Last Name 1 Name Professor Subject Date Calculus Question No. 1 Given ???? = ???????????? (???????????? −????(???? ????????))find ???? ′ ???????????????? ???????? ???????? ???????? . And simplify your answer. Solution ???? = ????????????(???????????? −1(???? 4????)) Using chain rule ???????? ???????? = ???????? ???????? . ???????? ???????? … . . … . . .. (1) Consider ???? = (???????????? −1(???? 4????)) ???????????? ???? = ????????????(????) ???????? ???????? = ???? ???????? (????????????(????)) = sec2(????) … … .. (2) Substituting back the value of “u” in equation 2 gives ???????? ???????? = sec2 (( ???????????? ???????????? (???? 4????))) … …… (3) ???????? ???????? = ???? ???????? ((???????????? ???????????? (???? 4????))) Using product rule in equation 1 = ???? ???????? (arccos(???? 4????)) ???? ???????? (???? 4???? ) As we know that ???? ???????? arccos(????) = (− 1 √1−(????) 2 ) Last Name 2 ???????? ???????? = (− 1 √1 − (???? 4????) 2 ) × 4(???? 4????) = − 4???? 4???? √1 − (???? 4????) 2 …… (4) Substituting equation 3 and 4 in equation 1 gives ???? ′ = ???????? ???????? = sec2 (( ???????????? ???????????? (???? 4????))) (− 4???? 4???? √1 − (???? 4????) 2 ) … … . . (5) Simplification: Using the mathematical rule sec (( ???????????? ???????????? (????))) = 1 ???? in equation 5 gives sec2 (( ???????????? ???????????? (???? 4????))) = ( 1 ???? 4???? ) 2 Simplified form becomes ???? ′ = ( 1 ???? 4???? ) 2 (− 4???? 4???? √1 − (???? 4????) 2 ) ???? ′ = − 4 ???? 4????√1 − (???? 8????) Question No. 2 Evaluate the following ∫ ???????????????? ???????????????? + ???????????? ???? ???? ???? ???? ???????? = ???? ???? Solution Let ???? = ∫ ???????????? ???? ???????????????? + ???????????? ???? ???? 2 0 ???????? … … … …(1) Multiplying dominator and denominator with factor (sin ???? − cos ????) in equation 1. Last Name 3 ???? = ∫ ???????????? ???? (sin ???? − cos ????) (???????????????? + cos ????)(sin ???? − cos ????) ???? 2 0 ???????? = ∫ ???????????? ???? (sin ???? − cos ????) (sin2 ???? − cos2 ????) ???? 2 0 ???????? As we know that (sin2 ???? − cos2 ????) = cos (2????) It becomes (cos2 ???? − sin2 ????) = − cos(2????) … … … … (2) ???? = ∫ ???????????? ???? (sin ???? − cos ????) (sin2 ???? − cos2 ????) ???? 2 0 ???????? = ∫ − ???????????? ???? cos ???? + sin2 ???? (sin2 ???? − cos2 ????) ???? 2 0 ???????? … … … (3) Inserting equation 2 in equation 3 ???? = ∫ [ − ???????????? ???? cos ???? − cos(2????) + sin2 ???? cos(2????) ] ???? 2 0 ???????? = ∫ [ ???????????? ???? cos ???? cos (2????) + sin2 ???? cos (2????) ] ???? 2 0 ???????? … … …… (4) ???? = ????1 + ????2 Inserting equation 5 in equation 4 ???????????? ???? cos ???? = sin (2????) 2 … … … … (5) ????1 = ∫ [ ???????????? ???? cos ???? cos (2????) ] ???? 2 0 ???????? = ∫ [ sin (2????) 2 cos (2????) ] ???? 2 0 ???????? Last Name 4 ????1 = 1 2 ∫ [ sin (2????) cos (2????) ] ???? 2 0 ???????? ???????????????????????????????????????????????? ???? = cos(2????) ????1 = 1 2 ∫ − 1 2???? ???????? = − 1 4 ∫ 1 ???? ???????? = − 1 4 ln|????| … … … (6) Substituting back in equation 6 ????1 = − 1 4 ln|cos(2????)| … … … (6) ????2 = ∫ [ sin2 ???? cos (2????) ] ???? 2 0 ???????? … … … … (7) Using the following identity sin2 ???? = 1 − 2 cos (2????) 2 Equation 7 becomes ????2 = ∫ [ 1− 2 cos (2????) 2 cos(2????) ] ???? 2 0 ???????? ????2 = 1 2 ∫ [ 1 − 2 cos (2????) cos (2????) ] ???? 2 0 ???????? = 1 2 ∫ [ 1 cos (2????) − cos (2????) cos (2????) ] ???? 2 0 ???????? ????2 = 1 2 ∫ [ 1 cos (2????) − 1] ???? 2 0 ???????? ????2 = 1 2 [∫ [ 1 cos(2????) ] ???????? − ∫[1] ????????] Last Name 5 ????2 = 1 2 [( 1 2 ln|tan(2????)| + sec(2????)) − ????]… … . … (8) Using equation 6 and equation 8 ???? = − 1 4 ln|cos(2????)| + 1 2 [( 1 2 ln|tan(2????)| + sec(2????)) − ????] Computing the boundaries lim ???? → 0 + (− 1 4 ln|cos(2????)| + 1 2 [( 1 2 ln|tan(2????)| + sec(2????)) − ????]) = 0 lim ???? → ???? 2 − (− 1 4 ln|cos(2????)| + 1 2 [( 1 2 ln|tan(2????)| + sec(2????)) − ????]) = ???? 4 ???? = ???? 4 − 0 = ???? 4 Hence proved ∫ ???????????? ???? ???????????????? + ???????????? ???? ???? 2 0 ???????? = ???? 4 Question No. 3 Evaluate the following limit. Use limit theorem not ???? − ???? techniques. If any of them fail to exist, say so and say why. ???????????????? → ???? + √???? − ???? ???? ???? − ???? Solution First simplifying the equation by factoring ???? 2 − 1 = (???? + 1)(???? − 1) and inserting in the equation. √???? − 1 ???? 2 − 1 = √???? − 1 (???? + 1)(???? − 1) = 1 (???? + 1)√???? − 1 Using x approaching to 1 from the right side the condition becomes Last Name 6 ???? > 1 → (???? + 1)√
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