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CHAPTER TWO: DERIVATIVES. Hare Krsna Hare Krsna Krsna Krsna Hare Hare Hare Rama Hare Rama Rama Rama Hare Hare Jaya Sri Sri Radha Vijnanasevara (Lord Krsna, the King of Math and Science) KRSNA CALCULUS™ PRESENTS:. INTRODUCTION. Welcome back, dear students!

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### CHAPTER TWO:DERIVATIVES

Hare Krsna Hare Krsna Krsna Krsna Hare Hare

Hare Rama Hare Rama Rama Rama Hare Hare

Jaya Sri Sri Radha Vijnanasevara (Lord Krsna, the King of Math and Science)

KRSNA CALCULUS™ PRESENTS:

• Welcome back, dear students!

• This chapter we will explore the first half of calculus known as differential calculus. We will study basic differentiation of various functions.

• We will encounter real life Krsna Conscious problems. Matter of fact, we will introduce derivatives in a Krsna Conscious manner with our first application!

• RECALL from last chapter: LIMITS. You need to know how to evaluate limits!

• Let’s go the Ratha Yatra everyone!

• Let’s say we come from Boston, MA and we want to go to New York City for the Ratha Yatra.

• Say that the distance is 100 miles and it takes us 2 hours to drive that distance.

• What speed did you drive at?

• You know that speed is the rate of distance and time. Let speed be vand let the distance be s and time be t.

• Therefore: v = Ds/Dt D (“delta”) means “change in.”

• So you would plug 100 miles for Ds and 2 hours for Dt. If you divide 100 miles into 2 hours, you will see that the speed, , is 50 miles/hour. You would say that is the speed.

• Be honest, but the moment you turned your keys at the Boston, did you instantly go southwest constantly at 50 miles and hour with cruise control, no bathroom or any breaks, and the moment you see the Ratha Yatra festival site, your car instantly halted?

• Would be nice if was true, but unfortunately, its not!

• Now, while driving, I plugged in a kirtan CD. Now, I want to know if the music made me speed up a little. I want to know how fast I was going when I plugged in the CD.

• Let’s assume that we were about 20 miles away from home when I turned on the CD player. This will mean that initial and final position will be the same. Initial and final times will be the same since its happening that instant. Let’s find the speed.

• We have a problem!!! 0/0 = ???

• Let’s say this is the graph of the position vs. time.

• Position, s, is a function of time, t, or in other words, s(t).

-The 100 mi/hr answer we gave was the average rate.

- Note the following.

-Very analogous to the slope formula learned in algebra when you graphed lines.

• Since the slope of the line is the rate, we have to find the slope of the curve at that particular point, since it is not linear.

• The most effective approach is to draw a line tangent to the point (just touching it only once) and find the line’s slope!

• Brown line is tangent line.

• Green line is AVERAGE line.

• As you can see from the green line, you will need to measure smaller intervals of time in order to get the instantaneous rate (the rate at that very instant).

• In other words, make the Dt very small, that it is so very very very close to zero!

• So you can say that as Dt approaches zero, you get the instantaneous rate.

• Sounds like a limit, doesn’t it? As a variable approaches a zero, what value will I get at?

• It can be written as:

• Consider that there are points x, and x+h, separated by the distance h. Thus Dx = (x+h)-x which is just h.

• Their respective f(x) values for x and x+h are simply f(x) and f(x+h) respectively. Remember how you work with functions? If its f(x) = 2x and if its f(3), its just 2(3) or 6. You merely replace the expression inside the parenthesis for the x. Thus, Dy is just f(x+h) – f(x).

x

• m= Dy/Dx

• Therefore, with the use of functions, the AVERAGE RATE is defined for f(x) as. This expression is known as the difference quotient.

• We want to make the distance between x and x+h to be really really small, very close to zero, so that x and x+h are “virtually on the same spot,” thus we will get the slope of the tangent line (what we wanted earlier).

• Therefore, it is important to set the limit as h, the distance between x and x+h, gets very very very very very very close to zero. The result is the expression for the instantaneous rate. The limit is known as the derivative.

• The definition of the derivative formula:

• The derivative is the slope of the line tangent to a point.

• The derivative is very helpful in finding slopes of ANY function.

• We can see how fast the Ratha Yatra car goes the moment Vaiyasaki starts kirtana!

• We can see the rate of how much the cost is changing for the temple gift shops!

• Basically:

• DERIVATIVE = slope = rate of change

• If f is a function of x, or f(x), then the derivative can be written as y’ (y-prime), f’(x), or dy/dx.

• Let’s try an example:

• Find the derivative of f(x) = x2 using the definition of the derivative.

• To make the process easier to understand, I shall show you statements and reasons.

Derivative formula

• Replace function with x2.

• Squaring x+h.

• x2 terms cancel. Factored h.

• Cancel h.

• Apply the limit (h=0)

• Yes.. I admit, tedious! I’d rather read the Gita, than compute derivatives. But you may ask, “Why that long?”

• Well imagine if we applied the limit first… f(x)-f(x)/0 = 0/0 is undefined. So, we had to get rid of the h in the denominator in order to get a derivative. But remember 0/0 is every possible answer. Since 2x(0) = 0. So, 0/0 is “known”, but we wanted an exact answer and not an ambiguous answer.

• HINTS:

• In many calculus textbooks, the first thing discussed it the slope of the line tangent to a point of the graph. Hence, the definition of the derivative using the difference quotient is introduced. Sometimes, teachers will ask you to differentiate (process of taking derivatives) using the definition. This helps reinforce algebra concepts. We saw from the last example that concepts like expanding binomials, factoring, canceling the denominator, and applying limits.

• Use all the algebra skills you have in order to make sure that the denominator is not zero.

• We’ll do one more example of how to differentiate with the definition and the difference quotient.

• Using the definition of the Derivative, find the derivative of….

• Definition of derivative

• Applying the square root function.

• To lose the radical on top, we multiply by its conjugate.

• After multiplication of the radical and its conjugate, you get x+h –x, thus the x cancels.

• The h cancels.

• Apply the limit as h0

• Simply.

• Determining the slope of the line tangent to a specific point x = c is not very difficult. After getting the derivative, you plug in the point. For example, find the slope of the line tangent to x = 3 for the graph y = x2.

• Find the derivative. The derivative is 2x.

• Then you put 3 in for x. 2(3) = 6.

• So the slope of the line tangent to x=3 for y = x2 is 6.

• Find the equation for the line tangent to x = 3 for y = x2.

• First you need to know the (x,y) points for x=3. If x = 3, then y = 9. We also know that the slope at x =3 is 6. So at point (3,9), the slope is 6. Remember the equation for a line?

• m = 6 (derivative of f(x) at x = 3.)

• x = 3 (given)

• y = 9 (f(3) is 9.)

• x0 and y0 are the points on the graph.

• m is the slope. x and y are the variables.

OR

• A function is known to be differentiable if you are able to take derivatives of the function within the range [a,b]. The following four are cases where the function is not differentiable at certain points.

DISCONTINUITIES

CORNERS

CUSPS

VERTICAL TANGENTS

• Therefore, if a function is differentiable, then it is continuous.

• However, if a function is continuous, it is not always differentiable. (y=|x| has a corner which is not differentiable at x=0.)

• ALWAYS REMEMBER THAT >=)

• But by Lord Sri Krsna’s grace, we have rules to rememorize on how to get derivatives even faster, so we would never ever have to use that derivative definition. It would be tedious and impractical when we discuss other types of functions.

• You may have to do a few of the derivative definition types, but your teacher will not be fanatic about it. It is just there to reinforce your algebra skills. The derivative is basically a limit as the change in x gets really really close to zero.

• Just remember that! >=) Jaya Jaya!

The derivative of a constant is zero. Duh! If everything is constant, that means its rate, its derivative, will be zero. The graph of a constant, a number is a horizontal line. y=c. The slope is zero.

The derivative of x is 1. Yes. The graph of x is a line. The slope of y = x is 1. If the graph of y = cx, then the slope, the derivative is c.

• y,u and v are functions of x. a,b,c, and n are constants (numbers).

• When you take the derivative of x raised to a power (integer or fractional), you multiply expression by the exponent and subtract one from the exponent to form the new exponent.

Example:

• The derivative of the sum or difference of the functions is merely the derivative of the first plus/minus the derivative of the second.

• The derivative of a product is simply the first times the derivative of the second plus second times the derivative of the first.

• The derivative of a quotient is the bottom times the derivative of the top, minus top times the derivative of the bottom….. All over bottom square..

• TRICK: LO-DEHI – HI-DELO

• LO2

• If you have constant multiplying a function, then the derivative is the constant times the derivative. See example below:

• The coefficient of the x6 term is 5 (original constant) times 7 (power rule.)

• You can take derivatives of the derivative. Given function f(x), the first derivative is f’(x). The second derivative is f’’(x), and so on and so forth.

• Using Leibniz notation of dy/dx

For math ponders, if you are interesting in the Leibniz notation of derivatives further, please see my article on that. Thank you. Hare Krishna >=) –Krsna Dhenu

• Find the derivative:

• Use the power rule and the rule of adding derivatives.

• Note 3/2 – 1 = ½. x½ is the square root of x.

• Easy eh??

• Find the equation of the line tangent to y = x3 +5x2 –x + 3 at x=0.

• First find the (x,y) coordinates when x = 0. When you plug 0 in for x, you will see that y = 3. (0,3) is the point at x=0.

• Now, get the derivative of the function. Notice how the power rule works. Notice the addition and subtraction of derivative. Notice that the derivative of x is 1, and the derivative of 3, a constant, is zero.

• Now find the slope at x=0, by plugging in 0 for the x in the derivative expression. The slope is -1 since f’(0) = -1.

• Now apply it to the equation of a line.

• Now, plug the x and y coordinate for x0 and y0 respectively. Plug the slope found in for m.

• And simplify

• On the AP, you can leave your answer as the first form. (point-slope form)

• Find all the derivatives of y = 8x5.

• Just use the power rule over and over again until you get the derivative to be zero.

• See how the power rule and derivative notation works?

• In addition to those rules, you will also need to know how to get derivatives of the six trigonometric functions.

• Getting the derivatives for any function starts from the definition of the derivative. However, deriving the six trig functions using this is very tedious and not even practical to discuss it in this course. Therefore you will merely need to memorize all six.

• If you are pretty good at trigonometry, you will know that sine and cosine functions are the only functions from which tangent, secant, cosecant and cotangent come from. To derive tangent, secant, cotangent, and cosecant, we will use their previous theorems we have learned, plus the derivative of the sine and cosine.

• After a very tedious and rigorous process of using the difference quotient and definition of derivative, you will get the derivative of sine and cosine. Not hard. Just be careful when you are positive or negative >=)

• You can find the derivative of tan x by knowing that tan x is no different than…

• (sin x)/(cos x).

• Thus, we can use the quotient rule to find the derivative of tan x.

sin2x+cos2x=1

TRIG IDENTITY

• You can find the derivative of the reciprocal functions (cosecant, secant, and cotangent.) Just know that the secant is the reciprocal of the cosine, the cosecant is the reciprocal of the sine and the cotangent is the reciprocal of tangent.

• At this stage of the game, you could use the quotient rule by sec x = 1/(cos x), csc x = 1/(sin x), and cot x = 1/(tan x).

• Cotangent is done easier using the identity that…

• cot x = (cos x)/(sin x)

• Find the derivative of: x (cot x)

• GIVEN

• PRODUCT RULE

• d(cot x) = -csc2x dx

• d(x)=1dx

• Simplified

• Find the derivative of:

• GIVEN

• QUOTIENT RULE

• d(xsinx) use of PRODUCT RULE

• Say for example, you are given a function, f(x) = (x+1)2. You are asked to find the derivative.

• You could FOIL it and get x2+2x+1 and differentiate term by term and get 2x+2.

• Say you had (x+1)5 and you wanted to find the derivative. Tedious, but you could use the binomial theorem and get x5+5x4+10x3+10x2+5x+1 and differentiate term by term to get 5x4+20x3+30x2+20x+5.

• Try to differentiate sin(5x). There is no real identity to help you with that, so you can’t rename this function in order to simplify it. The argument (the thing inside the parenthesis) is forbidding us to use the typical rules. How are we able to differentiate?

• You do NOT differentiate this way:

• y=(2x+1)2. y’=2(2x+1)=4x+2. (Power Rule)

• If you actually expand the binomial to 4x2+4x+1 and differentiate term by term, you will see that you get 8x+4.

• You can see that between 8x+4 and 4x+2, there is a factor change of 2. The 2, as you could see, is the derivative of 2x+1.

• You can find derivatives of composite functions like y=(x-3)-2 or y=2sin(3x+1) using the chain rule.

• If you consider there are really two functions. f(x) and g(x) and you have a composition of f(g(x)). y=(x-3)-2, g(x), the inner function would be x-3, while the outer function, f(x), would be x-2. In the form of f(g(x)), you would get the original equation.

• The chain rule says that in order to take the derivative of a composition of functions f(g(x)), you differentiate f first, and multiply f’ with g’.

• Looks like this. Let h(x) be a composite differentiable function f(g(x)).

• h’(x) = f’(g(x)) g’(x)

• For y=(2x-3)-2 say that you let 2x-3 = u. you would get u-2.

• Take the derivative of u-2. You will get -2u-3.

• Multiply that with the derivative of u. The derivative of u is 2.

• Given that u is a differentiable function of x, and y is a composition function with u and x, then..

• Find the derivative of y=2sin(2x2-3)

• Given and declaring u.

• y(u): y is written in terms of u.

• CHAIN RULE

• dy/du and du/dx found

• Replacing u and simplifying.

• Most of the time, your argument will be u. If you have a function y(x).

• Then the derivative of y(x) is the derivative of y with respect to u (the argument) multiplied with the derivative of u (the argument) with respect to x.

• A function y(x) which is solved for y in terms of x is called explicit functions. y=sin x is an example of an explicit function since it is solved for y and the function is only in terms of x.

• However, there will be relations which are not functions. They will have a combination of x and y. For example: x2-3y-8=y2. These functions with both x and y together is considered an implicit function. Consider that y is a hidden (implicitly defined) function of x.

• The process of finding derivatives, the slope of the line tangent to a point, of an implicitly defined relation is called implicit differentiation.

• Remember that dy/dx really means “Derivative of y (dy) with respect to x (dx).”

• The derivative of 2x with respect to x is 2.

• But the derivative with respect to y is dy/dx.

• The derivative with the respect to y2 is (2y)(dy/dx). Remember, y is a hidden function of x. We are differentiating with respect to x, not y.

• Remember chain rule? The derivative of u2 with respect to x is 2u (du/dx).

• Differentiate x2+y2=1

• Sum of derivatives

• Differentiate.. y is a function of x, so you have to do chain rule

• Solve for dy/dx algebraically.

• Yes, its okay to have a y in the derivative.

• You could have solve the problem by solving for y.

• Using the chain rule have 1-x2 be u, you can find the derivative.

• Since the entire denominator is equal to y, you can replace the denominator with y.

• We haven’t forgotten our Boston car, Transcendence, and Sriman Vaiyasaki Prabhu! >=) Jaya!

• We wanted to know our speed the moment Vaiyasaki sang, Antara Mandire Jago Jago (track one on the CD).

• Say that one of my devotee friends (Syamasundara) was bored and instead of doing japa, he was graphing the position of the car with respect to the time it took. With the knowledge Sri Krsna gave him, he figure out that the position s(t). {NOTE: The arrow over s just means that direction matters, aka vector quantity}

• Remember that the rate of change of position with respect to time is the speed or velocity. Since it is the rate of change of position versus time, the first derivative of position is the velocity. Plus/minus sign indicate the direction of the speed.

• The rate of the change in speed or velocity is how we accelerate or decelerate.

• Thus, acceleration is the first derivative of velocity and the second derivative of position.

• The plus and minus sign indicate if someone is increasing or decreasing in speed.

• Now, let’s look at Boston trip to the New York Ratha Yatra example.

• Let t be in hours and let s(t) be in miles. In half an hour (t=0.5 hr, when KD listened to the CD), what was the speed and acceleration of the car.

• Velocity is the derivative of position

• Acceleration is the derivative of velocity.

Note, that the acceleration changes with respect to time. That means the entire trip, I drove very jerkily changing the acceleration a lot.

• To find the speed at t=0.5 hr, plug 0.5 hr in for t in the derivative and acceleration formulas. You will see that v = 2.25 mi/hr. a = -8 mi/hr2.

• We are going either right or upwards and a speed of 2.25 mi/hr and we are decelerating at 8 mi/hr2.

• Not too fast for Ratha Yatra spirit! >=(

• Consider that we were looking at a small scale of miles. If we wanted to talk about hundred of miles, you would multiply s(t) by 100 and differentiate thus.

• In order to find the instantaneous rate, the slope of the line tangent to the graph at a certain point, you will need to find the derivative.

• Here are the definitions of average rate using difference quotients.

• The derivative is the limit as the change in x gets infinitesimally close to zero for the difference quotient.

• The equation of a line using the derivative of the function f(x) is at point c.

• Remember, differentiability implies continuity, however, continuity does not imply differentiability.

• Use chain rule whenever you have composite functions.

• Use implicit differentiation when you have more than two variables (x and y).

• Velocity is the first derivative of position with respect to time.

• Acceleration is the first derivative of velocity and second derivative of position both with respect to time.

• As we close the first chapter, I would like to add my final statements.

• You have just attacked one of the most important points of calculus. Derivatives and differentiation. You will be expected to know the concept of the difference quotient.

• Continue practicing how to do derivatives.

• Do not wait till calculus cause you trouble. Continually, practice problems from textbook and continually visit our site for the exam. If you need help please e-mail me at [email protected] I am very good at responding to e-mail at a very timely fashion. I will be more than happy to help out.

• I will pray that you will become successful in this subject. I will tell you honestly, when I took AP calculus AB in high school, it was really hard. Calculus is very integrated with a lot of algebra, geometry, trigonometry, and even previously learned calculus theorems. Make sure you understand everything before moving on. Calculus is NOT meant for rushing!

• Without a graphing calculator, how do I sketch a function to the best possible?

• How do you differentiate exponential and logarithmic functions?

• What about inverse trig functions and inverse functions in general?

• If we are designing a new temple, what dimensions are reasonable to make the Deities visible to everyone?

• If the demigods of math and science calculated the amount of Ganges that fell from the Lotus Feet of Lord Visnu with respect to time, how fast and how much force did the Ganga come to earth?

• Sri Krsna Caitanya Prabhu Nityananda