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CHAPTER TWO: DERIVATIVES

CHAPTER TWO: DERIVATIVES. INTRODUCTION. 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.

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CHAPTER TWO: DERIVATIVES

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  1. CHAPTER TWO:DERIVATIVES

  2. INTRODUCTION • 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!

  3. 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.

  4. BUT… • 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!

  5. AND….. • 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.

  6. Another approach?? • 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 = ???

  7. Time to Spill the Calculus! • 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).

  8. SLOPES -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.

  9. SLOPES • 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.

  10. SLOPES • 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:

  11. SLOPE CALCULATIONS • 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

  12. CALCULATING RATE • 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.

  13. THE INSTANTANEOUS RATE • 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.

  14. DEFINITION OF THE DERIVATIVE • The definition of the derivative formula:

  15. THE DERIVATIVE • 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.

  16. DERIVATIVE • 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.

  17. PROOF OF DERIVATIVE Derivative formula • Replace function with x2. • Squaring x+h. • x2 terms cancel. Factored h. • Cancel h. • Apply the limit (h=0)

  18. TEDIOUS? • 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.

  19. DEFINITION OF DERIVATIVE • 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.

  20. EXAMPLE #2 • Using the definition of the Derivative, find the derivative of….

  21. ANSWER PROCESS • 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.

  22. SLOPES • 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.

  23. TANGENT LINES • 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?

  24. TANGENT LINES • 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

  25. DIFFERENTIABILITY • 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

  26. LOGIC • 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 >=)

  27. SHORTCUTS!!!! • 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!

  28. DIFFERENTIATION RULES 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).

  29. MORE RULES • 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:

  30. OPERATIONS OF DERIVATIVES • 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

  31. JUST GENERAL RULES • 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.)

  32. SECOND DERIVATIVES • 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

  33. EXAMPLE 4: • 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??

  34. EXAMPLE 5 • 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.

  35. EX 5 (continued) • 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.

  36. EX 5. (continued) • 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)

  37. EXAMPLE 6 • 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?

  38. TRIG DERIVATIVES • 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.

  39. DERIVATIVE OF 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 >=)

  40. DERIVATIVE OF TANGENT • 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

  41. DERIVATIVE OF RECIPROCAL FUNCTIONS • 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).

  42. DERIVATION OF DERIVATIVES OF THE RECIPROCAL FUNCTIONS

  43. COTANGENT • Cotangent is done easier using the identity that… • cot x = (cos x)/(sin x)

  44. PRACTICE PROBLEMS • Find the derivative of: x (cot x) • GIVEN • PRODUCT RULE • d(cot x) = -csc2x dx • d(x)=1dx • Simplified

  45. Another example: • Find the derivative of: • GIVEN • QUOTIENT RULE • d(xsinx) use of PRODUCT RULE

  46. CHAIN 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?

  47. WRONG WAY • 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.

  48. CHAIN RULE • 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.

  49. THE CHAIN RULE • 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)

  50. CHAIN RULE • 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. • So your answer will be -4(x-3)-3.

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