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Calculus II

by: Luz Huels

Calculus II MATH 2644

Luz Huels

GPA 3.71

William Faucette

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William Faucette
Class Notes
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This 8 page Class Notes was uploaded by Luz Huels on Tuesday October 27, 2015. The Class Notes belongs to MATH 2644 at University of West Georgia taught by William Faucette in Fall. Since its upload, it has received 16 views. For similar materials see /class/230222/math-2644-university-of-west-georgia in Mathematics (M) at University of West Georgia.

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Date Created: 10/27/15
Derivatives to Know d dx a sinx cosx xn nxnil cosx sinx dx EtanOc seczx E cotx 0502 x secx secxtanx dx cscx cscx cotx dx d sin 1 x 1 dx xl x2 d 1 1 Wm 2 d t 1 iix an x 1 1 651x cot x l xz 1 a sec x x2 1 d 1 1 Ecsc x x x2 1 d 1 1 dx 110C x d 1 1 dx 0g x xlna d x x dxe e C xax axlna sinhx coshx 651x coshx sinhx dx d tanhx sech2x 651x cothx csch2x 651x sechx sechx tanhx dx Hyperbolic Trigonometric Functions Just as trigonometry can be performed on the unit circle it can also be performed on the unit hyperbola x 7 y2 1 De ne the hyperbolic cosine function by x 7 coshx and the hyperbolic sine function by ex 7 e71 nh s1 x 2 A simple computation then shows that cosh2 x 7 sinh2 x l which is the hyperbolic trigonometric identity analogous to the Pythagorean identity cos2x sin2x 1 from ordinary circular trigonometry Next as in the case of ordinary trigonometry we de ne the remaining four hyperbolic trigonometric functions from these rst two De ne the hyperbolic Iangenl mcz ion by sinhx t nh a x coshx and the hyperbolic cotangenl mcz ion by cothoc 7 coshx 7 sinhx 39 De ne the hyperbolic secanl function by l sechx m and the hyperbolic cosecanl mcz ion by l h csc x sinhx Hyperbolic Pythagorean Identities In addition to the identity derived earlier cosh2 x 7 sinh2 x 1 there are two other hyperbolic identities obtained by dividing this one by sinh2 x and cosh2 x respectively cosh2 x 7 sinh2 x l cosh2 x sinh2 x 7 l sinh2 x sinh2 x 7 sinh2 x ltcoshxgt2 17 l sinhx 7 sinh2 x coth2 x 7 l csch2 x l cosh2 x 7 sinh2 x l cosh2 x sinh2 x 7 l cosh2 x 7 cosh2 x 7 cosh2 x 17 sinhxgt2 7 1 coshx cosh2 x l 7 tanh2 x sech2 x We summarize these here cosh2 x 7 sinh2 x l l 7 tanh2 x sech2 x coth2 x 7 l csch2 x Derivatives of Hyperbolic Functions By using the de nition of the hyperbolic sine function we have d d equot7e C equote C smhx 7 lt 2 gt 7 2 7 coshx Sim ilarly 2 2 The derivatives of the remaining hyperbolic trigonometric functions are obtained from their de nitions using the Pythagorean hyperbolic trigonometric identities the de nitions of the hyperbolic functions and the quotient rule d d sinhx t nh dx a x dx coshx 7 7 sinhx coshx 7 sinhx 7 coshx d d e C e C e C 7 e C EQIOShOC 7 lt gt 7 7 Sinhx cosh2 x 7 coshx coshx 7 sinhx sinhx 7 cosh2 x 7 cosh2 x 7 sinh2 x 7 cosh2 x 7 cosh2 x sech2 x Similarly d d cosh x E mthoc sinhx 7 coshx sinhx 7 coshx 7 sinhx 7 sinh2 x 7 sinhx sinhx 7 coshx coshx 7 sinh2 x 7 sinh2 x 7 cosh2 x 7 sinh2 x 7 7l 7 sinh2 x 7 csch2 x The remaining two hyperbolic trigonometric functions are handled similarly d h 7 d 1 Esec 35 i g 7 cosh2 x 7 cosh2 x 7 cosh2 x 7 fl sinhx 7 ltcoshxgt ltcoshxgt 7 sechx tanhx and d h 7 d 1 E6350 35 7 7 sinh2 x 7 fl coshx smhltxgtgt sinhltxgtgt 7 cschx cothx Inverse Hyperbolic Trigonometric Functions The four functions sinh 1x y tanh 1x y coth 1x y csch 1 x are onetoone on their domains While the functions cosh 1x y sech 1 x are onetoone on the interval x 3 0 So we can de ne six inverse hyperbolic trigonometric functions Since the hyperbolic functions are de ned in terms of the exponential function it is only natural to expect that the inverse hyperbolic functions can be de ned in terms of the natual logarithm function We derive these six formulas next Logarithmic form for sinh 1 x Let y sinh 1 x Then x sinh so ey 6 x 2 Clearing fractions and moving everything to one side we get ey 7 2x 7 67y 0 and multiplying by ey we get 6 7 2xey 7l 0 This is a quadratic equation in ey so we may solve it by the quadratic formula y72xj 72x274l7l 72xix4x2472xj2 x2l 72xivx21 7 2 7 2 7 2 7 2 e x j x2 1 Since ey must be positive we can discard the negative square root here to get ey xx2 1 so sinh71x ln x x2 1 Logarithmic form for cosh71 x x 3 1 Let y cosh71 x Then x coshy so 7 ey 67y 7 2 Clearing fractions and moving everything to one side we get ey 7 2x 67y 0 and multiplying by ey we get 6 7 2xey 1 0 This is a quadratic equation in ey so we may solve it by the quadratic formula 6y 7 in 72x274ll 7 inx4x274 7 2xi2xx27l 7 2xivx217x7 x2 1 7 2 7 2 7 2 7 2 7 7 Since ey must be positive we can discard the negative square root here to get eyxx27l cosh71 x ln x x2 7l Logarithmic form for tanh71 x Let y tanh71 x Then x tanhO 4 ey 67y feye y39 Clearing fractions and moving everything to one side we get xeye yeyie y xezy1ezy71 xe2yxezy71 1xe2y7xezy 1xezy17x 1x 17x 1x 2 1 y HM 711n1x y z 17x 1 1 tanh 1 x 5 1n ezy 7x Logarithmic form for coth 1 x Let y coth 1 x Then x cothy so 7 ey e y 7 ey 7 eiy39 Clearing fractions and moving everything to one side we get xey7e yeye y xezy71ezy1 xezyixezy1 x1xezy7e2y x1e2yx71 x1 xil 2y 1nltgt x 711x1 y zn x71 coth 1 x l1n 2 x e Logarithmic form for sech 1 x y sech 1 x x sechy 7 l 7 2 7 costh 7 e e y39 Clearing fractions and moving everything to one side we get xeye y720 xe2y172ey 0 xezy 72ey x 0 This is a quadratic equation in ey so we can solve this using the quadratic formula y 2 i 32 7 40006 e 2x 7 2 i v 22 7 40006 7 2x 7 2 i m 2x 2 j Zxl 7 x2 2x 2 lt1 i 1 7 x2 2x 1 j l 7 x2 x Since ey must be positive we must have 6 7 1 W x sech 1 x ln Taking the natural logarithm we get Logarithmic form for csch 1 x y csch 1 x x cschy 7 l 7 2 7 siJith 7 e 7 e y39 Clearing fractions and moving everything to one side we get xey7e y720 xe2y7l726y0 xezy 72ey 7x 0 6 Integrals to Know xn1 n 1 sinx dx cosx C xquotdx Cf0rn7E 1 cosx dx sinx C sec2x dx tanx C csc2x dx cotx C secxtanx dx secx C cscx cotx dx cscx C r dx sin 1x C m dx cos 1x C 11x2 dx tan 1x C 1 x2dx cot 1x C 1 h dx sec 1x C dx csc 1x C dx 1n x C ex dx 2 ex C a dx i sinhx dx coshx C coshx dx sinhx C sech2x dx tanhx C csch2 x dx cothx C sechx tanhx dx sechx C


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