IV TOPIC CALCULUS
[1] Limit
Suppose that we have a function y = f(x). What would happen to y as x xo?
Definition: Let y = f(x). Then, the limit value of y as x a is denoted by limxaf(x). EX 1: y = 1 + 2x. y = 1/x, x 0.
As x 0, y 1. limx0y = 1. EX 2: As x 0, y or - Limit does not exists. As x 0 from the right, y . limxo 1 . x
As x 0 from the left, y -. limx0 1 . x
IV-1
EX 3:
y=
1 . x1
limx2 y = 1; limx1y ; limx1y . limxy = 0; EX 4: y= limx-y = 0.
1x 2 , x 1. 1x 0 = 0 (?) 0
limx1y =
Nope!!!. 1x 2 (1x)(1x) limx1 = limx1 = limx1(1+x) = 2!!! 1x 1x EX 5: x2 y= , x 0. x
limx0y = limx0x = 0. limxy = limxx = . Note: Before computing a limit, better to simplify the function f. Although f(x) may not be defined at x = xo, f(x) could have a limit.
IV-2
EX 6:
y=
xa . x a ) = 1. x
limxy = limx(1 +
Theorem: Suppose that lim f(x) and lim g(x) exists. (Here, lim = limxx .) Then,
o
(R1) lim [f(x) g(x)] = lim f(x) lim g(x). (R2) lim[f(x)g(x)] = [lim f(x)][lim g(x)] (R3) lim[f(x)/g(x)] = [lim f(x)]/[lim g(x)], if lim g(x) 0. EX: y = ex/x2. limxy = [limxex]/[limxx2] (?) Nope!!!, since limxex and limxx2 do not exist. In fact, limxex/x2 = . limx0y = [limx0ex]/[limx0x2] (?) Nope!!!, since limx0x2 = 0. In fact, limx0y = . EX: y = (x2-1)/(x-1), x 1. limx1y = [lim (x2-1)]/[lim (x-1)] (?) Nope!!!, since limx1(x-1) = 0. In fact, limx1y = limx1(x+1) = 2.
IV-3
EX: y = (x2+2)/(x2+2x+2). limxy = [limx(x2+2)]/[limx(x2+2x+2)] (?) Nope!!!, since limx(x2+2) does not exists. limx x 22 x 2 2x 2 = limx 1 2/x 2 1 2/x 2/x 2
= [lim (1+2/x2)]/[lim (1+2/x+2/x2)] = 1.
L'Hpital's Theorem Suppose that we have two functions f(x) and g(x). Suppose: lim f(x) = 0 and lim g(x) = 0 , or, lim f(x) = and lim g(x) = . f(x) f (x) Then, lim = lim . g(x) g (x)
IV-4
[2]
Derivative
y = f(x) We want to know changes in y (y) when x changes by x. rate of change = y x
f(x 0 x) f(x 0) y x x x 0 x dy f(x x) f(x) . y f (x) = lim dx x x0
dy = d x x x 0
derivative evaluated at x0 = f(x0)
Note: If lim lim , then, we say that derivative does not exist.
x0 x0
Question: What is derivative?
IV-5
y y=f(x)
x0
x
/ : tangent line at x = x0 f(x0) : measures slope of tangent line
IV-6
[3]
Continuity vs. differentiability
Definition: Consider a function y = f(x). Suppose: 1) f is defined at x0; 2) lim f(x) lim f(x) limf(x) ;
xx 0
xx 0
xx 0
3) lim f(x) f(x 0) .
xx 0
Then, f is called continuous at x = x0.
EX 1) y = f(x) = 6x, for all x. Is this function continuous at x = 1?
y
y = 6x
x
IV-7
1) f(1) = 6; 2) lim y 6 ;
x1
3) lim y f(1)
x1
Continuous. 1 x
EX 2)
y
Is this function continuous at x = 0 1) f(0) not defined.
y
y = 1/x
x
y is not continuous at x = 0.
IV-8
y x+1 2 1 x
y=x+2 -2
EX 3) y =
x1 if x 1 x2 if x < 1
1) f(1) = 2. 2) lim f(x) 2 ; lim f(x) 3
x1 x1
Not continuous.
Definition: Suppose that at x0, f(x0) = lim f(x 0 x) f(x 0) x
x0
exists.
IV-9
Then, we say thaty f is differentiable at x0. Note: If lim lim , we say that derivative does not exist.
x0 x0
Theorem: If f is differentiable at x0, then, f is continuous at x 0. <proof> Note that: f(x) - f(x0) = f(x) f(x 0) x x0 (x x 0) .
Then,
xx 0
lim [f(x) f(x 0)] = lim
f(x) f(x 0) x x0
xx 0
(x x 0) = f(x0) 0 = 0.
Thus, lim f(x) f(x 0) 0 lim f(x) = f(xo).
xx 0 xx 0
Corollary: If f is not continuous, then f is not differentiable. EX 1) y = x1 x 1 x2 x < 1
Not continuous at x = 1 IV-10
not differentiable.
Note: If a function is continuous, it may or may not be differentiable.
Definition: |a| = a if a 0 a if a < 0
EX 1) |-4| = 4; |4| = 4 EX2) y = f(x) = |x-2| + 1 y= x1 if x 2 x3 if x < 2
Continuous? 1) f(2) defined and f(2) = 1 2) lim f(x) lim f(x) 1
x2 x2
3) limx2f(x) = f(2). So, f is continuous. Differentiable?
x0
lim
f(2 x) f(2) (2 x) 1 1 = lim x x x0 = lim x =1 x
x0
IV-11
x0
lim
f(2 x) f(2) (2 x) 3 1 = lim x x x0 = lim
x0
x = -1 x
not differentiable
IV-12
[4]
Rules for Derivatives y = f(x) = k. dy 0. dx
(1)
(2)
y = axn y = anxn-1
EX 1) y = 3x6 y = 3 6 x6-1 = 18x5
1 2
EX 2) y = 3 x 3x
1 x y 3 2
1 1 2
3 2 3 1 3 x = = 2 2 1 2 x x2
1
(3)
d d d (f(x) g(x)) = f(x) g(x) f (x) g (x) dx dx dx
EX) y = 2x2 + x dy 4x 1 . dx
IV-13
(4)
d (f(x) g(x)) = dx
d d f(x) g(x) f(x) g(x) dx dx d d f(x) g(x) f(x) g(x) dx dx [g(x)]
2
(5)
d f(x) dx g(x)
=
.
EX) f(x) = 6x2 + 6x g(x) = 3x3 + 6x2 f(x) = 12x + 6; g(x) = 9x2 + 12x d f(x) g(x) = (12x + 6)(3x3 + 6x2) + (6x2 + 6x)(9x2 + 12x) dx d f dx g = (12x 6)(3x 3 6x 2) (6x 2 6x)(9x 2 12x) (3x 3 6x 2) 2
IV-14
[5]
Chain Rule
xyz z = f(y) y = g(x) dz dz dy f (y) g (x) dx dy dx
xyzw dw dw dz dy . dx dz dy dx
EX 1) z = 3y2; y = 2x + 5 xyz dz dz dy = (6y) 2 = 12y = 12(2x + 5) = 24x + 60 dx dy dx dz 24 1 60 84 dx x 1
EX 2) y = 3(2x2+1)3. Set z = 2x2 + 1. Then, y = 3z3; z = 2x2 + 1. dy/dx = (dy/dz)(dz/dx) = (9z2)(4x) = 36xz2 = 36x(2x2+1)2. 3) EX TR = f(Q); Q = g(L) dTR dTR dQ = MR MPL = MRPL dL dQ dL IV-15
[6]
Inverse Function
(EX 1) Consider the following example.
This is a function.
Looking at the reversed relation:
(EX 2) Consider the following example.
IV-16
Now, look at the reversed relation:
Definition: If the reversed relation is also a function, we say that there exists the inverse function, and we denote it by f-1.
Definition: A function y = f(x) is strictly increasing (decreasing) iff f(x1) > f(x2), whenever x1 > (<) x2.
Theorem: If a function y = f(x) is strictly increasing or strictly decresing, then x = f -1(y) (inverse function) exists.
Note:
IV-17
(1)
"Function": If you know what happened yesterday (cause), you can predict what will happen today (result).
(2)
"Inverse function": If you know what happens today (result), you can figure out what happened yesterday (cause).
EX) y = f(x) = 5x + 25, - < x < . x = y/5 - 5 = f--1(y).
Theorem: Suppose that a function y = f(x) has an inverse function. Then, dx 1 . dy dy dx EX) For the above example, dy/dx = 5 ; dx/dy = 1/5. Clearly, dx/dy = 1/(dy/dx).
IV-18
[7]
Partial Differentiation
Consider a function: y = f(x1,x2,...,xn), where xi's are independent variables and y is a dependent variables. Suppose we would like to know the rate of change of y when x1 changes.
Definition: dy y y . ceteris paribus lim dx1 x1 x1 x1 (Here, we treat all other variables as constant!!!!)
EX 1) y = 3x12 + x1x2 + 4x22 y/x1 = 6x1 + x2 . y/x2 = x1 + 8x2 .
EX 2) y = (x1 + 4)(3x1 + 4x2) . y/x1 = [(x1+4)/x1](3x1+4x2) + (x1+4)[(3x1+4x2)/x1] = 3x1 + 4x2 + (x1+4) 3 = 6x1 + 4x2 + 12. y/x2 = 4x1 + 16 (Show this at home.)
2
EX 3)
y
x1x2
x1 x2
. IV-19
(x1x2 ) y x1
2
2
x1
(x1 x2) (x1x2 ) (x1 x2)2
2
(x1 x2) x1 x2
3 2
x2 (x1 x2) x1x2 (1) (x1 x2)
2
2
(x1 x2)
.
Simliarly, y/x2 = (2x12x2 - x1x22)/(x1-x2)2.
IV-20
[8]
Jacobian Determinant y1 = f1(x1, x2, ... xn, xn+1, xn+2, ... xm) y2 = f2(x1, x2, .... , xn, xn+1, xn+2, ... , xm) yn = fn(x1, x2, .... , xn, xn+1, xn+2, ... , xm).
There are n equations:
Let's treat xn+1, ... xm as given. (Treat them as constants) Sometimes, we wish to know whether all the functions are distinctive, that is, whether there are redundant functions or not. For this case, we use Jacobian matrix: y 1 y 1 x 1 x 2 y 2 y 2 J= x 1 x 2 y n y n x 1 x 2 ... y n x n y 1 x n y 2 x n
...
...
If |J| = 0, we say that the functions are functionally dependent. If |J| 0, we say that the functions are functionally independent.
IV-21
EX 1)
y 1 2x 1 x 2 ; y 2 x 1 x 2 y 1 x 1 y 2 x 1 y 1 x 2 y 2 x 2
2
3
2
2 2x 1 0 ;
3x 2
2
2x 1x 2
;
x1
2
4x 3 6x x 3 0 |J| 1 1 2 2x x x 2 1 1 2 EX 2) y1 = x1 + 2x2 y 2 x 1 4x 1x 2 4x 2 = (x1 + 2x2)2 1 2 4x 8x 2(2x 4x ) 0 |J| 1 2 1 2 2x 1 4x 2 4x 1 8x 2
2 2
4x 1
3x 2
2
IV-22
[9]
Differentials dy : derivatives dx
dy, dx : differentials (small changes) dy f (x) dy f ...