Handout 3, Summer 2014
1.
Math 1823-171
27 May 2014
d
3x2 − 2 cos(x)
dx
Solution. We have
d
3x2 − 2 cos(x) = 6x − 2(− sin(x)) = 6x + 2 sin(x)
dx
2.
d
(4 sec(t) + tan(t))
dt
Solution. We have
d
(4 sec(t) + tan(t)) = 4 sec(t) tan(t) + sec2 (t)
dt
d
3.
dθ
sec θ
1 + sec θ
Solution. Using the quotient rule, we have
d
d
(1 + sec(θ)) dθ
(sec(θ)) − (sec(θ)) dθ
(1 + sec(θ))
d
sec θ
=
2
dθ 1 + sec θ
(1 + sec(θ))
(1 + sec(θ))(sec(θ) tan(θ)) − (sec(θ))(sec(θ) tan(θ))
=
(1 + sec(θ))2
sec(θ) tan(θ)
=
(1 + sec(θ))2
4.
d
x2 sin(x) tan(x)
dx
1
2
Solution. This takes a bit of work; we must use the product rule a
couple times. We have:
d 2
d 2
[x sin(x) tan(x)] =
[x (sin(x) tan(x))]
dx
dx
d
d
= (sin(x) tan(x)) x2 + x2 (sin(x) tan(x))
dx
dx
d
d
= 2x(sin(x) tan(x)) + x2 [tan(x) sin(x) + sin(x) tan(x)]
dx
dx
2
2
= 2x sin(x) tan(x) + x [tan(x) cos(x) + sin(x) sec (x)]
= 2x sin(x) tan(x) + x2 tan(x) cos(x) + x2 sin(x) sec2 (x)
We have the solution.
d
5. Assuming that dx
sin(x) = cos(x) and
d
prove that dx csc(x) = − csc(x) tan(x).
d
dx
cos(x) = − sin(x),
1
Solution. Recall first that csc(x) = sin(x)
. Therefore, using the quotient
rule, we have
d
d
1
csc(x) =
dx
dx sin(x)
d
d
sin(x) dx
1 − dx
sin(x)
=
sin2 (x)
0 − cos(x)
=
sin2 (x)
− cos(x)
1
=
·
sin(x) sin(x)
= − tan(x) csc(x)
We have the result.
6. If f (t) = csc(t), find f 00 (π/6).
Solution. We first determine f 00 (t). We have
f 0 (t) = − csc(x) tan(x)
3
Next
f 00 (t) = (f 0 (t))0
= tan(x)(− csc(x))0 + (− csc(x))(tan(x))0
= tan(x)(csc(x) tan(x)) − csc(x) sec2 (x)
= tan2 (x) csc(x) − csc(x) sec2 (x)
√
Now, we know
that
sin(π/6)
=
1/2
and
cos(π/6)
=
3/2, so that
√
2
2
tan (π/6) = 3/3, csc(π/6) = 2, and sec (π/6) = 4/3. Therefore,
using our work above, we have:
√
2 3−8
00
f (π/6) =
3
This gives the result.
7. For what values of x does y = x + 2 sin(x) have a horizontal
tangent?
Solution. Recall that the slope of the line tangent to the graph of
f (x) = x + 2 sin(x) is given by f 0 (x) = 1 + 2 cos(x). Now, the tangent
line is horizontal if and only if its slope is 0, so to find the x values
which correspond to a horizontal tangent line, we set f 0 (x) = 0. That
is, 1 + 2 cos(x) = 0, or cos(x) = −1/2. The x-values for which this is
true are all points x = 2π
+ 2πn and x = 4π
+ 2πn for all integers n.
3
3
This gives the result.
90
d
8. Find dx
90 sin(x) by computing the first few derivatives and looking for a pattern.
Solution. We have:
(sin(x))0 = cos(x)
(sin(x))00 = − sin(x)
(sin(x))000 = − cos(x)
(sin(x))(4) = sin(x)
At this point, it is easy to see that the cycle begins anew. Therefore,
d88
every fourth derivative of sin(x) is itself. Note, then, that dx
88 sin(x) =
d90
sin(x), so that dx90 = − sin(x).