CHAPTER TWO: DERIVATIVES

CHAPTER TWO:DERIVATIVES

INTRODUCTIONWelcome 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!

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

Therefore: v = s/t (“delta”) means “change in.”

So you would plug 100 miles for s and 2 hours for t. 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.

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!

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.

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 = ???

hr 0

mi 0

hr 5.0hr 5.0

mi 100mi 100

0

0

tt

ss

t

sv

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).

SLOPES

t

sv

x

ym

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

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

t

sv

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.

SLOPESAs 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 t very small, that it is so very very very close to zero!

So you can say that as t 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:t

s

t

lim

0

SLOPE CALCULATIONS

Consider that there are points x, and x+h, separated by the distance h. Thus x = (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, y is just f(x+h) – f(x).

x

CALCULATING RATE

m= y/xTherefore, with the use of functions, the

AVERAGE RATE is defined for f(x) as. This expression is known as the difference quotient.

h

xfhxfr

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.

DEFINITION OF THE DERIVATIVE

The definition of the derivative formula:

h

xfhxfr

h

0lim

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 changeIf f is a function of x, or f(x), then the derivative

can be written as y’ (y-prime), f’(x), or dy/dx.

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.

PROOF OF DERIVATIVEDerivative formula

Replace function with x2.

Squaring x+h.

x2 terms cancel. Factored h.

Cancel h.

Apply the limit (h=0)

h

xfhxfr

h

0lim

h

xhxr

h

22

0lim

h

xhhxxr

h

222

0

2lim

h

hxhr

h

2lim

0

hxrh

2lim0

xr 2

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.

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.

EXAMPLE #2

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

xx f

ANSWER PROCESS

Definition of derivativeApplying 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 h0Simply.

h

xfhxfr

h

0lim

xhxh

xhxxhxr

h

0

lim

h

xhxr

h

0lim

xhxh

hr

h

0lim

xhxr

h

1lim

0

xxxr

2

11

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.

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?

)( 00 xxmyy

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.

)3(69 xy

96 xyOROR

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.

CORNERS CUSPS DISCONTINUITIESVERTICAL TANGENTS

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 >=)

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!

DIFFERENTIATION RULES

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

0cdx

d

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.

1xdx

d

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.

1 nn nxxdx

d

23 3xxdx

d

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.

dx

duv

dx

dvuuv

dx

d

dx

dv

dx

duvu

dx

d

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

2vdxdvu

dxduv

v

u

dx

d

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

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.)

67 355 xxdx

d

dx

dvccv

dx

d

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

2

2

dx

yd

dx

dy

dx

d

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

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??

22

35 22 xxxy

xxxy 435 4

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 23 xxxy

1103 2 xxdx

dy

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.

10

xdx

dy

)( 00 xxmyy

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)

)0(13 xy

3 xy

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?

0

960

960

480

160

40

8

6

6

5

5

4

4

23

3

32

2

4

5

dx

yd

dx

yd

xdx

yd

xdx

yd

xdx

yd

xdx

dy

xy

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.

DERIVATIVE OF SINE AND COSINE

xdx

dy

xy

cos

sin

xdx

dy

xy

sin

cos

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 >=)

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.

xxdx

dyx

xx

dx

dy

x

xxxx

dx

dyx

xxy

22

2

22

2

seccos

1cos

sincos

cos

)sin)((sin))(cos(coscos

sintan

sin2x+cos2x=1TRIG IDENTITY

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).

DERIVATION OF DERIVATIVES OF THE RECIPROCAL FUNCTIONS

xxx

x

xdx

dyx

x

dx

dyx

xx

dx

dyx

xy

tanseccos

sin

cos

1cos

sincos

)sin(10)(coscos

1sec

2

2

xxx

x

xdx

dyx

x

dx

dyx

xx

dx

dyx

xy

cotcscsin

cos

sin

1sin

cossin

)(cos10)(sinsin

1csc

2

2

COTANGENTCotangent is done easier using the identity that…cot x = (cos x)/(sin x)

xxdx

dyx

xx

dx

dy

x

xx

dx

dy

x

xxxx

dx

dyx

xxy

22

2

22

2

22

2

cscsin

1sin

)cos(sin

sin

cossin

sin

))(cos(cos)sin)((sinsin

coscot

PRACTICE PROBLEMS

GIVENPRODUCT RULE

d(cot x) = -csc2x dxd(x)=1dxSimplified

xxxdx

dy

xxxdx

dy

xdx

dxx

dx

dx

dx

dy

xxy

cotcsc

)1)((cot)csc(

)(cotcot)(

cot

2

2

•Find the derivative of: x (cot x)

Another example:

Find the derivative of:

GIVEN

QUOTIENT RULE

d(xsinx) use of PRODUCT RULE

2

322

2

3232

32

sin

sincos)()32)(sin(

sin

sin)()sin(

sin

xx

xxxxxxxxx

dx

dy

xx

xxdxd

xxxxdxd

xx

dx

dy

xx

xxy

xx

xxy

sin

32

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?

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.

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.

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)

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.

CHAIN RULE FORMULA

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

dx

du

du

dy

dx

dy

EXAMPLEFind 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.

)32cos(8

4cos2

sin2

32 )32sin(2

2

22

xxdx

dy

xudx

dydx

du

du

dy

dx

dy

uy

xuxy

CHAIN RULE

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.

IMPLICIT DIFFERENTIATION

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.

IMPLICIT DIFFERENTIATION.

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).

EXAMPLE:Differentiate x2+y2=1Sum 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.

y

x

dx

dy

yx

dx

dy

xdx

dyy

dx

dyyx

dx

d

du

yd

dx

xd

yx

2

12

22

022

)1()()(

122

22

IMPLICIT DIFFERENTIATION

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. y

x

x

x

dx

dyx

x

dx

dy

xy

yx

2

2

2

22

1

12

2

1

1

ORIGINAL PROBLEM

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).

POSITION

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}

153)( 234 ttttts

VELOCITY

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.

dt

sdtv

)(

ACCELERATION

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.

2

2

)(dt

sd

dt

vdta

BACK TO RATHA YATRA PROBLEM

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.

DIFFERENTIATION

Velocity is the derivative of position

Acceleration is the derivative of velocity.

21812)(

5294)(2

23

ttta

ttttv

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

VELOCITY AND ACCELERATION

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.

SUMMARY

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.

h

xfhxfr

x

yr

SUMMARYThe derivative is the limit as the change in x gets

infinitesimally close to zero for the difference quotient.

h

xfhxfxfr

h

0lim'

dx

dy

x

y

x

lim

0

SUMMARY

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.

)()(')( cxxfcfy

dx

du

du

dy

dx

dy

xxdx

d

xxxdx

d

xxxdx

d

xxdx

d

xxdx

d

xxdx

d

2

2

csccot

cotcsccsc

tansecsec

sectan

sincos

cossin

dx

duccu

dx

d

nxxdx

d

vdxdvu

dxduv

v

u

dx

d

dx

duv

dx

dvuuv

dx

ddx

dv

dx

duvu

dx

d

xdx

d

cdx

d

nn

1

2

1

0

SUMMARY

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.

END OF CHAPTER ADVICE 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!

SNEAK PREVIEW OF CHAPTER FOUR.

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

How do you differentiate exponential and logarithmic How do you differentiate exponential and logarithmic functions?functions?

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

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

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

END OF CHAPTER THREE