In Class Problems
9/15/2014
d
12
2
1. Find
13x + πx +
dx
x
Solution:
d
d 12
12
d
d
2
(13x) +
(πx2 ) +
( )
13x + πx +
=
dx
x
dx
dx
dx x
12
= 13 + 2πx − 2
x
2. Find
d 2 2
x (x + x + 1)
dx
Solution:
d 2 2
d 2 2
d
x (x + x + 1) =
(x )(x + x + 1) + x2 [x2 + x + 1]
dx
dx
dx
= 2x(x2 + x + 1) + x2 (2x + 1)
Here we have used the product rule. You could also have distributed first and then taken the derivative.
d x2 + x + 1
3. Find
.
dx
x2
Solution:
By the Quotient Rule:
d
dx
x2 + x + 1
x2
d
d
x2 dx
(x2 + x + 1) − (x2 + x + 1) dx
(x2 )
(x2 )2
2
x (2x + 1) − 2x(x2 + x + 1)
=
x4
=
4. Find values of m and b that make the following function both continuous and differentiable.
(
(x − 1)2 + 2 if x ≥ 3
f (x) =
mx + b
if x < 3
Solution:
In order for the function to be differentiable, the limit
lim
h→0
f (x + h) − f (x)
h
1
must exist. Therefore the left and right handed limits must exist and be equal. Therefore we must
have
f (x + h) − f (x)
f (x + h) − f (x)
d lim+
=
(x − 1)2 + 2 x=3 = lim−
=m
h
dx
h
h→0
h→0
(2x − 2)
=m
x=3
4=m
Intuitively what is going on is that we have to make the tangent line from the left agree with the
tangent line from the right so there is no kink in the graph.
Now we use that we want f to be continuous. The only possible place f could not be continuous is at
x = 3. In order to make it so, we must have that the left and right handed limits exist and are equal.
Therefore
(3 − 1)2 + 2 = 4(3) + b
b = −6
5. A function is called even if f (x) = f (−x) for all x. What can you say about the derivative of an even
function?
Solution:
Graphically, an even function is one that is symmetric about the y-axis (which means that if you draw
the graph, then fold the paper along the y-axis, the two sides of the graph will line up). Examples of
even functions include y = x2 , cos(x), and y = |x|.
If the derivative f 0 (x) of an even function exists at x, then f 0 (−x) = −f 0 (x). This property is called
being an odd function. Thus the derivative of an even function is odd. Similarly the derivative of an
odd function is even.
Some examples of odd functions are y = x3 , sin(x), y = x.
Wednesday 9/17/14
1. Find
d
dx [sin(x) cos(x)].
Solution:
This is one application of the product rule
d
d
[sin(x)] cos(x) + sin(x) [cos(x)]
dx
dx
= cos2 (x) − sin2 (x)
=
2. Find
d
3
dx [cos (x)].
Solution:
Later the chain rule will give us a quicker way to compute this, but for now we can use the product
rule twice:
d
d
[cos3 (x)] =
[cos(x) cos(x) cos(x)]
dx
dx
d
d
=
[cos(x)] cos2 (x) + cos(x) [cos(x) cos(x)]
dx
dx
= − sin(x) cos2 (x) + cos(x)[−2 sin(x) cos(x)]
= −3 cos2 (x) sin(x)
sin(12x)
√
.
x→0 sin( 8x)
3. Find lim
Solution:
The trick here is to ”multiply by 1”.
"
√ !#
sin(12x) x 12
sin(12x)
8
√
√
√
= lim
lim
x→0 sin( 8x) x
x→0 sin( 8x)
12
8
!
√
sin(12x)
8x
12
√
lim
= √
lim
x→0
x→0 sin( 8x)
12x
8
12
=√
8
4. Find
d cot(x) + sec(x)
.
dx
csc x
Solution:
The simplest way is to simplify the trig functions first and then take the derivative. You can also
directly apply the quotient rule.
d cot(x) sec(x)
d cot(x) + sec(x)
=
+
dx
csc x
dx csc x
csc(x)
d
=
(cos(x) + tan(x))
dx
= − sin(x) + sec2 (x)
5. Find lim
t→0
tan(6t)
.
sin(5t)
Solutions:
We first rewrite tangent in terms of sine and cosine.
1 sin(6t)
tan(6t)
= lim
t→0 cos(6t) sin(5t)
t→0 sin(5t)
1
sin(6t)
= lim
lim
t→0 cos(6t)
t→0 sin(5t)
lim
Now proceed as in problem 3 to get
= (1)
6
6
=
5
5