Find the Derivative
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Calculus BC Sections 5/6 – 5/8 Review Worksheet answers
Find the derivative. 1 = sinh x cosh x 1. y = arcsin(4x) sinh x dy 1 d(4x) = = cosh x dx 1 (4x) 2 dx = tanh x 4 = 6. y = sinh x ∙ ecosh x 116x 2 dy cosh x 3 = sinh x ∙ (e ∙sinh x) 2. y = arctan dx x + ecosh x ∙cosh x 1 d(3x 1 ) = sinh2x ∙ ecosh x + cosh x ∙ ecosh x cosh x 2 dy 2 = e (sinh x + cosh x) = 3 dx dx 1 x 7. y = 2x∙arctan x – ln(x2 + 1) 1 3 dy 1 1 2 = 2x arctan x 2 2x = 9 x dx x 2 1 x 2 1 1 2 x 2x 2x 3 = 2 2 arctan x 2 x 1 x 1 = 2 x 9 = 2∙arctan x
3. y = arcsec(2x+1) Find the indefinite integral. 1 d(2x 1) dy = 2 dx 1 dx 2x 1 2x 1 1 dx 8. I = 2 1 4 9x = 2 1 2x 1 4x 2 4x dx = 2 2 2 (2) (3x) = 3x = 2 2x 1 2 x 2 x 3∙dx = 2d 1 dx = 2/3d = 2 1 2 2x 1 x x dx I = 2 2 (2) (2 ) 3 2 2 4. y = sinh x + cosh x 2 1 dx dy = 2 = 2∙sinhx∙coshx + 2∙coshx∙sinhx 3 4 4 dx 2 1 1 = 4∙sinh x∙cosh x dx = 2 = 2(2∙sinh x∙cosh x) 3 2 1 = 2∙sinh(2x) 1 = ∙arcsin() + C 3 5. y = ln(cosh x) Since 3x = 2, = 3x/2 dy 1 d(cosh x) = 1 3x dx cosh x dx I = = ∙arcsin + C 3 2 Calculus BC Sections 5/6 – 5/8 Review Worksheet answers
1 1 9. I = dx 11. I = dx 25x 2 4 x 2 4x 13 1 Completing the square: = dx 2 2 (5x)2 22 x + 4x + 13 = (x + 4x ) + 13 = (x2 + 4x + 22 ) + 13 – 22 5x = 2 = (x + 2)2 + 9 5dx = 2d = (x + 2)2 + (3)2 dx = 2/5 d 1 1 2 I = dx I = dx (x 2) 2 (3) 2 (2 )2 22 5 x + 2 = 3 2 1 = dx dx = 3d 2 5 4 4 1 2 1 1 I = 3d = dx (3 )2 (3)2 5 4 2 1 1 1 = 3 d = ∙arctan() + C 9 2 9 10 1 1 Since 5x = 2, = 5x/2. = 3 d 9 2 1 1 5x I = ∙arctan + C 1 10 2 = ∙arctan() + C 3 Since x + 2 = 3, = (x+2)/3 1 dx 1 x 2 10. I = 2 4x 16x 9 I = = ∙arctan + C 3 3 1 dx = 2 4x 4x (3) 2 1 dx 12. I = 2 4x = 3 40 6x x 4dx = 3d Completing the square: dx = 3/4 d 40 – 6x – x2 = -(x2 + 6x – 40) 1 3 = -[(x2 + 6x ) – 40] dx 2 2 2 I = 2 = -[(x + 6x + 3 ) – 40 - 3 ] 3 3 (3)2 4 = -[ (x+3)2 – 49] 3 1 1 2 2 dx = -[ (x+3) – (7) ] = 2 2 2 4 3 9 9 = (7) – (x+3) 1 3 1 1 1 dx dx I = 2 2 = 2 4 3 3 1 (7) (x 3) 1 x + 3 = 7 = ∙arcsec|| + C 12 dx = 7d 1 4x 7d Since 4x = 3, = I = 2 2 3 (7) (7 ) 1 4x 1 I = ∙arcsec + C 7 dx = 2 12 3 49 49 Calculus BC Sections 5/6 – 5/8 Review Worksheet answers
1 1 1 1 7 dx 2d = 2 I = 0 2 2 7 1 (2 ) (2) = 1∙arcsin() + C 1 1 = 2 d Since x + 3 = 7, = (x + 3)/7 0 4 2 4 x 3 1 1 1 I = arcsin + C = 2 d 7 4 0 2 1 1 = arctan 1 sinh x 2 0 13. I = dx cosh x 1 = ½ [arctan(1) – arctan(0)] 1 1 1 = sinh x dx = 0 cosh x 1 2 4 2 4 8 u = coshx + 1 du = sinhx∙dx 0.5 12 dx 1 16. I = 0.2 2 I = du 1 4x u 0.5 1 = ln | u | + C 12 dx = 0.2 2 2 = ln(cosh x + 1) + C (1) (2x) 2x = When x = 0.5, = 1 sinh1 2dx = d When x = 0.2, = 0.4 14. I = x dx x 2 dx = ½ d 1 2 1 1 1 = sinh(x ) x dx 12 dx = 0.4 2 u = x-1 1 2 du = -1∙x-2∙dx 1 1 -2 6 dx = 0.4 2 -du = x ∙dx 1 I = sinh(u)(du) 1 = 6arcsin 0.4 = sinh(u) du = 6[ arcsin(1) – arcsin(0.4) ] = -cosh(u) + C = 6.9557 1 ln(2) = -cosh + C 17. I = [sinh(x) cosh(x)]dx x 0 ln(2) =cosh x sinh x0 Evaluate the definite integral. ln(2) e x e x e x e x = 2 1 2 2 0 15. I = 2 dx 0 x 4 ln(2) 2e x 2 1 = = dx 2 0 (x)2 (2) 2 0 ln(2) x = 2 When x = 2, = 1 = e x dx = 2d When x = 0, = 0 0 = eln(2) – e0 = 2 – 1 = 1