Instructor: Dr. R.A.G. Seely
Test 1
(Sept 2016)
(Version A)
Cal I (S) (Maths 201–NYA)
Answers
1.
(a) lim
h→0
1
1
− 2x+1
2(x+h)+1
√
(b) lim
h→0
√
4−x−h− 4−x
h
2.
(2x+1)−(2x+2h+1)
h→0 h(2x+2h+1)(2x+1)
= lim
(4−x−h)−(4−x)
√
√
h→0 h( 4−x−h+ 4−x)
= lim
= lim
h
= lim
−h
h→0 2h(2x+2h+1)(2x+1)
=
−2
(2x+1)2
−h √
√
h→0 h( 4−x−h+ 4−x)
=
√−1
2 4−x
y
6
a
x
c
b
24 3 9 −4
x − x
(b) y 0 = ex (6x5 − x1/3 + 1 + 30x4 − 13 x−2/3 )
5
5
(3x2 − 6x)(x − 3x−2 ) − (x3 − 3x2 + 2)(1 + 6x−3 )
(c) y 0 =
(x − 3x−2 )2
3. (a) y 0 = 75x2 −
(d) y 0 = sec(x2 ) tan(x2 ) · 2x · cos(2x3 + 1) − sec(x2 ) sin(2x3 + 1)(6x2 )
(3x7 + 2x3 − 1)9
4. (a) y = q
0
0
5x21 −
(b) y = (2x + 1)
5. y 0 =
−5
sin(x2 +3)
21x6 + 6x2
1 105x20 + 5x−2
9 7
−
3x + 2x3 − 1 2 5x21 − 5/x − 5
#
2
2x cos(x + 3) ln(2x + 1) + sin(x + 3)
2x + 1
2
2
(12x + 10)(2x3 + 5x2 − 7) − (6x2 + 10x)2
6x2 + 10x
00
so
y
=
2x3 + 5x2 − 7
(2x3 + 5x2 − 7)2
6. (a) y 0 =
(b)
5
x
"
1
, so slope =
(2x+1)2
3 +y 4 −y
y 0 = − 3x2xy
2 y 2 +4xy 3 −x , so
1/9. Equation: y = 19 x + 29 .
slope= −2/9. Equation: y = − 29 x +
7. (a) y 0 = 6x−1/3 (x − 5) + 9x2/3 = 0 if 5x = 10, so x = 2
(b) x = ±1
8. f 0 (3) = 7
Remark: The graph of y = 9x2/3 (x − 5) looks like this:
13
9 .