Math 125
Quiz 3
2/21/2017
This quiz is over the sections 3.3 and 3.7 of the textbook. You have 30 minutes to
complete it. Full marks is 10. This is a group quiz. You may work in groups of 2-3 but
do not talk to people in other groups. Please show your work and leave comments; your
grade is based upon the steps you take completing a problem, not solely on the final
answer. Each group only submit one quiz with all the names of the group members.
The lines in red are essential steps and rules that we expect you to remember.
1. (4 Points) Differentiate the following functions:
√
(a) (x2 + 8)( x2 + 1)
√
Sin( x2 + 1)
(b)
Cos(x + 1)
(c) 6π
e
+x9
(d) [x + (x + e2x )2 ]6
Solution:
1
(a) f 0 (x) = (x2 + 8). 12 .(x2 + 1)− 2 (2x) +
√
x2 + 1.(2x)
√
√
1
Cos(x + 1).Cos( x2 + 1). 12 (x2 + 1)− 2 .(2x) − Sin( x2 + 1)(−Sin(x + 1))
(b) f (x) =
Cos2 (x + 1)
0
(c) 6π
e
+x9
.ln(6).(9x8 )
(d) 6[x + (x + e2x2 ]5 [1 + 2(x + e2x ).(1 + e2x .2)]
2. (4 Points) Assume that f (x) and g(x) are such that f (5) = 1, f 0 (5) = 9, g(5) = −5 and g 0 (5) = 5.
Find the following:
(a) (f g)0 (5)
0
f
(b)
(5)
g
d g(x2 ) (c)
√
dx
x
x= 5
d
(d)
(f (2x + g(x)))
dx
x=5
Solution:
(a) (f g)0 (x) = f (5).g 0 (5) + f 0 (5).g(5) = −40
Math 125
(b)
Quiz 3
Page 2 of 2
0
f
−45 − 5
g(5).f 0 (5) − f (5).g 0 (5)
=
(5) =
= −2
g
(g(5))2
25
√ 0
√
x.g 0 (x2 ).(2x) − g(x2 ).1 d g(x2 ) 5.g (5).2 5 − g(5)
(c)
= 11
√ =
√ =
dx
x
x2
5
x= 5
x= 5
(d)
d
= f 0 (2.5 + g(5)).(2 + g 0 (5)) = 63
= f 0 (2x + g(x)).(2 + g 0 (x))
(f (2x + g(x)))
dx
x=5
x=5
3. (2 Points) Find out the equation of the tangent line to the graph of the function f (x) = Cot(2x) at
x = π8 .
Solution:
Step 1: Find out f 0 ( π8 ).
Cos(2x)
Sin(2x)
Sin2 (2x) + Cos2 (2x)
Sin(2x)(−Sin(2x)).2
− Cos(2x).Cos(2x).2
=
−2.
= −2.Csc2 (2x)
Hence f 0 (x) =
Sin2 (2x)
Sin2 (2x)
Hence f 0 (5) = −2Csc2 (2. π8 ) = −4
Note that Cot(2x) =
Step 2: Write the equation of tangent in point slope form.
The equation of the tangent is: (y − f ( π8 )) = (−4)(x − π8 ).
Note that f ( π8 ) = Cot( π4 ) = 1.
Hence The equation of the tangent at x =
π
8
is y − 1 = (−4)(x − π8 ).
The End.