MATH 205–03
Quiz #3
d
ds
−5
1. (3 points) Calculate
We rephrase
4
s5
2s7 −
4
s5
Name:
+ 3es .
as 4s , and then invoke the power rule and exponential rule:
d
2s7 − 4s−5 + 3es = 14s6 + 20s−6 + 3es
ds
2. (5 points) If y = 3 sec x + 2, calculate its second derivative
d2 y
.
dx2
dy
First we calculate the derivative dx
= 3 sec x tan x; then, to find the second derivative, we will
dy
invoke the product rule, since dx is a product:
d2 y
d
(3 sec x tan x)
=
dx2
dx
d
d
= ( 3 sec x) tan x + 3 sec x tan x
dx
dx
= 3 sec x tan x tan x + 3 sec x sec2 x = 3 sec x tan2 x + 3 sec3 x
3. (4 points) If f (t) =
t3 +et
,
t2 −3t
calculate f 0 (t).
This is an application of the quotient rule:
d t3 + et
dt t2 − 3t
(t2 − 3t) dtd (t3 + et ) + (t3 + et ) dtd (t2 − 3t)
=
(t2 − 3t)2
(t2 − 3t)(3t2 + et ) + (t3 + et )(2t − 3)
=
(t2 − 3t)2
f 0 (t) =
4. (4 points) Calculate
d
dx
cos(x3 − 2x).
This is an application of the chain rule. If we denote x3 − 2x as u, then:
d
d
cos(x3 − 2x) =
cos(u)
dx
dx
du d
=
cos(u)
dx du
d 3
=
(x − 2x)(− sin u)
dx
= −(x2 − 2) sin(x3 − 2x)
5. (4 points) Find the equation of the tangent line to the curve y = (2x2 − 7x + 5)ex at (0, 5).
Using the product rule, we can determine that
dy
d
d
2
=
(2x − 7x + 5) ex + (2x2 − 7x + 5) ex = (4x − 7)ex + (2x2 − 7x + 5)ex
dx
dx
dx
dy
so when x is 0, dx
= −7e0 + 5e0 = −2. Thus, the tangent line to this curve at (0, 5) has slope
of −2; since it passes through (0, 5), its equation can then be easily seen to be y = −2x + 5.
Friday, February 17, 2012