Midterm 2 Review
Material from the previous midterm
The midterm is not cumulative but you are expected
to know the following important facts:
d
dx
(1)
=
0
d
dx
(
x
a
)
=
ax
a

1
d
dx
(
f
(
x
) +
g
(
x
)
)
=
f
0
(
x
) +
g
0
(
x
)
d
dx
(
kf
(
x
)
)
=
kf
0
(
x
)
d
dx
(
f
(
x
)
g
(
x
)
)
=
f
0
(
x
)
g
(
x
) +
f
(
x
)
g
0
(
x
)
This midterm will
not
test on limits, tangent lines,
Squeeze Theorem, or other facts from the first midterm
(though they may make a comeback for the final!)
Derivatives of trigonometric functions
Using the limit of sin(
h
)
/h
as
h
→
0 and some basic
trigonometric identities we were able to find deriva
tives for the trigonometric functions.
d
dx
(
sin
x
)
=
cos
x
d
dx
(
cos
x
)
=

sin
x
d
dx
(
tan
x
)
=
(
sec
x
)
2
d
dx
(
sec
x
)
=
sec
x
tan
x
The important thing to remember is to get the signs
on the derivative of sine and cosine correct. Since we
can now be tested on these derivatives it is also help
ful to know some of the values of the trigonometric
functions.
θ
0
1
6
π
1
4
π
1
3
π
1
2
π
sin
θ
0
1
2
1
2
√
2
1
2
√
3
1
cos
θ
1
1
2
√
3
1
2
√
2
1
2
0
tan
θ
0
1
3
√
3
1
√
3
DNE
Chain rule
The chain rule deals with the situation of how to
take the derivative of a function inside of another func
tion.
Often a hint that we will use the chain rule is
parentheses “(” and “)” or if when reading it out loud
we use the word “of”. For example sin(
x
2
) (read “sine
of
x
squared”) is the function
x
2
inside the sine func
tion.
d
dx
(
f
(
g
(
x
)
))
=
f
0
(
g
(
x
)
)
g
0
(
x
)
Common mistakes when using the chain rule include
forgetting to multiply by
g
0
(
x
) or putting
f
0
(
g
0
(
x
)). So
for example
d
dx
(
sin
(
x
2
)
)
= cos(
x
2
)
·
2
x
and not cos(2
x
) or cos(
x
2
). Also make sure you have
clearly identified what the inside and outside func
tions are, i.e., sin(
x
2
) is a very different function than
(sin
x
)
2
.
An important application of the chain rule is the
power rule, which is the special case when
f
(
x
) =
x
a
.
d
dx
((
g
(
x
)
)
a
)
=
a
(
g
(
x
)
)
a

1
g
0
(
x
)
.
Implicit differentiation
An explicit function is when you have
y
=
stuff with
x
.
An
implicit
function is when we are given a relation
ship involving
x
and
y
but we do not (or can not) write
y
as a simple function of
x
. For example,
yx
2
+ 3
x
cos
y
= sin
x
+
y
gives a relationship between
x
and
y
, and for a given
value of
x
we can solve for what possible value(s)
y
can
be. So even though
y
is not given as a function of
x
we still can think of it as a function of
x
(simply being
the case that we don’t know exactly which function it
is).
So to take the derivative we start with the equation
defining the implicit relationship and take the deriva
tive of both sides with respect to
x
. There are a few
things to observe
1. If we take the derivative of two functions which
are equal the derivatives are still equal.
2. We treat
y
as a function of
x
, in particular when it
comes to expressions involving
y
we use the chain
rule to take the derivative and when we finally
get to the point where we need to write down the
derivative of
y
we put
dy/dx
or
y
0
(both mean the
derivative of
y
, how easy is that!).
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 Winter '10
 BUTLER
 Calculus, Derivative, Mathematical analysis

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