Calculus Review
AGEC 317
Material from MATH 142
The ability to manipulate equations using calculus is critical in economic analysis.
The following problems review single-variable calculus. In AGEC 317, the singlevariable calculus will be extended to include more than a single variable.
1.
Informally, the concept of a limit is what value does one variable (say y)
approach as another variable (say x) approaches a specific value (say n). For a
limit to make sense, y must be a function of x. The variable x can approach n
from values either greater than or less than n. The idea of a limit will be used in
AGEC 317 will only to illustrate the concept of a derivative.
(a)
Graph the following linear line, y = 1 + 3x, in x, y space. What is the limit
of y = f(x) when x approaches 2?
(b)
Graph the following function, y = 5 if x is less than 3 and y = 10 if x is
greater than or equal to 3. What is the limit of this function as x approaches 3?
2.
The following rules of differentiation must be memorized for AGEC 317. A
derivative finds the slope of the function at a point. The first order derivative of y
with respect to x is denoted as either dg/dx or g’(x), where the prime denotes the
derivative with respect to x.
Constant function rule if g(x) = c, dg/dx = 0. Derivative of a constant is zero.
Power function rule if g(x) = cxn, dg/dx = cnxn-1.
Sum rule if y = f(x) + g(x), dy dx  f ' (x)  g' (x) .
Difference rule if y = f(x) - g(x), dy dx  f ' (x)  g' (x) .
Product rule if y = f(x) g(x), dy dx  f ' (x)g(x)  f (x)g' (x) .
Quotient rule if y 
f (x)
f ' ( x )g ( x )  f ( x )g ' ( x )
, dy / dx 
.
g( x )
(g( x )) 2
Chain rule if z  f ( y) where y  g( x ) then dz / dx 
dz dy
 .
dy dx
Log rule (natural or base 10) if y = ln f(x), dy dx  f ' (x) f (x)
Exponential rule if y  e f ( x ) , dy / dx  e f ( x ) f ' ( x).
Find dy/dx (first derivative) for the following functions.
(a ) y  3  4 x  6 x 5
(b) y  3x 4 ln 2 x  x 6
(c) y  e 3 x  2 x
2
(d ) y  6z 22 , z  ln 6 x
(e) y  a  bx  cx 2
(f ) y  (9 x 2  2)(3x  1)
(g ) y  ln( 2 x  1)
ln x
(h ) y  2
x
 2x 2  x 

(i) y  ln 
 x 1 
(j)
What does the first derivative measure?
3.
Because the first order derivative is, in general, a function of the variable x, the
first derivative is itself differentiable with respect to x, if it is continuous and
smooth. Higher order derivatives are the derivative of the derivatives. For
example, the second derivative is the derivative of the first derivative with respect
 dy 
d 
d2y
dx
to x:    2  f ' ' ( x ). Note, higher order derivatives are denoted by a
dx
dx
superscript number, which is the same as the order of the derivative or by the
number of primes, which denote the order. The same rules of differentiation used
in question 2 apply to higher order derivatives.
Find f ' ' ( x ) for the following functions:
(a ) f ( x)  a  bx 3  ln x  cx
x3
x2
(c) f ( x)  x ( x  3)  2
(b) f ( x) 
(d ) f ( x)  f ( x) g ( x)
(e)
Find the first six derivatives of the function f(x) = a + 2x2 + 3x5.
(f)
What does the second derivative measure?
4.
Examples of the use of calculus in economics include comparative statics,
elasticities, marginal analysis, and optimization. MATH 142 introduced some of
these concepts. AGEC 317 will introduce, review, reinforce and use these
concepts, whereas higher-level AGEC courses will expand and assume
knowledge of these basic concepts and the use of calculus.
df ( x ) x
, where f(x) is the
dx f ( x )
function of interest and x is the variable of interest.
Elasticities, in general, can be calculated as  
Marginal analysis is accomplished by taking the derivative of the function, for
example the marginal product of a production function is the first derivative with
respect to x.
To find the relative extreme points of a function, the first derivative is set equal to
zero and the resulting equation solved. The second derivative test indicates
whether the point is a maximum or minimum. If the first derivative, f ' ( x )  0 at
x = x0, then the value of the function at the point x0 will be a
(i)
(ii)
relative maximum if the second derivative is f ' ' (x 0 )  0 ,
relative minimum if the second derivative is f ' ' (x 0 )  0 .
An inflection point is on the function where the rate of change of the slope
changes. At this point, the second derivative equals zero, that is, f ' ' ( x 0 )  0.
For the following functions, find the elasticity with respect to x, marginal function
with respect to x, and if the function contains a maximum or minimum.
(a) y  2  3x 2
(b) y  4  4 x 2  8 x
(c) y  5  6 ln x
5.
(d)
What is the elasticity for the function in (a) at x = 2 and x = 3?
(e)
For the function, y = x3 - 3x + 5, find the stationary values (maximum,
minimum, and inflection points).
Show that the function y = x + 1/x (with x not equal to zero) has two relative
extrema, one being a maximum and the other being a minimum. Is the y value
associated with the “minimum” larger or smaller than the maximum. How is this
paradoxical result possible? Hint: plot the function and the graph will help to
explain.
6.
What is the difference between relative and global extreme points? Does
calculus find relative or global extreme points? How must global points be
found?
Answers
1.
(a)
When graphed it is clear that the function approaches 7 when x approaches
2 from either side of 2.
7
Approach x=2 from either
side, y =7
2
(b)
No limit exists, in this case as the function f(x) does not approach the
same number from both sides of x → 3.
y = 10
Approach x = 3 from
greater than three, y = 10.
Approach x = 3 from less
than three, y = 5.
y=5
3
2.
(a ) dy / dx  4  30x 4
(b) dy / dx  12x 3 ln 2 x  3x 4 2( 1
2x
)  6x 5  12x 3 ln 2x  3x 3  6x 5
(c) dy / dx  e 3 x  2 x (3  4x )
1
(d) dy / dx  132z 21  132 (ln 6x ) 21 x 1
x
(e) dy / dx  b  2cx
2
(f ) dy / dx  18x (3x  1)  3(9x 2  2)  18x  81x 2  6
2
(g) dy / dx 
2x  1
1 2
x  2x ln x
1  2 ln x
(h ) dy / dx  x

2 2
(x )
x3
(4x  1)( x  1)  1(2x 2  x )
2x 2  4x  1
( x  1) 2
(i) dy / dx 
 3
(2x 2  x )
2x  3x 2  x
( x  1)
(j)
The first derivative measures the instantaneous rate of change of the
function or the slope of the function.
3.
(a ) f ' ' ( x)  6bx  x 2
2 x  18
x4
12  2 x
(c ) f ' ' ( x ) 
( 3  x ) 4
(d ) f ' ' ( x)  2 g ' ( x) f ' ( x)  f ' ' ( x) g ( x)  g ' ' ( x) f ( x)
(b) f ' ' ( x) 
(e) f ' ( x)  4 x  15 x 4
f ' ' ( x)  4  60 x 3
f ' ' ' ( x)  180 x 2
f ' ' ' ' ( x)  360 x
f ' ' ' ' ' ( x)  360
f ' ' ' ' ' ' ( x)  0
(f)
The second derivative measures the instantaneous rate of change of the
instantaneous rate of change of the original function.
4.


dy x
x
6x 2


(a )  
 6x

2 
2
dx y
 (2  3x )  2  3x
m arg inal y  (call it MY )  MY  6x
f ' ( x )  6 x , f ' ( x )  0  6 x , if x  0
f ' ' ( x )  6  0 min


x

(b)   (8x  8)
2
(
4

4
x

8
x
)


MY  8x  8
f ' ( x )  8x  8, f ' ( x )  0  8x  8, if x  1
f ' ' ( x )  8  0 max
6
(c)  
5  6 ln x
6
MY 
x
f ' ( x )  0 for all x no extremes
(d)
6 * 22
12

2
7
2  3* 2
56
3 
29
(e)
y’ = 3x2 - 3 = 3(x2 - 1) set this equal to zero and solve for x gives two
extremes at x = 1 and x = -1.
2 
y’’ = 6x. At x = 1, y’’(x) = 6(1) = 6 > 0 so have minimum at x = 1
y’’ = 6x. At x = -1, y’’(x) = 6(-1) = -6 < 0 so have maximum at x = -1
y’’(x) equals zero at x = 0, inflection point at x = 0, y = 5.
Graph of the function
12.5
10
7.5
5
2.5
-4
-2
2
-2.5
4
5.
y’ = x + 1/x2 set this equal to zero and solve for x gives two extremes at x = 1 and
x = -1.
y’’ = 2/x3.
At x = 1, y’’(x) = 2/(1) = 2 > 0 so have minimum at x = 1.
At x = 1, y(1) = 2.
y’’ = 2/x3.
At x = -1, y’’(x) = 2/(-1) = -2 < 0 so have maximum at x = -1
At x = -1, y(-1) = -2.
At the minimum y = 2, but at the maximum y = -2. This seems paradoxical, but a
graph of the function reveals why this result is true. Graph of the function
15
10
5
-4
-2
2
4
-5
-10
-15
6.
Relative or local maximums or minimums are extreme points, in the sense they
represent an extreme in the immediate neighborhood of the point. Global
extremes are the absolute maximum or minimum of the function. Calculus finds
all relative extremes. To find the global, all relative extremes must be evaluated
to find the absolute maximum or minimum.