Derivative Practice By: Prof. Cesar Aguilar Department of Mathematics, SUNY Geneseo Find the derivative y ′ . The right column has the correct answer. You may need to simplify your work so that your solution matches the right column. Function Answer 1. y = (x + 1)2 (x2 + 2x) y ′ = 2(x + 1)(x2 + 2x) + 2(x + 1)3 2. y = √ x √ 1+ x y′ = √ 3. y = sin( 2θ) 4x −1 4. y = x+1 y′ = y ′ = −(4x2 )−1 5. y = cos4 (1 − 2x) q 2 6. y = x +x+7 x2 7. y = (cos(x))x 1 √ √ 2 2 x(1+ x) √ cos( 2θ) √ 2θ y ′ = 8 cos3 (1 − 2x) sin(1 − 2x) y′ = 2 −(x+14) q x2 +x+7 x2 x2 2x3 y ′ = (cos(x)) (2x ln(cos(x)) − x2 tan(x)) 8. y = ln(x2 ex ) y′ = 9. y = ecos(x) + cos(ex ) y ′ = − sin(x)ecos(x) − ex sin(ex ) 10. y = tan2 (sin(θ)) √ 11. y = sin(tan( x3 + 1)) y ′ = 2 tan(sin(θ)) sec2 (sin(θ)) cos(θ) 12. y = [tan(f (x)) + (g(x))2 ] 13. y = (sin(x)) √ y′ = 3 √ √ 3x2 cos(tan( x3 +1)) sec2 ( x3 +1) √ 3 2 x +1 2 y ′ = 3 tan(f (x)) + (g(x))2 [sec2 (f (x))f ′ (x) + 2g(x)g ′ (x)] x 14. y = 7x2 + sec(1 + p 15. y = 3 csc(ln(x)) x+2 x y ′ = sin(x) √ 3x) √ x h ln(sin(x))+2x cot(x) √ 2 x y = 14x + sec(1 + √ 3x) tan(1 + 3x y′ = 17. xy 4 + x2 y = x + 3y y′ = 1−y 4 −2xy x2 +4xy 3 −3 18. sin(xy) = x2 − y y′ = 2x−y cos(xy) 1+x cos(xy) 19. x2 cos(y) + sin(2y) = xy y′ = y−2x cos(y) 2 cos(2y)−x2 sin(y)−x y′ = 30x4 sin(3x5 ) cos(3x5 ) 1+sin2 (3x5 ) 20. y = ln(1 + sin2 (3x5 )) + ecot(3x)+14x 3x) 2√33x csc (ln(x)) x2 +x 2x2x √ √ cot(ln(x)) √ y ′ = − csc(ln(x)) 3 2 2x (1+ln(2)(2x2 +2x)) 16. y = i 2 √ (x+1) x2 +x 2 + ecot(3x)+14x (−3 csc2 (3x) + 28x)