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FUNCTION :

Physical laws are nothing but statement regarding the way in which the certain physical quantities depend onother physical quantity / quantities. ex energy of a bullet depends on its mass and its speed. so we need tolearn something we call function

Function is a rule of relationship between two variables in which one is assumed to be dependentand the other independent variable, for example :

e.g. The temperatures at which water boils depends on the elevation above sea level (the boiling point drops asyou ascend). Here elevation above sea level is the independent & temperature is the dependent variable

e.g. The interest paid on a cash investment depends on the length of time the investment is held. Heretime is the independent and interest is the dependent variable.

In each of the above example, value of one variable quantity (dependent variable) , which we might call y,depends on the value of another variable quantity (independent variable), which we might call x. Since thevalue of y is completely determined by the value of x, we say that y is a function of x and represent it math-ematically as y = f(x).Here f represents the function, x the independent fx

Input (Domain)

Ouput(Range)

f(x)

variable & y is the dependent variable.

All possible values of independent variables (x) are called domain of function.All possible values of dependent variable (y) are called range of function.

Think of a function f as a kind of machine that produces an output value f(x) in its range whenever we feed it aninput value x from its domain (figure).

When we study circles, we usually call the area A and the radius r. Since area depends on radius, wesay that A is a function of r, A = f(r) . The equation A = r2 is a rule that tells how to calculate a unique(single) output value of A for each possible input value of the radius r.A = f(r) = r2 . (Here the rule of relationship which describes the function may be described as square& multiply by ).If r = 1 A = if r = 2 A = 4 if r = 3 A = 9The set of all possible input values for the radius is called the domain of the function. The set of all outputvalues of the area is the range of the function.

We usually denote functions in one of the two ways :

1. By giving a formula such as y = x2 that uses a dependent variable y to denote the value of the function.

2. By giving a formula such as f(x) = x2 that defines a function symbol f to name the function.

Strictly speaking, we should call the function f and not f(x),

y = sin x. Here the function is sine, x is the independent variable.

A mathematical function is just a law which governs the inter dependence of variable quantities it does not implythe existence of of cause and effect between theme.g. Boyles law PV = constant may be solved either for P or V as a function of other variable

P = VC or V = P

C

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One Such Interdependence of Two Physical Quantities :

Galileo developed a formula for a body’s velocity during free fall by

rolling balls from rest down increasingly steep inclined planks andlooking for a limiting formula that would predict a ball’s behaviourwhen the plank was vertical and the ball fell freely(part a of the accompanying (figure). He found that, for any givenangle of the plank, the ball’s velocity t seconds into the motion wasa constant multiple of t. That is, the velocity was given by a formulaof the form v = kt.The value of the constant k depended on the inclination of the plank.

In modern notation (part b of the figure), with distance in meters and time in seconds, what Galileo determinedby experiment was that, for any given angle , the ball’s velocity t seconds after it is set into the roll was

v = 9.8 (sin ) t m/sec.thus he established the interdependence of two variablesvelocity and time . 9.8 sin becomes the proportionality factordeciding the change of v with respect to t. Say the angle of inclinationis 30° then 9.8 sin = 9.8 × 1/2 = 4.9 .The graph shows the st line behaviour of the relation / function v = 4.9t

vt

is constant and is equal to tan30 ( here it is the slope of the st line)

Ex. 10 The volume V of a ball (solid sphere) of radius r is given by the function V(r) = 3)r()3/4(

The volume of a ball of radius 3m is ?

Sol. V(3) = 3)3(3/4 = 36 m3 .

Ex. 11 Suppose that the function F is defined for all real numbers r by the formula F(r) = 2(r – 1) + 3.Evaluate F at the input values 0, 2, x + 2, and F(2).

Sol. In each case we substitute the given input value for r into the formula for F :F(0) = 2(0 – 1) + 3 = – 2 + 3 = 1 ;F(2) = 2(2 – 1) + 3 = 2 + 3 = 5F(x + 2) = 2(x +2 – 1) + 3 = 2x + 5 ;F(F(2)) = F(5) = 2(5 – 1) + 3 = 11.

Ex. 12 A function (x) is defined as (x) = x2 + 3, Find 0) , (1), x2), (x+1) and 1)).Sol. (0) = 02 + 3 = 3 ;

(1) = 12 + 3 = 4 ;(x2) = (x2)2+3 = x4+3(x+1) = (x + 1)2 + 3 = x2 + 2x + 4 ;((1)) = 4) = 42+3 = 19

Ex. 13 If function F is defined for all real numbers x by the formula F(x) = x2 .Evaluate F at the input values 0,2, x + 2 and F(2)

Sol. F(0) = 0 ;F(2) = 22 = 4 ;F(x+2) = (x+2)2 ;F(F(2)) = F(4) = 42 = 16

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AVERAGE RATES OF CHANGE :

Lets take a case where a disk sliding along an inclined plane is found to have its position at anyinstant given by x = 3t2 + 1, where x is in meters and t is in second.Here x is a function of time ; x = f(t) = 3t2 + 1.Now, we find the average velocity for the following three time intervals(a) 2 and 3 s (b) 2 and 2.1 s (c) 2 to 2.00001

For t = 2 and 3 s, we have t = 1 sFor t = 3 s, x = 3(3)2 + 1 = 28 mFor t = 2 s, x = 3(2)2 + 1 = 13 m

and x = 28 m – 13 m = 15 m

Thus, vavg = x 15t

m/1 s = 15 m/s

For t = 2 and 2.1 s, we have t = 0.1 sFor 2.1 s, x = 3(2.1)2 + 1 = 14.23 mFor t = 2 s, x = 3(2)2 + 1 = 13 m

and x = 1.23 m

Thus vavg = x 1.23t

m/0.1 s = 12.3 m/s

You may verify that for t = 2.00001 s, vavg = 12.00003 m/s

But still these are average velocities over and interval. But what is the velocity AT the instant t = 2 s ??

FROM FINITE TO INFINITELY SMALL CHANGE :

The infinitely small difference means very-very small difference. And this difference is represented by ‘d’ notationinstead of ‘’.For example infinitely small difference in the values of y is written as ‘dy’if y2 = 100 and y1 = 99.99999999........then dy = 0.000000...................00001Now, back to the function f(t) = 3t2 + 1

FROM AVERAGE RATE TO RATE HERE AND NOW :

First lets find the average velocity over time interval t in generalx = 3t2 + 1x + x = 3 (t + Dt)2 + 1 = 3t2 + 3t2 + 3.2tt + 1 x = 3t2 + 3.2t t

tx = 3t + 3.2t (For any general interval t this is the average velocity)

0tlim t2.3t

x

(here as t tends to 0 the first term 3t tends to 0 and so the limiting value of the ratio x/t becomes 3.2t)

dxdt = 6t

at the instant t = 2 s dxdt = 6.2 = 12 m/s

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Similarly at t = 3 s dxdt = 6.3 = 18 m/s

In other words, the instantaneous velocity v equals the limiting value of the ratio x/T as Tapproaches zero but it cannot be zero as t being zero will mean x also becoming zero and velocity becomingundefined.

v = 0tlim t

x (read it as the limiting value of the ratio t

x as t tends to zero)

In calculus notation, this limit is called the derivative of x with respect to t, written as dx/dt.

v = 0tlim dt

dxtx

The idea here was to make an infinitesimal distance and the corresponding infinitesimal time form the ratioand watch what happen to the ratio as the time gets smaller and smaller and smaller. This was somethingunknown to you till now. and This idea was invented by Newton and leibinitz which is calculus.

Calculus is thus simply very advanced algebra and geometry that has been tweaked to solve moresophisticated problems.

Question. How much energy does the man use to push the crate up the incline?The regular way :

For the straight incline, the man pushes with an unchanging force, and the crate goes up the incline at anunchanging speed. With some simple physics formulas and regular math (including algebra and trig), youcan compute how many calories of energy are required to push the crate up the incline.

The “calculus” way :

For the curving incline, on the other hand, things are constantly changing. The steepness of the incline is changing — and notjust in increments like it’s one steepness for the first 10 feetthen a different steepness for the next 10 feet — it’s constantlychanging. And the man pushes with a constantly changing force — thesteeper the incline, the harder the push. As a result, the amount ofenergy expended is also changing, not every second or everythousandth of a second, but constantly changing from one

moment to the next. That’s what makes it a calculus problem.

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It is a mathematical way to express something that is ……CHANGING! It could be anything??But here is the cool part :

Calculus allows you to ZOOM in on a small part of the problem and apply the “regular” math tools.

DEFINING DIFFERENTIATION IN GENERAL :

Given an arbitrary function y = f(x) we calculate the average rate of change of y with respect to x over theinterval (x , x + x) by dividing the change in value of y, i.e.y = f(x + x) – f(x), by length of interval x overwhich the change occurred.

The average rate of change of y with respect to x over the interval [x, x + x] = x

)x(f)xx(fxy

Geometrically, xy

= PRQR

= tan = Slope of the line PQ

x

y

x + x

y+ y

P

Q

y

xR

In triangle QPR tan = yx

therefore we can say that average rate of change of y with respectto x is equal to slope of the line joining P & Q.

We know that, average rate of change of y w.r.t. x is xy

= x

)x(f–)xx(f .

If the limit of this ratio exists as x 0, then it is called the derivative of given function f(x) and is denoted as

f ’(x) = dxdy

= lim

0x x

)x(f–)xx(f

GEOMETRICAL MEANING OF DIFFERENTIATION :

The geometrical meaning of differentiation is very much useful in the analysis of graphs in physics. To understandthe geometrical meaning of derivatives we should have knowledge of secant and tangent to a curve

Secant and tangent to a curve :

Secant : A secant to a curve is a straight line, which intersects the curve at any two points. P

q Secant

x

y

Tangent:-A tangent is a straight line, which touches the curve at a particular point.

x

y

x + x

y+ y

P

Q

y

xR

Figure - 1

Tangent is a limiting case of secant which intersects the curve at twooverlapping points. In the figure-1 shown, if value of x is gradually reducedthen the point Q will move nearer to the point P. If the process is continuouslyrepeated (Figure - 2) value of x will be infinitely small and secant PQ to thegiven curve will become a tangent at point P .

Therefore

xy

0x = dx

dy= tan

6

we can say that differentiation of y with respect to x,i.e.

dxdy

is

x

y

x+ x

y+ y

P

Q

Q

Q

Q

y

xR

Figure - 2

equal to slope of the tangent at point P (x, y) or tan = dxdy

(From fig. 1, the average rate of change of y from x to x + x isidentical with the slope of secant PQ.)

QUESTION TO CONCLUDE :

Suppose we have functions of one variable f(x). What does derivatives df(x)dx

do for us ?

THE FINAL ANSWER :

It tells us how rapidly the function varies when x is changed by tiny amount dx

df = df(x)dx dx here the derivative is the proportionality factor which decides the change of function f

when x is changed by dx. Lets take an exampleThe area A of a square of length L is A = L2. if we change L to L + L, the area will change from A to A + A

A + A = (L + L)2

= L2 + 2L L + (L)2

or, A = 2L(L) + (L)2

or,A 2L LL

Now if L is made smaller and smaller, 2L + L will approach 2L.

Thus, dAdL = 0L

lim L2L

A

ONE PROCESS MANY NOTATION :There are many ways to denote the derivative of a function y = f(x). Besides f ’(x), the most common notationsare these :

y´ “y p rim e” o r “y dash ” N ice and brie f bu t does no t nam e the ind epende nt variab le

dxdy

“dy b y dx” N am es the variab les a nd u ses d fo r deriva tive

dxdf

“d f by dx” E m phas izes the func tion ’s nam e

)x(fdxd

“d b y dx o f f” E m phas izes the ide a tha t d iffe ren tia tion is an opera tio n perfo rm ed on f.

D xf “dx o f f” A com m on opera tor n o ta tio n y “y do t” O ne o f N ew ton ’s no ta tio ns , no w com m on fo r

tim e deriva tives i.e . d tdy .

f´(x ) f dash x M ost com m on nota tion , it n am es the ind epende nt variab le a nd E m phas ize the func tion ’s nam e

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RULES FOR DIFFERENTIATION :

Rule No. 1 : Derivative of a constant

The first rule of differentiation is that the derivative of every constant function is zero.

If c is constant, then cdxd

= 0.

Ex. 14 0)8(dxd

, 021

dxd

, 03dxd

Rule No. 2 : Power rule

If n is a real number, then 1nn nxxdxd .

To apply the power Rule, we subtract 1 from the original exponent (n) and multiply the result by n.

Ex. 15 f x x2 x3 x4 ....

f' 1 2x 3x2 4x3 ....

Ex. 16 (i)

x1

dxd

= dxd

(x–1) = (–1)x–2 = 2x

1

(ii)

3x4

dxd

= dxd4 (x–3) = 4(–3)x–4 =

4x12

.

Ex. 17 (a) )x(dxd 2/1 = 2/1x

21 =

x21

| | Function defined for x 0 derivative defined only for x > 0

(b) )x(dxd 5/1 = 5/4–x

51

| | Function defined for x 0 derivative not defined at x = 0

Rule No. 3 : The constant multiple rule

If u is a differentiable function of x, and c is a constant, then )cu(dxd = dx

duc

In particular, if n is a positive integer, then )cx(dxd n = cn xn–1

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Ex. 18 The derivative formula

)x3(dxd 2 = 3(2x) = 6x

says that if we re-scale the graph of y = x2 by multiplying each y–coordinate by 3, then we multiply the slope ateach point by 3.

Ex. 19 A useful special caseThe derivative of the negative of a differentiable function is the negative of the function’s derivative. Rule 3 with c= –1 gives.

)u(dxd

= )u1(dxd

= –1 )u(dxd

= dxd

(u)

Rule No. 4 : The sum rule

The derivative of the sum of two differentiable functions is the sum of their derivatives.

If u and v are differentiable functions of x, then their sum u + v is differentiable at every point where u and v areboth differentiable functions is their derivatives.

)vu(dxd

= ]v)1(u[dxd

= dxdv

dxdu

dxdv)1(

dxdu

The Sum Rule also extends to sums of more than two functions, as long as there are only finitely manyfunctions in the sum. If u1, u2,.........un are differentiable at x, then so is u1 + u2 + ........+ un , and

)u.....uu(dxd

n21 = dxdu.......

dxdu

dxdu n21 .

Ex. 20 (a) y = x4 + 12x (b) y = x3 + 34

x2 – 5x + 1

)x12(dxd)x(

dxd

dxdy 4 )1(

dxd)x5(

dxdx

34

dxdx

dxd

dxdy 23

= 4x3 + 12 = 3x2 + 34

.2x – 5 + 0 = 3x2 + 38

x – 5.

Notice that we can differentiate any polynomial term by term, the way we differentiated the polynomials in aboveexample.Rule No. 5 : The product rule

If u and v are differentiable at x, then so is their product uv, and dxd

(uv) = dxduv

dxdvu .

The derivative of the product uv is u times the derivative of v plus v times the derivative of u. In prime notation(uv)’ = uv’ + vu’.

While the derivative of the sum of two functions is the sum of their derivatives, the derivative of the product of twofunctions is not the product of their derivatives. For instance,

dxd

(x.x) = dxd

(x2) = 2x, while dxd

(x) . dxd

(x) = 1.1 = 1.

Ex. 21 Find the derivatives of y = (x2 + 1) (x3 + 3).Sol. From the product Rule with u = x2 + 1 and v = x3 + 3, we find

9

)]3x)(1x[(dxd 32 = (x2 + 1) (3x2) + (x3 + 3) (2x)

= 3x4 + 3x2 + 2x4 + 6x = 5x4 + 3x2 + 6x.

Example can be done as well (perhaps better) by multiplying out the original expression for y and differentiatingthe resulting polynomial. We now check : y = (x2 + 1) (x3 + 3) = x5 + x3 + 3x2 + 3

dxdy

= 5x4 + 3x2 + 6x.

This is in agreement with our first calculation.There are times, however, when the product Rule must be used. In the following examples. We have onlynumerical values to work with.

Ex. 22 Let y = uv be the product of the functions u and v. Find y’(2) if u’(2) = 3, u’(2) = –4, v(2) = 1, and v’(2) = 2.Sol. From the Product Rule, in the form

y’ = (uv)’ = uv’ + vu’ ,we have y’(2) = u(2) v’(2) + v(2) u’ (2) = (3) (2) + (1) (–4) = 6 – 4 = 2.

Rule No. 6 : The quotient rule

If u and v are differentiable at x, and v(x) 0, then the quotient u/v is differentiable at x,

and

vu

dxd

= 2vdxdvu

dxduv

Just as the derivative of the product of two differentiable functions is not the product of their derivatives,the derivative of the quotient of two functions is not the quotient of their derivatives.

Ex. 23 Find the derivative of y = 1t1t

2

2

Sol. We apply the Quotient Rule with u = t2 – 1 and v = t2 + 1 :

dtdy

= 22

22

)1t(t2).1t(t2).1t(

2v)dt/dv(u)dt/du(v

vu

dtd

= 22

33

)1t(t2t2t2t2

= 22 )1t(

t4

.

Rule No. 7 : derivative of sine function

xcos)x(sindxd

Ex. 24 (a) y = x2 – sin x : dxdy

= )x(sindxdx2 Difference Rule

= 2x – cos x

(b) y = x2 sin x : dxdy

= x2

dxd

(sin x) + 2x sin x Product Rule

= x2 cos x + 2x sin x

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(c) y = x

xsin: dx

dy =

2x

1.xsin)x(sindxd.x

Quotient Rule

= 2xxsinxcosx

.

RULE NO. 8 : DERIVATIVE OF COSINE FUNCTION

xsin)x(cosdxd

Ex. 25 (a) y = 5x + cos x

dxdy

= dxd

(5x) + dxd

(cosx) Sum Rule

= 5 – sin x(b) y = sinx cosx

dxdy

= sinx dxd

(cosx) + cosx dxd

(sinx) Product Rule

= sinx (– sinx) + cosx (cosx) = cos2 x – sin2 x

RULE NO. 9 : DERIVATIVES OF OTHER TRIGONOMETRIC FUNCTIONS

Because sin x and cos x are differentiable functions of x , the related functions

tan x = xcos xsin

; sec x = xcos1

cot x = xsin xcos

; cosec x = xsin1

are differentiable at every value of x at which they are defined. There derivatives. Calculated from the QuotientRule, are given by the following formulas.

dxd

(tan x) = sec2 x ; dxd

(sec x) = sec x tan x

dxd

(cot x) = – cosec2 x ; dxd

(cosec x) = – cosec x cot x

Ex. 26 Find dy / dx if y = tan x .

Sol.dxd (tan x) =

dxd

xcos xsin

= xcos

x)(cos dxd x sin – )x(sin

dxdxcos

2

= xcos

x)(–sin x sin x – cos x cos2

= xcos

xsinxcos2

22 =

xcos1

2 = sec2 x

11

Ex. 27 (a) dxd

(3x + cot x) = 3 + dxd

(cot x) = 3 – cosec2 x

(b) dxd

x sin2

= dxd

(2cosec x) = 2 dxd

(cosec x)

= 2 (–cosec x cot x) = – 2 cosec x cot x

RULE NO. 10 : DERIVATIVE OF LOGARITHM AND EXPONENTIAL FUNCTIONS

x1xlog

dxd

e xx eedxd

Ex. 28 y = ex . loge (x)

xedxd

dxdy

. log (x) +. dxd

[loge (x)] ex dxdy

= ex . loge (x) + xex

RULE NO. 11 : CHAIN RULE OR “OUTSIDE INSIDE” RULE

dxdy = du

dy.

dxdu

It sometimes helps to think about the Chain Rule the following way. If y = f(g(x)),

)x('g)].x(g['fdxdy

.

In words : To find dy/dx , differentiate the “outside” function f and leave the “inside” g(x) alone ; then multiply bythe derivative of the inside.

We now know how to differentiate sin x and x2 – 4, but how do we differentiate a composite like sin (x2 – 4)? Theanswer is, with the Chain Rule, which says that the derivative of the composite of two differentiable functions isthe product of their derivatives evaluated at appropriate points. The Chain Rule is probably the most widely useddifferentiation rule in mathematics. This section describes the rule and how to use it. We begin with examples.

Ex. 29 The function y = 6x – 10 = 2(3x – 5) is the composite of the functions y = 2u and u = 3x – 5. How are thederivatives of these three functions related ?

Sol. We have dxdy

= 6 , dudy

= 2 , dxdu

= 3.

Since 6 = 2 . 3 ,dxdy = du

dy.

dxdu

Is it an accident thatdxdy = du

dy.

dxdu

?

If we think of the derivative as a rate of change, our intuition allows us to see that this relationship is reasonable.For y = f(u) and u = g(x) , if y changes twice as fast as u and u changes three times as fast as x, then we expecty to change six times as fast as x.

Ex. 30 Let us try this again on another function.y = 9x4 + 6x2 + 1 = (3x2 + 1)2

is the composite of y = u2 and u = 3x2 + 1 .

12

Sol. Calculating derivatives. We see that

dudy .

dxdu = 2 u . 6x = 2(3x2 +1) . 6x = 36x3 + 12 x

and dxdy

= dxd

(9x4 + 6x2 + 1) = 36x3 + 12 x

Once again,dudy .

dxdu =

dxdy

The derivative of the composite function f(g(x)) at x is the derivative of f at g(x) times the derivative of g at x.

Ex. 31 Find the derivative of y = 1x2 Sol. Here y = f(g(x)) , where f(u) = u and u = g (x) = x2 + 1. Since the derivatives of f and g are

f ’ (u) = u21

and g ’(x) = 2x ,

the Chain Rule gives

dxd

dxdy

f(g(x)) = f ’(g(x)) . g ’(x) = )x(g2

1. g’(x) =

1x2

12

. (2x) = 1x

x2

.

Ex. 32

dxd sin(x +x) = cos(x + x) . (2x + 1) 2 2

outsidederivative ofthe outside

Insideleft alone

Inside derivativeof the Inside

Ex. 33 We sometimes have to use the Chain Rule two or more times to find a derivative.Here is an example. Find the derivative of g(t) = tan (5 – sin 2t)

Sol. g ’(t) = dtd

(tan(5– sin 2t)

= sec2 (5– sin 2t) . dtd

(5 – sin 2t)

Derivative of tan u with u = 5 – sin 2t

Derivative of 5 – sin u with u = 2t

= sec2 (5 – sin 2t) . (0 – (cos 2t) . dtd

(2t) = sec2 (5 – sin 2t) . ( – cos 2t) . 2 = –2(cos 2t) sec2 ( 5 – sin 2t)

Ex. 34 (a)

Function definedon [–1 , 1]

dxd (1 – x ) = (1–x ) (–2x) 2 1/4 2 –3/4

41

u = 1 – x2 and n = 1/4

=

derivative defined only on (–1 ,1)

4/32 )x–1(2x–

(b) –1/5 x)(cosdxd

= – 5/6)x(cos51

dxd

(cos x)

= – 5/6)x(cos51 (– sin x) = 5

1 sin x (cos x)–6/5

13

(c)

xsine

dxd

= xsindxd·e xsin = x

dxd·xcose xsin

= x2

1·xcose xsin

= xcos·ex2

1 xsin

(d) dxd

5xsin 2 = cos 5x2 dx

d5x2

= cos 5x2 5x2

12

· dxd

(x2 + 5)

= cos 5x2 · x2·5x2

12

= 5x

x2

cos 5x2

(e) dxd

sin 2x = cos 2x dxd

(2x) = cos 2x . 2 = 2 cos 2x

(f) dtd

(A sin (t + )

= A cos (t + ) dtd

(t + ) = A cos (t + ). .

= A cos (t + )

RULE NO. 12 : POWER CHAIN RULE

If u(x) is a differentiable function and where n is a Real number , then un is differentiable and

nudxd

= dxdunu 1n , n R

Ex. 35 (a) xsindxd 5 = 5 sin4 x dx

d (sin x)

= 5 sin4 x cos x

(b) dxd

(2x + 1)–3 = – 3(2x + 1)–4 dxd

(2x + 1)

= – 3(2x + 1)–4 (2) = – 6 (2x + 1)–4

(c) dxd

(5x3 – x4)7 = 7(5x3 – x4)6 dxd

(5x3 – x4)

= 7(5x3 – x4)6 (5 . 3x2 – 4x3)= 7(5x3 – x4)6 (15x2 – 4x3)

(d) dxd

2–x31

= dxd

(3x – 2)–1 = – 1(3x – 2)–2 dxd

(3x – 2) = –1 (3x – 2)–2 (3) = – 2)2–x3(3

In part (d) we could also have found the derivative with the Quotient Rule.

14

Ex. 36 dxd

(Ax + B)n

Sol. Here u = Ax + B , dxdu

= AA

dxd

(Ax + B)n = n(Ax + B)n–1 . AA

RULE NO. 13 : RADIAN VS. DEGREES

dxd

sin(x°) = dxd

sin

180

x = 180

cos

180

x = 180

cos(x°) .

DOUBLE DIFFERENTIATION :

If f is differentiable function, then its derivative f ’ is also a function, so f ’ may have a derivative of its own, denotedby (f ’ )’ = f ’’ . This new function f ’’ is called the second derivative of f because it is the derivative of the derivativeof f . Using Leibniz notation, we write the second derivative of y = f (x) as

2

2

dxyd

dxdy

dxd

Another notation is f ’’ (x) = D2 f (x) = D2f(x)

INTERPRETATION OF DOUBLE DERIVATIVEWe can interpret f ’’ (x) as the slope of the curve y = f ’(x) at the point (x, f ’(x)). In other words, it is the rate ofchange of the slope of the original curve y = f (x) .In general, we can interpret a second derivative as a rate of change of a rate of change. The most familiarexample of this is acceleration, which we define as follows.If s = s(t) is the position function of an object that moves in a straight line, we known that its first derivativerepresents the velocity v(t) of the object as a function of time :

v (t) = s’ (t) = dtds

The instantaneous rate of change of velocity with respect to time is called the acceleration a(t) of the object.Thus, the acceleration function is the derivative of the velocity function and is therefore the second derivative ofthe position function :

a (t) = v ’ (t) = s’’ (t) or in Leibniz notation, a = dtdv

= 2

2

dtsd

Ex. 37 If f (x) = x cos x , find f ’’ (x).Sol. Using the Product Rule, we have

f ’ (x) = x dxd

(cos x) + cos x dxd

(x)

= – x sin x + cos xTo find f ’’ (x) we differentiate f ’ (x) :

f ’’ (x) = dxd

(–x sin x + cos x)

= – x dxd

(sin x) + sin x dxd

(– x) + dxd

(cos x)

= – x cos x – sin x – sin x = – x cos x – 2 sin x

15

Ex. 38 The position of a particle is given by the equation

s = f (t) = t3 – 6t2 + 9twhere t is measured in seconds and s in meters.

Find the acceleration at time t. What is the acceleration after 4 s ?Sol. The velocity function is the derivative of the position function :

s = f (t) = t3 – 6t2 + 9t v(t) = dtds

= 3 t2 – 12 t + 9

The acceleration is the derivative of the velocity function :

a (t) = 2

2

dtsd

= dtdv

= 6t – 12 a(4) = 6(4) – 12 = 12 m/s2

APPLICATION OF DERIVATIVES :

Differentiation as a rate of change

dxdy

is rate of change of ‘y’ with respect to ‘x’ :

For examples :

(i) v = dtdx

this means velocity ‘v’ is rate of change of displacement ‘x’ with respect to time ‘t’

(ii) a = dtdv

this means acceleration ‘a’ is rate of change of velocity ‘v’ with respect to time ‘t’ .

(iii) F = dtdp

this means force ‘F’ is rate of change of momentum ‘p’ with respect to time ‘t’ .

(iv) = dtdL

this means torque ‘ ’ is rate of change of angular momentum ‘L’ with respect to time ‘t’

(v) Power = dtdW

this means power ‘P’ is rate of change of work ‘W’ with respect to time ‘t’

(vi) = dtdq

this means current ‘’ is rate of flow of charge ‘q’ with respect to time ‘t’

Ex 39. From the curve given in figure, find dydx at x = 2, 6 and 10.

Sol. The tangent to the curve at x = 2 is AC. Its slope is tan 1 = AB 5BC 4

Thus, dy 5 at x 2dx 4

The tangent to the curve at x = 6 is parallel to the X - axis.

16

Thus, dy tan 0dx

at x = 6.

The tangent to the curve at x = 10 is DF. Its slope is

tan2 = DE 5–EF 4

Thus, dy 5–dx 4

at x = 10

If we are given the graph of y versus x, we can find dydx

at any point of the curve by drawing the tangent at that

point and finding its slope. Even if the graph is not drawn and the algebraic relation between y and x is given in

the form of an equation, we can find dydx algebraically..

Ex. 40 Does the curve y = x4 – 2x2 + 2 have any horizontal tangents ? If so, where ?Sol. The horizontal tangents, if any, occur where the slope dy/dx is zero. To find these points. We

Calculate dy/dx : dxdy

= dxd (x4 – 2x2 + 2) = 4x3 – 4x

and then solve the equation : dxdy = 0 for x : 4x3 – 4x = 0

4x(x2 – 1) = 0x = 0,1, –1

The curve y = x4 – 2x2 + 2 has horizontal tangents at x = 0,1, and –1.

The corresponding points on the curve are(0,2) (1,1) and (– 1,1).

See figure.

Ex. 41 The area A of a circle is related to its diameter by the equation A = 4

D2.

How fast is the area changing with respect to the diameter when the diameter is 10 m?

Sol. The (instantaneous) rate of change of the area with respect to the diameter is

dDdA

= D24

= 2D

When D = 10 m, the area is changing at rate (/2) 10 = 5 m2/m. This means that a small change D m in thediameter would result in a change of about 5 D m2 in the area of the circle.

Ex. 42 The volume of a sphere is given by

3R34V

where R is the radius of the sphere. (a) Find the rate of change of volume with respect to R. (b) Find the changein volume of the sphere as the radius is increases from 20.0 cm to 20.1 cm. Assume that the rate does notappreciably change between R = 20.0 cm to R = 20.1 cm.

Sol. V = 34R3

or dRdV

= 34 dr

d(R)3 = 3

4. 3R2 = 4 R2.

(b) At R = 20 cm, the rate of change of volume with the radius is

dRdV

= 4R2 = 4 (400 cm2)

17

= 1600 cm2.The change in volume as the radius changes from 20.0 cmt o 20.1 cm is

V = RdRdV

= (1600 cm2) (0.1 cm) = 160 cm3.

Ex. 43 Changing voltage. The voltage V (volts), current (amperes) and resistance R (ohms) of an electric circuit likethe one shown here are related by the equation V = R. Suppose that V is increasing at the rate of 1 volt/secwhile is decreasing at the rate of 1/3 amp/sec. Let t denote time in seconds.(a) What is the value of dV/dT ?(b) What is the value of dI/dt ?(c) Find the rate at which R is changing when V = 12 volts and = 2 amp. Is R increasing, or decreasing ?

Sol. V = R (This is the function which shows the interdependence of variable V, and R)We have to find the relation between their respective rates of changedV d dRRdt dt dt

dV 1 volt / sdt

( V is increasing therefore it is positive)

d 1– amp/ sdt 3 ( I is decreasing therefore it is negative)

1 = – 1 dRR 23 dt

V 12R R 62

1 = – 1 dR6 23 dt

3 = dR 2dt

3 dR2 dt (It is positive R is increasing at the rate of 3/2 ohm/s)

Ex. 44 Experimental and theoretical investigations revealed that the distancea body released from rest falls in time t is proportional to the squareof the amount of time it has fallen. We express this by saying that 0

51015202530354045

s (meters)t = 0

t (seconds)

t = 2

t = 3

A ball bearing falling from rest

s = 21

gt2 ,

where s is distance and g is the acceleration due to Earth’sgravity. This equation holds in a vacuum, where there is no air resistance,but it closely models the fall of dense, heavy objects in air. Figure shows thefree fall of a heavy ball bearing released from rest at time t = 0 sec.(a) How many meters does the ball fall in the first 2 sec?(b) What is its velocity, speed, and acceleration then?

Sol. (a) The free–fall equation is s = 4.9 t2.During the first 2 sec. the ball falls

s(2) = 4.9(2)2 = 19.6 m,

18

(b) At any time t, velocity is derivative of displacement :

v(t) = s’(t) = dtd

(4.9t2) = 9.8 t.

At t = 2, the velocity is v(2) = 19.6 m/secin the downward (increasing s) direction. The speed at t = 2 is

speed = |v(2)| = 19.6 m/sec. a = 2

2

dtsd

= 9.8 m/s2

Ex. 45 A body hanging from a spring (fig.) is stretched 5 units beyond its rest position and released at time t = 0 tooscillate up and down. Its position at any later time t is

s = 5 cos tWhat are its velocity and acceleration at time t ?

–5

0

5 Position at t = 0

Rest position

Sol. We havePosition : s = 5 cos t

Velocity : v = dtds

= dtd

(5 cos t) = dtd5 (cos t) = – 5 sin t

Acceleration : a = dtdv

= dtd

(– 5 sin t) = –dtd5 (sin t) = – 5 cos t

Ex. 46 A hot air balloon rising straight up from a level field is tracked by a range finder 500 ft from the lift-off point. At themoment the range finder’s elevation angle is /4, the angle is increasing at the rate of 0.14 rad/min. How fast isthe balloon rising at the moment ?

Sol. We answer the question in six steps.

Rangefinder

dydt

dt

y = ?

500 feet

when =

4

d0.14 rad / min=

when =

4

Ballon

Step 1 : Draw a picture and name the variables and constants (Figure ) . The variables in the picture are = the angle the range finder makes with the ground (radians)y = the height of the balloon (feet).We let t represent time and assume and y to be differentiable functions of t.The one constant in the picture is the distance from the range finder to the lift-off point (500 ft.) There isno need to give it a special symbol s.

Step 2 : Write down the additional numerical information.

dtd

= 0.14 rad/min when = / 4

Step 3 : Write down what we are asked to find. We want dy/dt when = /4 .

Step 4 : Write an equation that relates the variables y and . 500y

= tan , or y = 500 tan

19

Step 5 : Differentiate with respect to t using the Chain Rule. The result tells how dy/dt (which we want) is relatedto d/dt (which we know).

dtdy

= 500 sec2 dtd

Step 6 : Evaluate with = /4 and d / dt = 0.14 to find dy/dt.

dtdy

= 500 22 (0.14) = (1000) (0.14) = 140 (sec 4

= 2 )

At the moment in question, the balloon is rising at the rate of 140 ft./min.

Ex. 47 A police cruiser, approaching a right-angled intersection from the north, is chasing a speeding car that hasturned the corner and is now moving straight east. When the Cruiser is 0.6 mi north of the intersection and thecar is 0.8 mi to the east, the police determine with radar that the distance between them and the car isincreasing at 20 mph. If the cruiser is moving at 60 mph at the instant of measurement, what is the speed of thecar?

Sol. We carry out the steps of the basic strategy.

Situations whenx=0.8, y=0.6

dsdt =20

xdxdt =?

dydt =–60

0

y

Step 1 : Picture and variables. We picture the car and cruiser in the coordinate plane, using the positive x-axisas the eastbound highway and the positive y-axis as the northbound highway (Figure) . We let t representtime and set x = position of car at time t.y = position of cruiser at time t, s = distance between car and cruiser at time t.We assume x, y and s to be differentiable functions of t.

x = 0.8 mi, y = 0.6 mi , dtdy

= – 60 mph , dtds

= 20 mph

(dy/dt is negative because y is decreasing.)

Step 2 : To find : dtdx

Step 3 : How the variables are related : s2 = x2 + y2 Pythagorean theorem

(The equation s = 22 yx would also work.)

Step 4 : Differentiate with respect to t. 2sdtds = 2x

dtdx

+ 2ydtdy

Chain Rule

dtds

= s1

dtdyy

dtdxx =

22 yx

1

dtdyy

dtdxx

Step 5 : Evaluate, with x = 0.8 , y = 0.6 , dy/dt = – 60 , ds/dt = 20 , and solve for dx/dt.

20

20 =

)60)(–6.0(

dtdx8.0

)6.0()8.0(

122

1

20 = 0.8 dtdx

– 36 dtdx

= 8.03620

= 70

At the moment in question, the car’s speed is 70 mph.

MAXIMA AND MINIMA :

Suppose a quantity y depends on another quantity x in a manner shown

x1 x2

x

y

in the figure. It becomes maximum at x1 and minimum at x2. At these pointsthe tangent to the curve is parallel to the xaxis and hence its slope is tan = 0.Thus, at a maximum or a minimum,

slope = dxdy

= 0.

Maxima :Def. : It is a point on the curve where the value of function is greater orhigher from the near by points.

Just before the maximum the slope is positive, at the maximum it is

zero and just after the maximum it is negative. Thus, dxdy

decreases

at a maximum and hence the rate of change of dxdy

is negative at a

maximum i.e. dxd

dxdy

< 0 at

maximum.

The quantity dxd

dxdy

is the rate of change of the slope. It is written as 2

2

dxyd

.

Mathematical Conditions for maxima are : (a) dxdy

= 0 (b) 2

2

dxyd

< 0

Maximum value :It exists either at one of the boundary point or one of the maxima of the function.

maxima

maximum value is at maxima

maximum value is at greatest of all the maxima

maximum value

maximum value

maximum value is at boundary point

21

Greatest of all the maxima : Global MaximaMINIMA :

Def. : It is a point on the curve where the value of function is less fromthe near by points.Similarly, at a minimum the slope changes from negative to positive.Hence with the increases of x. the slope is increasing thatmeansthe rate of change of slope with respect to x is positive

hence dxd

dxdy

> 0.

Mathematical Conditions for minima are: (a) dxdy

= 0 (b) 2

2

dxyd

> 0

Note: Quite often it is known from the physical situation whether the quantity

is a maximum or a minimum. The test on 2

2

dxyd

may then be omitted.

Minimum value :It exists either at one of the boundary point or one of the minima of the function.

minimaminimum value is at minima

minimum value

minimum value is at boundary point

Ex. 48 Smallest of all the minima : Global MinimaProve that the product of two numbers will be maximum, if the sum of two numbers is constant and the numbersare equal.

Sol. Let the two numbers be x and yAs given :- x + y = C .... (i)

xy = P .... (ii)x (c – x) x = P cx – x2 = P

22

dxdp

= c – 2x odxdp

c – 2x = 0 x = 2c

2–dx

pd2

2 (maxima)

Ex. 49 Particle’s position as a function of time is given as x = 5t2 9t + 3. Find out the maximum value of positionco-ordinate? Also, plot the graph.

Sol. x = 5t2 9t + 3

dtdx

= 10t 9 = 0

t = 9/10 = 0.9

Check, whether maxima or minima exists. 2

2

dtxd

= 10 > 0

there exists a minima at t = 0.9

Now, Check for the limiting values.When t = 0 ; x = 3

t = ; x = So, the maximum position co-ordinate does not exist.

Graph

Putting t = 0.9 in the equationx = 5(0.9)2 9(0.9) + 3 = 1.05

NOTE : If the coefficient of t2 is positive, the curve will open upside.

Ex. 50 The speed of light depends on the medium through which it travels and tends to be slower in denser medium. In

a vacuum, it travels at the famous speed c = 3 × 108 m/sec, but in the

earth’s atmosphere it travels slightly slower than that, and in glassslower still (about two - thirds as fast). Fermat’s principle in opticsstates that light always travels from one point to another along thequickest route. This observation enables us to predict the path lightwill take when it travels from a point in one medium(air, say) to a point in another medium (say, glass or water).

Find the path that a ray of light will follow in going from a point A in a medium where the speed of light is c1 acrossa straight boundary to a point B in a medium where the speed of light is c2.

Sol. Since light traveling from A to B will do so by the quickest route, we look for a path that will minimize the traveltime.We assume that A and B lie in the xy-plane and that the line separating the two media is the x-axis.In a uniform medium, where the speed of light remains constant, ‘’shortest time’’ means ‘’shortest path,’’ andthe ray of light will follow a straight line. Hence the path from A to B will consist of a line segment from A to aboundary point P, followed by another line segment from P to B. From the formula distance equals rate timestime, we have

time = ratecetandis

23

The time required for light to travel from A to P is therefore

t1 = 1

22

1 cxa

cAP

From P to B the time is

t2 = 2

22

2 c)x–d(b

cPB

The time from A to B is the sum of these :

t = t1 + t2 = 2 22 2

1 2

b (d – x)a xc c

We want find the absolute minimum value of t

222

221 )x–d(bc

)x–d(–xac

xdxdt

(Sorry ! you will have to do differentiation your self by applying the chain rule)

to get the minimum t we have dxdt = 0

2 2 2 21 2

(d – x)x0 –c a x c b (d – x)

In terms of the angles 1 and 2

2

2

1

1c

sin–csin0

or2

2

1

1c

sinc

sin

This equation is Snell’s law or the law of refraction. (I know you know it ! )