f ) AVERAGE RATE OF CHANGE p. 87 DEFINITION OF DERIVATIVE p. 99
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1 AVERAGE RATE OF CHANGE p. 87 The verge rte of chnge of fnction over n intervl is the mont of chnge ivie by the length of the intervl. DEFINITION OF DERIVATIVE p. 99 f ( h) f () f () lim h0 h Averge rte of chnge of f over the intervl [, b] is (Note: instntneos rte of chnge of f is f ) f ( b) f ( ) b Derivtive fils to eist: 1. corner. csp 3. verticl tngent 4. iscontinity i. removble (hole) ii. bre (jmp) iii. verticl symptote iv. oscillting AVERAGE RATE OF CHANGE PRODUCT/QUOTIENT/CHAIN RULES p. 119/p. 10 PRODUCT RULE p. 119 ( v ) v v QUOTIENT RULE p. 10 v v v v CHAIN RULE p. 149 If f ( ) g( h( )), then f ( ) g( h( )) h( ) In wors: The erivtive of the otsie evlte t the insie, times the erivtive of the insie Hey! PRODUCT/QUOTIENT CHAIN RULES DEFINITION OF DERIVATIVE DERIVATIVES OF FUNCTIONS FORMULAS p Derivtive Formls ( is fnction of ) n ( n1) n e e ln 1 ln sin cos tn sec sec sec tn DERIVATIVES OF FUNCTIONS 1 log ln cossin cot csc csc csccot
2 INVERSE TRIG DERIVATION FORMUALS p sin cos tn cot sec 1 csc 1 1 INVERSE TRIG DERIVATION 1 MEAN VALUE THEOREM p. 196 If y f ( ) is continos on [, b] n ifferentible on (, b) t every point in its interior, then there is t lest one vle, c, t which: Inst. rte of chnge f() c f ( b) f ( ) b 1. Fin the slope from = to = b. Fin erivtive of fnction t c 3. Set nswers to #1 n # eql n solve for c MEAN VALUE THEOREM v. rte of chnge IMPLICIT DIFFERENTIATION p. 157 Use when eqtion is ifficlt or impossible to solve for y. 1. Differentite both sies of the eqtion with respect to.. Collect the terms with y 3. Fctor ot y. 4. Solve for y. on one sie. FIRST DERIVATIVE TEST p. 05 IMPLICIT DIFFERENTIATION y f ( ) is continos fnction Use sign chrt; smmrize in complete sentence*: (*reqire on AP em for creit) 1. If f chnges sign from positive to negtive t c, then f hs locl mimm vle of f() c t = c.. If f chnges sign from negtive to positive t c, then f hs locl minimm vle of f() c t = c. 3. If f oes not chnge sign t criticl point c, then f hs no locl etreme vles t c. 4. If f0 ( f0) for where is left enpoint in the omin of f, then f hs locl mimm (minimm) vle, f() t. 5. If f0 ( f0) for bwhere b is right enpoint in the omin of f, then f hs locl minimm (mimm) vle, f(b), t b. Note: Sign chrt of f is lso se to fin where f is incresing or ecresing incle enpoints in inc./ec. intervls. FIRST DERIVATIVE TEST
3 SECOND DERIVATIVE SIGN CHART p If secon erivtive of f is positive, then f is concve p. If secon erivtive of f is negtive, then f is concve own.. Inflection points occr when concvity chnges. SECOND DERIVATIVE TEST p If f( c) 0n f( c) 0, then f hs locl mimm of f(c), t = c.. If f( c) 0n f( c) 0, then f hs locl minimm of f(c), t = c. OPTIMIZATION p Write n eqtion for the fnction to be minimize or mimize.. Te the erivtive of this fnction. 3. Me sign chrt of the erivtive (incling enpoints). 4. Write complete sentence smmrizing reslts. 5. Fin fnction vle incling nits which gives optimiztion. RELATED RATES p Write n eqtion relting the vribles.. Differentite both sies with respect to time. 3. Sbstitte vles for the prticlr moment in time. SECOND DERIVATIVE TEST OPTIMIZATION AVERAGE VALUE OF A FUNCTION p. 87 If f is integrble on [, b], its verge vle on [, b] is: 1 b v( f ) f ( ) b 4. Solve for the rte in qestion. 5. Answer in complete sentence with nits RELATED RATES AVERAGE VALUE OF A FUNCTION
4 FUNDAMENTAL THEOREM OF CALCULUS p. 94 FTC PART I DERIVATIVE OF AN INTEGRAL If f is continos on [, b], then the fnction, F( ) f ( t) t hs erivtive t every point in [, b]: F f ( t) t f ( ) FUNDAMENTAL THEOREM OF CALCULUS p. 99 FTC PART II EVALUATION OF AN INTEGRAL If f is continos t every point of [, b], n if F is ny ntierivtive of f on [, b], then: b f ( ) F ( b ) F ( b ) Note: Lower limit is ny nmber. If pper limit is fnction of, then pply the chin rle. FTC PART I INTEGRAL FORMULAS AND PROPERTIES P. 33 Properties of Inefinite Integrls f ( ) f ( ) f g f g ( ( ) ( )) ( ) ( ) Inefinite Integrl Formls n1 n C, n 1 n 1 1 ln C ln ln C cos sin C sin cos C sec tn C e e C csc cot C C sec tn sec C ln csc cot csc C INTEGRAL FORMULAS/ PROP. FTC PART II INTEGRATION BY SUBSTITUTION METHOD p. 333 Metho is se to trn n nfmilir integrl into one we cn evlte. Integrl is chnge into one of the integrl on p. 33. Let = =. Sbstitte so integrl is in terms of. Don t forget to chnge the limits for to limits for. SUBSTITUTION METHOD
5 SEPARATION OF VARIABLES p. 350 Use to solve ifferentil eqtions. 1. Seprte the vribles.. Integrte both sies. Don t forget the + C. 3. Solve for y. 4. Use the initil vle for fin C. 5. Write soltion sing the C vle. SEPARATION OF VARIABLES
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