Review 1
Derivatives/Antiderivatives
Procedure. In the table below, start in row 1 and complete the sentence on
the ‘Differentiation’ side. Then move to the “Integration’ side and complete
the corresponding sentence (if possible). Follow this procedure through row
13 (row 18, too).
Checking. One helpful aspect to the problem of integrating a function is that
one can always check to see if one has an antiderivative ‘just by
differentiating’ the proposed antiderivative; of course, this check does
depend on determining the derivative correctly! Starting with row 17,
complete the “Integration’ side first, and then check your answer by
differentiating and show that process on the ‘Differentiation’ side.
Alternate Notations.
d
[ f (x)] =
dx
D x f (x)
= f '(x) , and if y = f(x),
d
dy
.
[ f (x)] =
dx
dx
Differentiation ‘Fact’
1. For any constant, a,
d a
=
dx x
ax a−1 .
Corresponding General Integral
Fact (also called ‘indefinite
integral)
For any constant, a,
∫ ax
a−1
a
2. For any positive constant, a,
dx = x + c
(In #15 below, write a more
‘usable’ form of this fact.)
For any positive constant, a,
d
x
(a ) =
dx
x
x
∫ (lna)a dx =a + c
(lna)a x .
3. As a specific case of #1 above:
d
1 = 0.
dx
4. As a specific case of #2 above:
d x
( )=
dx e
ex .
5. For any constant, a,
6.
d
a=0
dx
d
1
ln(x) =
dx
x
7. For any positive constant, a,
d
Log a(x) =
dx
1
xlna
(Not in textbook table, but note
that Log a x =
ln(x)
.)
ln(a)
(In item #16 below, write a more
‘usable’ form of this fact.)
An antiderivative of 0 is the
constant function y=1.
x
x
∫ e dx =e + c .
An antiderivative of 0 is any
constant function y=a.
1
∫ x dx = ln x + c
ln x
1
∫ x lna dx = loga x + c (= lna + c)
8.
∫ cos(x)dx = sin(x) + c
d
sin(x) = ε 1cos(x) and
dx
d
cos(x) = ε 2sin(x) where
dx
ε 1 and ε 2
are either 1 or negative 1.
Determine, using only the graphs
of the sine and cosine, which
value goes with which function.
These graphs appear below this
differentiation/integration table.
(See next page.)
∫ −sin(x)dx = cos(x) + c , which is more usually and
usefully written as ∫ sin(x)dx = − cos(x) + c .
d
sin(x) = cos(x), since the slope of
dx
the sine function at x=0 is
positive.
d
cos(x) = -sin(x), since the slope
dx
of the cosine function at x=0 is
negative.
9. For constants a and b, provided
(af '(x) + bg'(x))dx = af (x) + bg(x) + c , which is
that f and g are differentiable
more usually and usefully written as
functions,
∫
d
[af (x) + b(g(x)] =
dx
af '(x) + bg'(x)
10. As a specific case of #9
above, for constant a,
d
[af (x)] =
dx
af '(x)
∫ (af (x) + bg(x))dx = a ∫ f (x)dx + b ∫ g(x)dx + c
∫ af '(x)dx = af (x) + c , which is more usually and
usefully written as ∫ af (x)dx = a ∫ f (x)dx + c
11. Provided that f and g are
differentiable functions,
The product rule, symbolized in column 1, leads
to what is called Integration by Parts (See 7.2):
d
[ f (x) ⋅ (g(x)] =
dx
∫ f (x)g'(x)dx = ∫
f (x)g'(x) + g(x) f '(x)
d
[ f (x)g(x)]dx −
dx
f (x)g(x) − ∫ g(x) f '(x)dx
∫ g(x) f '(x)dx =
12. Provided that f and g are
Rules can be derived from the quotient rule,
differentiable functions, and that symbolized in column 1, but they don’t seem to
d f (x)
be especially useful.
]=
g(x) is not zero, [
dx g(x)
g(x) f '(x) − f (x)g'(x)
[g(x)]2
13. As a specific case of #12
above,
d
tan(x) =
dx
sec2 (x)
14. For suitably defined and
differentiable functions f and g,
d
d
d
[ f (g(x))] =
[ f (g(x))] ⋅ [g(x)] ,
dx
dg(x)
dx
which, in simpler notation, and
letting y=f(u) and u=g(x),
dy
=
dx
= dy du
du dx
∫ sec
2
(x)dx = tan(x) + c
The differentiation rule on the left is known as
the ‘Chain Rule,’ which applies to a composite
(chain) function. The corresponding integration
procedure (It’s more a procedure than a rule.)
is called the method of ‘basic substitution,’ the
most critical and widely applicable technique of
integration and the subject of section 7.1. More
to come! (No written response needed in this
column of row #14.)
15. (See instructions given in row 1.)
a
∫ x dx =
x a+1
a +1
+ c, if a is not 1.
16. (See instructions given in row 2.)
x
∫ a dx =
ax
+ c for positive a.
(ln a)
17. Consider the following
d
1
(ln(x + 17) + c) =
dx
x + 17
d 1
1
1
1
( ln(17x + 38) + c) =
(17) =
dx 17
17 17x + 38
17x + 38
d −1
d
1
( + c) = (−x −1 + c) = 2
dx x
dx
x
integrals:
∫ x +1 17 dx
∫ 17x1+ 38 dx , ∫ 12 dx
x
∫ 21+
x
1
,
, and
dx . One of these can
be determined directly from
one of the formulas above,
two of these can be done by
guessing an antiderivative,
differentiating the guess, and
then revising the guess if the
differentiation check fails;
the fourth integral requires
special methods, and is
addressed later in this review.
Do the three “easy” ones!
1
∫ x +17 dx = ln(x + 17) + c
1
1
dx
=
∫ 17x + 38 17 ln(17 x + 38) + c
1
x −1
−1
−2
∫ x 2 dx = ∫ x dx = −1 + c = x + c
18.
d
d
1
2
ln( x + 1 ) =
ln(x 2 + 1) = ( 2 )(2x) by the
dx
dx
x +1
Chain Rule
d 1
1 1
x
(by the
( ln(x 2 + 1) + c) =
(2x) = 2
2
dx 2
2 x +1
x +1
Chain Rule)
∫(x
19.
2x
)dx = ln(x 2 + 1) + c
+1
2
∫ 2x+
x
1
2
dx = 2 ln(x + 1) + c
1
Portion of the graph of
the (periodic) sine
function
Portion of the graph of
the (periodic) cosine
function