1 Extra differentiation problems with solutions.

These problems were prepared using Maple. There are sometimes 2 answers; the second is just Maple’s “simplification” of the first answer. In most cases there are other reasonable ways of writing the answer. d −x2 + 4 x − 1 = −2 x + 4 dx d 4 x8 − 5 x3 − x2 + 11 = 32 x7 − 15 x2 − 2 x dx d 1 + 4 x6 x−3 − 4 x3 = −3 x−4 − 12 x2 = −3 dx x4 d 1/3 x3 − 1/2 x2 + 3 x = x2 − x + 3 dx d 5 x−1 − 3 x−3 − 6 x−5 = −5 x−2 + 9 x−4 + 30 x−6 dx d dx d dy x + 3 x−1 x2 − 1 − x−1
10

1. 2.

3. 4. 5.

6. 7.

=

3 x5 + 2 x 3 + 3 x + 6 x3
9

y3 + 2 y2 − 1

= 10 y 3 + 2 y 2 − 1

y (3 y + 4)

√ 8. g(z) = 2 z 2 + 1 z−1 √ 2 z2 + 1 10 z 2 − 8 z 3/2 + 1 √ z − 1 + 1/2 √ = 1/2 g (z) = 4 z z z 9. f (u) = u3 + 1 f (u) = 12 u3 + 1 d dx d dx d du 3x − 2 2x + 7 x2 − 1 x−4 u−4 u2 − 1
4 3

u2 + 3 u2 + 3

5 5

(3 u − 2) u2 +10 u3 + 1 3x − 2

(3 u − 2)

6 4

6

u2 + 3

4

(3 u − 2) u+18 u3 + 1

6

4

u2 + 3

5

(3 u − 2)

5

10.

= 3 (2 x + 7)−1 − 2 =

(2 x + 7)2 =

= 25 (2 x + 7)−2

11.

2 x (x − 4) − x2 + 1 (x − 4)
2

x2 − 8 x + 1 (x − 4)2 u2 + 1 − 8 u (u2 − 1)
2

12.

=

u2 − 1 − 2 (u − 4) u (u2 − 1)
2

=−

13. f (t) = f (t) =

t2 − 2 t + 2 t2 − 1 (2 t − 2) t2 − 1 − 2 t2 − 2 t + 2 t (t2 − 1)2 1 (x − 1) (x − 2) 3 − 2x
2

=2

−3 t + t2 + 1 (t2 − 1)2 2x − 3
2

14.

d dx

=

(x − 1) (x − 2)

2

=−

(x − 1) (x − 2)

2

2 d dx (x − 1) x2 − 3 x2 + 1 x2 − 3 + 2 (x − 1) x x2 + 1 − 2 (x − 1) x2 − 3 x
2

15.

=

(x2 + 1)

=

x4 − 3 − 8 x + 6 x 2 (x2 + 1)
2

16.

√ √ d sin x2 − 2 = 2 cos x2 − 2 x dx d dx d dx sin (5 x2 ) = 5 x+1 √ x2 + 1 cos 5 x2 x sin (5 x2 ) 1 x2 +1 − (x + 1) x (x2 + 1)
3/2

17.

18.

=√

=−

(x2

x−1

+ 1)3/2

19.

d tan (5 x) = 5 (sec (5 x))2 dx d cot x2 = −2 csc x2 dx d dx cos (x) x2 =
2

20.

x

21.

sin (x) x + 2 cos (x) − sin (x) x2 − 2 cos (x) x =− 4 x x3

22.

d (cot (x) − 3 csc (5 x)) = − (csc (x))2 + 15 csc (5 x) cot (5 x) dx d dt cos (t) 1 − sin (t) = sin (t) (−1 + sin (t)) + (cos (t))2 (−1 + sin (t))
2

23.

= − (−1 + sin (t))

−1

24.

−2 + (cos (2 x))2 d 3 2 (tan (2 x) sec (2 x)) = 2 (sec (2 x)) + 2 (tan (2 x)) sec (2 x) = −2 3 dx (cos (2 x)) d dx 1/5 x5 − 3 x + x−2
5

25.

= 5 1/5 x5 − 3 x + x−2

4

x4 − 3 − 2 x−3

26.

d (y csc (5 y)) = csc (5 y) − 5 y csc (5 y) cot (5 y) dy
2/5

27. f (x) = x5 − 4 x + 1 5 x4 − 4 f (x) = 2/5 3/5 (x5 − 4 x + 1) 28.

d 4 3 (cos (x)) = −4 (cos (x)) sin (x) dx d dx d du tan √ x
2

29.

=

√ √ 2 tan ( x) (sec ( x)) √ x
2

30.

sec u−1

2

− tan u−1

=0

3 31. g(t) = g (t) = sin (3 t) + cos (3 t)
2

(cos (3 t)) + 1 (3 cos (3 t) − 3 sin (3 t)) (cos (3 t))2 + 1 + 6 (sin (3 t) + cos (3 t)) cos (3 t) sin (3 t) (cos (3 t))2 + 1
2

=3

sin (3 t) (cos (3 t))2 − sin (3 t) − (cos (3 t))3 + 3 cos (3 t) (cos (3 t)) + 2 (cos (3 t)) + 1
4 2