Derivative Practice

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)