|
|
|
|
|
| Author |
Message |
![[Post New]](/templates/default/images/icon_minipost_new.gif) 9 Jun 2007 23:50:44 IST
|
|
|
DIFFERENTIATION The Definition and Limit Method The derivative of a function expresses its rate of change with respect to an independent variable. The derivative is also the slope of the tangent line to the curve. As you might recall. Many students start out differentiating with the Limit Method. The Limit Method utilizes the definition of a derivative of a function. The Limit Method, as seen below, is imperative because some problems may not work or may take longer with the differentiation rules.
lim f(x + c) - f(x)
c-->0 c
(Where c denotes a small distance on the x-axis usually written as the lower case delta sign next to an x)
Derivative Rules The Constant Rule
The derivative of a constant function is 0. That is, if c is a real number, then d/dx[c] = 0.
The Sum and Difference Rules
The sum(or difference) of two differentiable functions is differentiable and is the sum(or difference) of their derivatives.
d/dx[f(x) + g(x)] = f'(x) + g'(x)
d/dx[f(x) - g(x)] = f'(x) - g'(x)
The Constant Multiple Rule
If f is a differentiable function and c is a real number, then cf is also differentiable and d/dx[cf(x)] = cf'(x)
The Power Rule
If n is a rational number, then the function f(x) = xn is differentiable and d/dx[xn] = nxn-1
The Product Rule
The product of two differentiable functions, f and g, is itself differentiable. Moreover, the derivative of fg is the first function times the derivative of the second, plus the second function times the derivative of the first.
d/dx[f(x)g(x)] = f(x)g'(x) + g(x)f'(x)
The Quotient Rule
The quotient f/g, of two differentiable functions, f and g, is itself differentiable at all values of x for which g(x) does not = 0. Moreover, the derivative of f/g is given by the denominator times the derivative of the numerator minus the numerator times the derivative of the denominator divided by the square of the denominator.
d/dx[ f(x)/g(x) ] = (g(x)f'(x) - f(x)g'(x)) / [g(x)]2 g(x) does not = 0
The Chain Rule
If y = f(u) is a differentiable function of u and u = g(x) is a differentiable function of x, then y = f(g(x)) is a differentiable function of x and d/dx[f(g(x))] = f'(g(x))g'(x)
The General Power Rule
If y = [u(x)]n, where u is a differentiable function of x and n is a rational number, then d/dx = [un] = nun-1u'.
Implicit Differentiation An implicit function of x and y is a function in which one of the variables is not directly expressed in terms of the other. In other words, you use implicit differentiation when both x and y are on the same side. To implicitly differentiate an implicit function follow the steps below.
1. Apply all differentiation rules. Except to y.
y3 - x2 = -5
y3 - 2x = 0
2. Then differentiate the y's in regard to x. In turn you add a y prime to the differentiated y's.
3y2y' -2x = 0
3. Now factor out any y primes. Hint. In this example there are none.
4. Lastly, place all the y primes on one side.
y' = (2x) / (3y2) Trigonometric Differentiation Trigonometric Derivatives | 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)2(tan x) | | d/dx(csc x) = -(csc x)2(cot x) | | d/dx(cot x) = -(csc x) | Inverse Trigonometric Differentiation Unlike the six basic trigonometric functions, the six inverse trigonometric functions must have restricted domains. The reason being that the six basic trigonometric functions are periodic and as a result have no on-to-one inverses. Of coarse, all Calculus students should know the frase if and only if. This page will denote the frase as iff.
Inverse Trigonometric Functions | Function | Domain | Range | | y = arcsin x iff sin y = x | -1 < x < 1 | -pi/2 < y < pi/2 | | y = arccos x iff cos y = x | -1 < x < 1 | 0 < y < pi | | y = arctan x iff tan y = x | -infinity < x < infinity | -pi/2 < y < pi/2 | | y = arccot x iff cot y = x | -infinity < x < infinity | 0 < y < pi | | y = arcsec x iff sec y = x | | x | > 1 | 0 < y < pi, y doesn't = 0 | | y = arccsc x iff csc y = x | | x | > 1 | -pi/2 < y | Higher Ordered DerivativesAs you may recall from class, a higher ordered derivative is nothing more than a second derivative or greater. In other words, it is a function that was differentiated more than once. Why would we differentiate again? Well, higher ordered derivatives are usually used in finding the acceleration from teh position function, as seen below.
Position function = s(t) Velocity function = s'(t) Acceleration function = s''(t) Exponential and Logarithmic DifferentiationExponential Functions to Base a
If a is a positive real number (a does not = 1) and x is any real number, then the exponential function to the base a is denoted by ax and is defined as x = e(ln a)x.
If a = 1, then y = 1x = 1 is a constant function.
Logarithmic Functions to Base a
If a is a positive real number (a does not = 1) and x is any real number, then the logarithmic function to the base a is denoted by logax and is defined as logax = (1/ln a) ln x
Derivatives for Bases other than e
Let a be a positive real number (a does not = 1) and let u be a differentiable function of x.
d/dx[ax] = (lna)ax
d/dx[au] = (ln a)au du/dx
d/dx[lobax] = 1/((ln a)x)
d/dx[logau] = 1/((ln a)u) du/dx
The Power Rule for Real Exponents
Let n be any real number and let be a differentiable function of x.
d/dx[xn] = nxn-1
d/dx[un] = nun-1 du/dx
Derivative of the Natural Logarithmic Function
Let u be a differentiable function of x.
1. d/dx[ln x] = 1/x, x > 0
But don't forget the Chain rule!
2. d/dx[ln u] = u'/u, u > The Mean Value Theorem The mean value theorem refers to the mean or average rate of the change of f in the interval [a,b]. Listed below are two variations of the mean value theorem. The first is the basic theorem and the second is an extended version.
The Mean Value Theorem
If f is continuous on the closed interval [a,b] and differentiable on the open interval (a,b), then there exists a number c in (a,b) such that f'(c) = (f(b) - f(a)) / (b - a)
The Extended Mean Value Theorem
If f and g are differentiable on the open interval (a, b) and continuous on [a b] such that g'(x) does not = 0 for any x in (a, b), then there exists a point c in (a, b) such that f'(c)/g'(c) = (f(b) - f(a)) / (g(b) - g(a))
Additional Theorems and Rules of Differentiable Functions If f(x) is differentiable at c, it is continuous at c.
The Extreme Value Theorem
If f(x) is continuous on the closed interval [a, b], then f has both a minimum and maximum on the interval.
Relative Extrema Occur only at Critical Numbers
If f has a relative minimum or relative maximum at x = c, then c is a critical number of f.
Rolle's Theorem
Let f be continuous on the closed interval [a, b] and differentiable on the open interval (a, b). If f(a) = f(b) then there is at least one number c in (a, b) such that f'(c) = 0.
Absolute Value Rule
If u is a differentiable function of x such that u does not = 0, then d/dx[ln | u |] = u'/u L'Hopital's Rule Let f and g be functions that are differentiable on an open interval (a, b) containning c, except possibly at c itself. Assume that g'(x) does not = 0 for all x in (a, b), except possibly at c itself. If the limit of f(x)/g(x) as x approaches c produces the indeterminate for 0/0, then
lim f(x) = lim f '(x)
x-->c g(x) x-->c g '(x)
provided the limit on the right exists or is infinite. This result also applies if the limit of f(x) / g(x) as x approaches c produces any one of the indeterminate forms infinity/infinity, (- infinity)/infinity, infinity/(- infinity), or (- infinity)/(- infinity). Applications of Differentiation Tangents and Normal Lines As you may recall, a line which is tangent to a curve at a point a, must have the same slope as the curve. Therefore, the slope of the tangent is
m = lim f(a + h) - f(a)
h-->0 h
Since the slope equation of the tangent line is exactly the same as the derivative definition, an easier way to find the tangent line is to differentiate using the rules on the function f. For example,
Find the slope of a line tangent to the function f(x) = x2 + 1.
f '(x) = 2x
The slope of the tangent line for all points on the graph is 2x.
To find the slope and equation of a line tangent to a certain point, you must:
First find the slope of the function by differentiation.
Second plug in the certain point's values for x and y
Finally plug both the slope and point values into a linear equation. Observe the example below.
Find the equation of the tangent line for the function f(x) = x2 + 1 at point (3,10).
Find the slope of the function by differentiation
f '(x) = 2x
Plug in the certain point's values Since this function does not have y we don't plug in y yet
f '(3) = 6 {6 is now the slope of the point 3,10}
Plug both slope and point values into a linear equation
(y - y1) = m(x - x1) {this is the linear equation}
(y - 10) = 6(x - 3) {Which can be simplified as below}
y = 6x -8
Just as we can find the slope and equation of a tangent line for a function, we can also do the same for a normal line. However, the normal line has two differences from the tangent line.
1. The slope of a normal line is perpendicular to the slope of the tangent line. Or in other words, the negative inverse of the tangent line.
2. The normal line is only defined if x does not = 0.
As a result, to find the slope and equation of the normal line, follow the steps above and convert the slope of the tangent line to the slope of the normal line. Finding Minimum and Maximum ExtremaDefinition of Extrema
Let f be defined on an interval I containing c. f(c) is the minimum of f on I if f(c) < f(c) for all x in I
f(c) is the minimum of f on I if f(c) > f(c) for all x in I
The minimum and max of a function on an interval are the extreme values or extrema of the function on the interval. The minimum and max of a function on an interval are also called the absolute min and absolute max on the interval, respectively.
Definition of Relative Extrema
If there is an open interval containing c on which f(c) is a maximum, then f(c) is called a relative maximum of f.
If there is an open interval containing c on which f(c) is a minimum, then f(c) is called a relative minimum of f.
Definition of Critical Number
Let f be defined at c. If f '(c) = 0 or if f ' is undefined at c, then c is a critical number of f.
One of the skills that a Calc student must know is how to find Extrema on a Closed Interval. Notice the table of steps below.
Finding Extrema on a Closed Interval | 1. | Find the critical numbers of f in (a, b). | | 2. | Evaluate f at each critical number in (a, b). | | 3. | Evaluate f at each endpoint of [a, b]. | | 4. | The least of these values is the min and the greatest is the max. |
Example:
Find the extrema of f(x) = 3x3 - 4x4 on the interval [-1, 2].
Find the critical numbers
f(x) = 3x3 - 4x4 {Problem}
f '(x) = 9x2 - 16x3 {Differentiate}
9x2 - 16x3 = 0 {Set f '(x) = 0}
x(9x - 16) = 0 {Factor}
x = 0, 16/9 {solve to get the critical numbers}
Evaluate f at each critical number in (a,b)
See table below
Evaluate f at each endpoint of [a, b]
See table below
Locate the min and max from the results
See table below
| Left Endpoint | Critical Number | Critical Number | Right Endpoint | | f(-1) = -7 | f(0) = 0 | f(16/9) = -23.1 | f(2) = -40 | | Maximum | | Minimum |
Another method of finding the extrema of a function is the First Derivative Test.
The First Derivative Test
Let c be a critical number of a function f that is continuous on an open interval I containing c. If f is differentiable on the interval, except possibly at c, then f(c) can be classified as follows.
If f '(x) changes from negative to positive at c, then f(c) is a relative minimum of f.
If f '(x) changes from positive to negative at c, then f(c) is a relative maximum of f.
Second Derivative Test
Let f be a function such that f '(c) = 0 and the second derivative of f exists on an open interval containing c.
If f ''(c) > 0, then f(c) is a relative minimum.
If f ''(c) < 0, then f(c) is a relative maximum.
If f ''(c) = 0, then the test fails. In such cases, you can use the First Derivative Test.
Example of the Second Derivative Test
Find the relative extrema for f(x) = -3x5 + 5x3
Since the theorem states the f '(c) must = 0, then c must be a critical number. So we get the critical numbers as
x = -1, 0, 1
Next we plug in the x values in f to get the y values which give us the coordinate of the points as
(-1, -2) (0,0) (1, 2)
Now we use the Second Derivative Test to determine the extrema.
Using the Second Derivative Test | Point | Sign of f '' | Conclusion | | (-1, -2) | f ''(-1) = 30 > 0 | Relative minimum | | (0, 0) | f ''(0) = 0 = 0 | Test Fails | | (1, 2) | f ''(1) = -30 < 0 | Relative max | Increasing or Decreasing IntervalsIncreasing/Decreasing Test Theorem
let f be a function that is continuous on the closed interval [a, b] and differentiable on the open interval (a, b).
1. if f '(x) > 0 for all x in (a, b), then f is increasing on [a, b].
2. if f '(x) < 0 for all x in (a, b), then f is decreasing on [a, b].
3. if f '(x) = 0 for all x in (a, b), then f is constant on [a, b].
Finding Intervals where a Function is Increasing or Decreasing | 1. | Locate the critical numbers of f in (a, b), and use these numbers to determine test intervals. | | 2. | Determine the sign of f '(x) at one value in each of the test intervals. | | 3. | Use the Increasing/Decreasing Test Theorem to decide whether f is increasing or decreasing on each interval. |
Example:
Find the open intervals on which f(x) = x3 - x2 is increasing or decreasing.
Get the Critical Numbers{ see Finding Minimum and Maximum Extrema}
x= 0, 2/3 see table for test intervals
Determine the sign of f '(x)
see table below
Use the Increasing/Decreasing Test Theorem
see table below
| Interval | -infinity < x < 0 | 0 < x < 2/3 | 2/3 < x < infinity | | Test Value | -1 | 1/3 | 3 | | Sign of f '(x) | f '(-1) = 5 > 0 | f '(1/3) = -1/3 < 0 | f '(3) = 21 >0 | | Theorem | Increasing | Decreasing | Increasing | Finding Concavity and Points of InflectionTo determining concavity of a graph is similar to the method of finding increasing/decreasing intervals of a graph. Note that concavity has nothing to do with teeth, but Calculus is another word for tartar.
Definition of Concavity
Let f be differentiable on an open interval, I. The graph of f is concave upward on I if f ' is increasing on the interval and concave downward on I if f ' is decreasing on the interval. In other words, if the graph arcs like a U then it is concave upwards. If it arcs like the golden arches of McDonald's then it arcs downward.
Concavity Test Theorem
Let f be a function whose second derivative exists on an open interval I.
If f ''(x) > 0 for all x in I, then the graph of f is concave upward.
If f ''(x) < 0 for all x in I, then the graph of f is concave downward.
If f ''(x) = 0 for all x in I, then the graph of f is a line, neither concave upward or downward.
Points of Inflection
If (c, f(c) is a point of inflection on the graph of f, then either f ''(c) = 0 or is undefined at x = c. Points of Inflection are the same as critical points except they use the second derivative of f. Finding Concavity | 1. | Locate the points of inflection and use these numbers to determine test intervals. | | 2. | Determine the sign of f ''(x) at one value in each of the test intervals. | | 3. | Use the Concavity Test Theorem to decide whether f is concave upward or downward on each interval. |
Example:
Determine the Concavity of the graph of f(x) = x4 - 4x3.
Locate the points of inflection and use them as test intervals
f '(x) = 4x3 - 12x2
f ''(x) = 12x2 - 24x
12x(x - 2) = 0
x = 0, 2
see table below for test intervals
Determine the sign of f ''(x)
see table below
Use the Concavity Test Theorem
see table below
| Interval | -infinity < x < 0 | 0 < x < 2 | 2 < x < infinity | | Test Value | x = -1 | x = 1 | x = 3 | | Sign of f ''(x) | f ''(-1) = 36 > 0 | f ''(1) = -12 < 0 | f ''(3) = 36 > 0 | | Use Theorem | Concave upward | Concave downward | Concave upward | Rate of Change By using the derivative, one can not only find the slope, but also the rate of change. These should be no surprise to any well taught Calc student. Here's why: if we were to find the rate of change of a graph wouldn't we find the slope. Therefore the rate of change is some what like the slope. However, this slope is called the rate. Which is expressed as
Rate = Distance = f(d) - f(c)
Time d - c
Many applications of rates of change are for describing the motion of an object moving in a straight line. In Calculus, one can find two types of rates, an average rate and an instantaneous rate.
Notes on the Position Function
Many rates of change problems and related rates problems will utilize the position function and its derivatives, which are seen below. Note that g is gravity if s is a function of a falling object. v0 is the initial velocity and s0 is the initial height or position.
Position function = s(t) = 1/2gt2 + v0t + s0 {for falling objects}
s '(t) = v(t) {The velocity function}
s ''(t) = a(t) {The acceleration function}
Average Rate
Change in distance
Change in time
Instantaneous Rate of fat point x
f '(x) {The derivative} Example Using Average Rate and Instantaneous Rate:
Situation: You are a prosecuter, which happens to have a degree in Calculus. You must prove that a truck exceeded the speed limit of 55 miles per hour at some point during a 4 minute period. The truck passed two stationary patrol cars, which were 5 miles apart, equipped with radar. As the truck passed the first patrol car, its speed was clocked at 55 miles per hour. Four minutes later, when the truck passes the second car, its speed is clocked at 50 miles per hour. A pretty smart driver seeing the second car, but not smarter than Calculus.
Solution:
First you're going to reconstruct the crime scene in mathematical terms.
Let t = 0 be the time in hours when the truck passes the first patrol car. The time when the truck passes the second is t = 4/60 = 1/15 hr.
Second you find the average velocity.
Average velocity = s(1/15) - s(0) = 5 = 75 mph.
(1/15) - 0 1/15
Since you find that the average velocity is above 55 miles per hour, you will then prove that at some point the truck was traveling at the average velocity. To do this you remember the Mean Value Theorem. Which proves that the truck was going at 75 mph at some point.
The Mean Value Theorem If f is continuous on the closed interval [a,b] and differentiable on the open interval (a,b), then there exists a number c in (a,b) such that f'(c) = (f(b) - f(a)) / (b - a) Related Rates A key part of solving related rates problems is translating the word problem to mathematical terms.
| Verbal Statement | Mathematical Model | | A gear is revolving at the rate of 30 revolutions per minute ( 1 rev = 2pi rad). | A = Angle of revolution dA/dt = 30(2pi) rad/min | | Water is being pumped into a swimming pool at the rate of 12 cubic feet per minute. | V = Volume of water in the pool dV/dt = 12 ft3/min | | The velocity of a car after traveling 2 hours is 50 miles per hour. | x = Distance traveled dx/dt = 50 when t = 2 |
Follow the steps below for solving related rates problems.
Solving Related Rate Problems | 1. | Identify all given quantities and quantities that must be found. Make a sketch and label the quantities. | | 2. | Write an equation involving the variables whose rates of change either are given or are to be found. | | 3. | Using the Chain Rule, implicitly differentiate both sides of the equation with respect to time t. | | 4. | Substitute into the resulting equation all known values for the variables and their rates of change. Then solve for the required rate of change. |
Example:
Situation: You're sitting at your computer trying to remember all of the things about Calculus that you should know. You then see a red balloon inflating on your screen. You some how find out that air is being pumped into the round red ballon at 4.5 cubic inches per minute. You job is to find the rate of change of the radius when the radius is 2 inches.
Solution: Let V be the volume of the ballon and let r be its radius. Since the volume is increasing at the rate of 4.5 cubic inches per minute, you know that at time t the rate of change of the volume is dV/dt = 9/2.
Identify all quantities
Given rate = dV/dt = 9/2
Find dr/dt = when r = 2
Write an equation involving the variables
Volume of a Sphere V = 4/3pi r3
Differentiate with respect to time t
dV/dt = 4pi r2 dr/dt
dr/dt = 1/(4pi r2) (dV/dt)
Subtitute all values for the variables and their rates of change
dr/dt = (1/(16pi)) (9/2) which = .09 in./min.
|
     
----------------------------------------------------
simpler@INDIAN |
this article: 2 points
(with 0 
in 1 votes ) [?]
|
|
You have to be logged on to rate
|
|
|
|
|
|
|
|
|
|
Moderation Team
![[Up]](/templates/default/images/icon_up.gif "[Up]"){kind=icon}
