equal then ()2limxgx - would have existed and had the same value. Some Continuous Functions Partial list of continuous functions and the values of x for which they are Polynomials for all x. 2. Rational function, except for x s that give division by zero. 3. nx(n odd) for all x. 4. nx(n even) for all 0x . 5. xe for all x. 6. lnx for 0x>. 7. ()cosx and ()sinx for all x. 8. ()tanx and ()secx provided 33,,,,,2222xpppp --LL 9. ()cotx and ()cscx provided ,2,,0,,2,xpppp --LL Intermediate Value Theorem Suppose that ()fx is continuous on [a, b] and let M be any number between ()fa and ()fb. Then there exists a number c such that acb<< and ()fcM=. Calculus Cheat Sheet Visit for a complete set of Calculus notes. 2005 Paul Dawkins Derivatives Definition and Notation If ()yfx= then the derivative is defined to be()()()0limhfxhfxfxh +- =. If ()yfx= then all of the following are equivalent notations for the derivative. ()()()()dfdydfxyfxDfxdxdxdx ===== If()yfx=all of the following are equivalent notations for derivative evaluated at xa=. ()()xaxaxadfdyfayDfadxdx=== ==== Interpretation of the Derivative If ()yfx= then, 1. ()mfa = is the slope of the tangent line to ()yfx= at xa=and the equation of the tangent line at xa= is given by ()()()yfafaxa =+-. 2. ()fa is the instantaneous rate of change of ()fx at xa=. 3. If ()fx is the position of an object at time x then ()fa is the velocity of the object at xa=. Basic Properties and Formulas If ()fx and ()gx are differentiable functions (the derivative exists), c and n are any real numbers, 1. ()()cfcfx = 2. ()()()fgfxgx = 3. ()fgfgfg =+ Product Rule 4. 2ffgfggg -= Quotient Rule 5. ()0dcdx= 6. ()1nndxnxdx-= Power Rule 7. ()()()()()()dfgxfgxgxdx = This is the Chain Rule Common Derivatives ()1dxdx= ()sincosdxxdx= ()cossindxxdx=- ()2tansecdxxdx= ()secsectandxxxdx= ()csccsccotdxxxdx=- ()2cotcscdxxdx=- ()121sin1dxdxx-=- ()121cos1dxdxx-=-- ()121tan1dxdxx-=+ ()()lnxxdaaadx=()xxddx=ee ()()1ln,0dxxdxx=>()1ln,0dxxdxx= ()()1log,0lnadxxdxxa=> Calculus Cheat Sheet Visit for a complete set of Calculus notes. 2005 Paul Dawkins Chain Rule Variants The chain rule applied to some specific functions. 1. ()()()()1nndfxnfxfxdx- = 2. ()()()()fxfxdfxdx =ee 3. ()()()()lnfxdfxdxfx = 4. ()()()()sincosdfxfxfxdx = 5. ()()()()cossindfxfxfxdx =- 6. ()()()()2tansecdfxfxfxdx = 7. []()[][]()()()()secsectanfxfxfxfxddx = 8. ()()()()12tan1fxdfxdxfx- = + Higher Order Derivatives The Second Derivative is denoted as ()()()222dffxfxdx == and is defined as ()()()fxfx =, the derivative of the first derivative,()fx . The nth Derivative is denoted as ()()nnndffxdx= and is defined as ()()()()()1nnfxfx- =, the derivative of the (n-1)st derivative,()()1nfx-. Implicit Differentiation Find y if ()2932sin11xyxyyx-+=+e. Remember()yyx= 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 . ()()()()()()2922329222929223329329292229 32cos1111232932cos1129cos29cos1123xyxyxy xyxyxyxyyxyxyyyyxyyxyxyyyyyxyyxyyyxy---- --- -++=+-- -++=+fi=-- --=--eeeeeee Increasing/Decreasing Concave Up/Concave Down Critical Points xc= is a critical point of ()fx provided either 1. ()0fc = or 2. ()fc doesn t exist. Increasing/Decreasing 1. If ()0fx > for all x in an interval I then ()fx is increasing on the interval I. 2. If ()0fx < for all x in an interval I then ()fx is decreasing on the interval I. 3. If ()0fx = for all x in an interval I then ()fx is constant on the interval I. Concave Up/Concave Down 1. If ()0fx > for all x in an interval I then ()fx is concave up on the interval I. 2. If ()0fx < for all x in an interval I then ()fx is concave down on the interval I. Inflection Points xc= is a inflection point of ()fx if the concavity changes at xc=. Calculus Cheat Sheet Visit for a complete set of Calculus notes. 2005 Paul Dawkins Extrema Absolute Extrema 1. xc=is an absolute maximum of()fx if()()fcfx for all x in the domain. 2. xc= is an absolute minimum of()fx if()()fcfx for all x in the domain. Fermat s Theorem If ()fx has a relative (or local) extrema at xc=, then xc= is a critical point of ()fx. Extreme Value Theorem If ()fx is continuous on the closed interval [],ab then there exist numbers c and d so that, 1. ,acdb , 2. ()fc is the abs. max. in [],ab, 3. ()fd is the abs. min. in [],ab. Finding Absolute Extrema To find the absolute extrema of the continuous function ()fx on the interval [],ab use the following process. 1. Find all critical points of ()fx in [],ab. 2. Evaluate ()fx at all points found in Step 1. 3. Evaluate ()fa and ()fb. 4. Identify the abs. max. (largest function value) and the abs. min.(smallest function value) from the evaluations in Steps 2 & 3. Relative (local) Extrema 1. xc= is a relative (or local) maximum of ()fx if()()fcfx for all x near c. 2. xc= is a relative (or local) minimum of ()fx if()()fcfx for all x near c. 1st Derivative Test If xc= is a critical point of ()fx then xc= is 1. a rel. max. of()fx if()0fx > to the left of xc= and()0fx < to the right of xc=. 2. a rel. min. of()fx if()0fx < to the left ofxc=and()0fx >to the right of xc=. 3. not a relative extrema of()fx if()fx is the same sign on both sides of xc=. 2nd Derivative Test If xc= is a critical point of ()fx such that ()0fc = then xc= 1. is a relative maximum of()fx if()0fc <. 2. is a relative minimum of()fx if()0fc >. 3. may be a relative maximum, relative minimum, or neither if ()0fc =. Finding Relative Extrema and/or Classify Critical Points 1. Find all critical points of ()fx. 2. Use the 1st derivative test or the 2nd derivative test on each critical point. Mean Value Theorem If ()fx is continuous on the closed interval [],ab and differentiable on the open interval (),ab then there is a number acb<< such that ()()()fbfafcba- =-. Newton s Method If nx is the nth guess for the root/solution of()0fx= then (n+1)st guess is ()()1nnnnfxxxfx+=- provided ()nfx exists. Calculus Cheat Sheet Visit for a complete set of Calculus notes. 2005 Paul Dawkins Related Rates Sketch picture and identify known/unknown quantities. Write down equation relating quantities and differentiate with respect to t using implicit differentiation ( 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. The bottom is initially 10 ft away and is being pushed towards the wall at 14ft/sec. How fast is the top moving after 12 sec? x is negative because x is decreasing. Using Pythagorean Theorem and differentiating, 22215220xyxxyy +=fi+= After 12 sec we have ()1410127x=-=and so 22157176y=-=. Plug in and solve for y . ()14771760 ft/sec4176yy -+=fi= Ex. Two people are 50 ft apart when one starts walking north. The angleq changes at rad/min. At what rate is the distance between them changing when rad? We have = rad/min. and want to find x . We can use various trig fcns but easiest is, secsectan5050xxqqqq =fi= We so plug in q and solve. ()()() ft/secxx = = 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 500 ft of fence material and one side of the field is a building. Determine dimensions that will maximize the enclosed area. Maximize Axy= subject to constraint of 2500xy+=. Solve constraint for x and plug into area. ()2500250025002Ayyxyyy=-=-fi=- Differentiate and find critical point(s). 5004125Ayy =-fi= By 2nd deriv. test this is a rel. max. and so is the answer we re after. Finally, find x. ()5002125250x=-= The dimensions are then 250 x 125. Ex. Determine point(s) on 21yx=+ that are closest to (0,2). Minimize ()()22202fdxy==-+- and the constraint is 21yx=+. Solve constraint for 2x and plug into the function. ()()22222121233xyfxyyyyy=-fi=+-=-+-=-+Differentiate and find critical point(s). 3223fyy =-fi= By the 2nd derivative test this is a rel. min. and so all we need to do is find x value(s). 23112221xx=-=fi= The 2 points are then ()3122, and ()3122,-. Calculus Cheat Sheet Visit for a complete set of Calculus notes. 2005 Paul Dawkins Integrals Definitions Definite Integral: Suppose ()fx is continuous on [],ab. Divide [],ab into n subintervals of width xD and choose *ix from each interval. Then ()()*1limibanifxdxfxx = =D . Anti-Derivative : An anti-derivative of ()fx is a function, ()Fx, such that ()()Fxfx =. Indefinite Integral :()()fxdxFxc=+ where ()Fx is an anti-derivative of ()fx. Fundamental Theorem of Calculus Part I : If ()fx is continuous on [],ab then ()()xagxftdt= is also continuous on [],ab and ()()()xadgxftdtfxdx == . Part II : ()fxis continuous on[],ab, ()Fx is an anti-derivative of()fx( ()()Fxfxdx= ) then()()()bafxdxFbFa=- .Variants of Part I : ()()()()uxadftdtuxfuxdx = ()()()()bvxdftdtvxfvxdx =- ()()()()[]()[]()()uxvxuxvxdftdtuxfvxfdx =- Properties ()()()()fxgxdxfxdxgxdx = ()()()()bbbaaafxgxdxfxdxgxdx = ()0aafxdx= ()()baabfxdxfxdx=- ()()cfxdxcfxdx= , c is a constant ()()bbaacfxdxcfxdx= , c is a constant ()bacdxcba=- ()()bbaafxdxfxdx ()()()bcbaacfxdxfxdxfxdx=+ for any value of c. If ()()fxgx onaxb then ()()bbaafxdxgxdx If()0fx on axb then ()0bafxdx If ()mfxM on axb then ()()()bambafxdxMba- - Common Integrals kdxkxc=+ 111,1nnnxdxxcn++=+ - 11lnxxdxdxxc-==+ 11lnaaxbdxaxbc+=++ ()lnlnuduuuuc=-+ uuduc=+ ee cossinuduuc=+ sincosuduuc=-+ 2sectanuduuc=+ sectansecuuduuc=+ csccotcscuuduuc=-+ 2csccotuduuc=-+ tanlnsecuduuc=+ seclnsectanuduuuc=++ ()11122tanuaaauduc-+=+ ()1221sinuaauduc--=+ Calculus Cheat Sheet Visit for a complete set of Calculus notes. 2005 Paul Dawkins Standard Integration Techniques Note that at many schools all but the Substitution Rule tend to be taught in a Calculus II class. u Substitution : The substitution ()ugx=will convert ()()()()()()bgbagafgxgxdxfudu = using ()dugxdx =. For indefinite integrals drop the limits of integration. Ex. ()23215cosxxdx 322133uxduxdxxdxdu=fi=fi= 33111::228xuxu=fi===fi== ()()()()()()232853118553315coscossinsin8sin1xxdxuduu===- Integration by Parts : udvuvvdu=- and bbbaaaudvuvvdu=- . Choose u and dv from integral and compute du by differentiating u and compute v using vdv= . Ex. xxdx- e xxuxdvdudxv--==fi==-ee xxxxxxdxxdxxc-----=-+=--+ eeeee Ex. 53lnxdx 1lnxuxdvdxdudxvx==fi== ()()()()55553333lnlnln5ln53ln32xdxxxdxxxx=-=-=-- Products and (some) Quotients of Trig Functions For sincosnmxxdx we have the following : 1. n odd. Strip 1 sine out and convert rest to cosines using 22sin1cosxx=-, then use the substitution cosux=. 2. m odd. Strip 1 cosine out and convert rest to sines using 22cos1sinxx=-, then use the substitution sinux=. 3. n and m both odd. Use either 1. or 2. 4. n and m both even. Use double angle and/or half angle formulas to reduce the integral into a form that can be integrated. Fortansecnmxxdx we have the following : 1. n odd. Strip 1 tangent and 1 secant out and convert the rest to secants using 22tansec1xx=-, then use the substitution secux=. 2. m even. Strip 2 secants out and convert rest to tangents using 22sec1tanxx=+, then use the substitution tanux=. 3. n odd and m even. Use either 1. or 2. 4. n even and m odd. Each integral will be dealt with differently. Trig Formulas : ()()()sin22sincosxxx=, ()()()212cos1cos2xx=+, ()()()212sin1cos2xx=- Ex. 35tansecxxdx ()()()35242424751175tansectansectansecsec1sectansec1secsecsecxxdxxxxxdxxxxxdxuuduuxxxc==-=-==-+ Ex. 53sincosxxdx ()2211222254333223222433sin(sin)sinsinsincoscoscossin(1cos)cos(1)12cossec2lncoscosxxxxxxxxxxxuuuuudxdxdxdxuxduduxxxc---+=====-=-=+-+ Calculus Cheat Sheet Visit for a complete set of Calculus notes. 2005 Paul Dawkins Trig Substitutions : If the integral contains the following root use the given substitution and formula to convert into an integral involving trig functions. 222sinababxxq-fi= 22cos1sinqq=- 222secabbxaxq-fi= 22tansec1qq=- 222tanababxxq+fi= 22sec1tanqq=+ Ex. 221649xxdx- 2233sincosxdxdqqq=fi= 22244sin4cos2cos49xqqq=-==- Recall 2xx=. Because we have an indefinite 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, if 0if 0xxxxx = -< In this case we have 22cos49xq=-. ()()23sin2cos222491612sincos12csc12cotdddcqqqqqqqq===-+ Use Right Triangle Trig to go back to x s. From substitution we have 32sinxq= so, From this we see that 2493cotxxq-=. So, 2221644949xxxxdxc--=-+ Partial Fractions : If integrating ()()PxQxdx where the degree of ()Px is smaller than the degree of ()Qx. Factor denominator as completely as possible and find the partial fraction decomposition of the rational expression. Integrate the partial fraction decomposition ( ). For each factor in the denominator we get term(s) in the decomposition according to the following table. Factor in ()Qx Term in Factor in ()Qx Term in axb+ Aaxb+ ()kaxb+ ()()122kkAAAaxbaxbaxb++++++L 2axbxc++ 2AxBaxbxc+++ ()2kaxbxc++ ()1122kkkAxBAxBaxbxcaxbxc++++++++L Ex. 2()()214713xxxxdx-++ ()()2222()()2132223164114431641447134ln1ln48tanxxxxxxxxxxxxdxdxdxxx-+--++-+++=+=++=-+++ Here is partial fraction form and recombined. 22224)()()()()()()21114414(713BxCxxxxxxxAxBxCAxx+++---++-+++=+= Set numerators equal and collect like terms. ()()227134xxABxCBxAC+=++-+- Set coefficients equal to get a system and solve to get constants. 713404316ABCBACABC+=-=-==== An alternate method that sometimes works to find constants. Start with setting numerators equal in previous example : ()()()2271341xxAxBxCx+=+++-. Chose nice values of x and plug in. For example if 1x= we get 205A= which gives 4A=. This won t always work easily. Calculus Cheat Sheet Visit for a complete set of Calculus notes. 2005 Paul Dawkins Applications of Integrals Net Area : ()bafxdx represents the net area between ()fx and the x-axis with area above x-axis positive and area below x-axis negative. Area Between Curves : The general formulas for the two main cases for each are, ()upper functionlower functionbayfxAdx =fi=- & ()right functionleft functiondcxfyAdy =fi=- 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. ()()baAfxgxdx=- ()()dcAfygydy=- ()()()()cbacAfxgxdxgxfxdx=-+- Volumes of Revolution : The two main formulas are ()VAxdx= and ()VAydy= . Here is some general information about each method of computing and some examples. Rings Cylinders ()()()22outer radiusinner radiusAp=- ()()radiuswidth / height2Ap= Limits: x/y of right/bot ring to x/y of left/top ring Limits : x/y of inner cyl. to x/y of outer cyl. Horz. Axis use()fx, ()gx,()Ax and dx. Vert. Axis use()fy, ()gy,()Ay and dy. Horz. Axis use()fy, ()gy,()Ay and dy. Vert. Axis use()fx, ()gx,()Ax and dx. Ex. Axis : 0ya=> Ex. Axis : 0ya= Ex. Axis : 0ya=> Ex. Axis : 0ya= outer radius :()afx- inner radius : ()agx- outer radius:()agx+ inner radius:()afx+ radius :ay- width : ()()fygy- radius :ay+ width : ()()fygy- These are only a few cases for horizontal axis of rotation. If axis of rotation is the x-axis use the 0ya= case with 0a=. For vertical axis of rotation (0xa=> and 0xa= ) interchange x and y to get appropriate formulas. Calculus Cheat Sheet Visit for a complete set of Calculus notes. 2005 Paul Dawkins Work : If a force of()Fxmoves an object inaxb , the work done is ()baWFxdx= Average Function Value : The average value of ()fx on axb is()1bavgabaffxdx-= Arc Length Surface Area : Note that this is often a Calc II topic. The three basic formulas are, baLds= 2baSAydsp= (rotate about x-axis) 2baSAxdsp= (rotate about y-axis) where ds is dependent upon the form of the function being worked with as follows. ()()21 if ,dydxdsdxyfxaxb=+= ()()21 if ,dxdydsdyxfyayb=+= ()()()()22 if ,,dydxdtdtdsdtxftygtatb=+== ()()22 if ,drddsrdrfabqqqq=+= 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 1. ()()limtaatfxdxfxdx = 2. ()()limbbttfxdxfxdx- - = 3. ()()()ccfxdxfxdxfxdx-- =+ provided BOTH integrals are convergent. Discontinuous Integrand 1. Discont. at a:()()limbbattafxdxfxdx+ = 2. Discont. at b :()()limbtaatbfxdxfxdx- = 3. Discontinuity at acb<< : ()()()bcbaacfxdxfxdxfxdx=+ provided both are convergent. Comparison Test for Improper Integrals : If ()()0fxgx on [),a then, 1. If()afxdx conv. then()agxdx conv. 2. If()agxdx divg. then()afxdx divg. Useful fact : If 0a> then 1apxdx converges if 1p> and diverges for 1p . Approximating Definite Integrals For given integral ()bafxdx and a n (must be even for Simpson s Rule) define banx-D= and divide [],ab into n subintervals []01,xx, []12,xx, ... , []1,nnxx- with 0xa= and nxb= then, Midpoint Rule : ()()()()***12bnafxdxxfxfxfx D+++ L, *ix is midpoint []1,iixx- Trapezoid Rule : ()()()()()()01212222bnnaxfxdxfxfxfxfxfx- D ++++++ L Simpson s Rule : ()()()()()()()0122142243bnnnaxfxdxfxfxfx fxfxfx--D ++++++ L

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