LIMITS AND DERIVATIVE - Lecture Notes

FUNCTIONS, LIMIT, DIFFERENTIATION AND INTEGRATION Functions Consider two variables x any y. Whenever there is a change in ‘x’ if there is a corresponding change in y we say the variable y is a function of the variable x. It is denoted by y = f(x) (we read it as ‘y’ is a function of ‘x’) Here ‘x’ is called the independent variable and ‘y’ is called the dependent variable Thus the function y  f (x) means when ever there is a change in the independent var iable ' x ' there is a corresponding change in the dependent var iable y For example we know that the area of circle is Area of circle is Area   r2 where ‘r is the radius. Whenever there is a change in radius ‘r’ there is a corresponding change in Area Independent var iable  r  Dependent var iable  Area 1

Hence we say Area of a circle is a function of its radius and is denoted by A  f (r) Example 2 : The mark of a student is a function of Hard work . ie Independent variable = Hard work Dependent variable = Mark  Mark = f (Hard work) Univariate function A dependent variable depending on one independent variable . In case of a circle Area  f radius Displacement  f  time are univariate function Bivariate function A dependent variable ‘u’ depends on two independent variables ‘x’ and ‘y’. u  f x, yis a Bi var iate function Ex: The area of Triangle is given by A  1 bh 2 b = base b = Altitude Area depends on base and Altitude  Area of   f b, h Is Marks of students a Bivariate function? Function as a production process A function can be regarded as a production process in which Input = Independent variable Output = Dependent variable whenever you give an input ‘x’, the production process makes some work and gives you the output y 2

y  f (x) Input(x) Pr oduction Output (y) Pr ocess For example whenever you give an input radius (r) of a circle the process makes the work  r2 and gives you the output Area of circle A  f (r) Pr ocess Area Output  Input( r ) r 2 Domain Range Set of values of the independent Set of values of the dependent variable variable 'x' or input y or output Increasing and Decreasing functions y = f(x) is an increasing function if dependent variable ‘y’ increases when there is increase in independent variable r = radius of circle A = Area of circle  A  f (r) is an increasing function Marks  f Hard work  is an increasing function. Decreasing function. y = f(x) is a decreasing function if the dependent variable decreases as the independent variable increases For example at constant temperature, as pressure of gas increases, the volume of gas decreases  Volume  f (pressure) is a decreasing function Also Marks  f laziness is a decreasing function 3

Example As speed increases, the time taken to travel from city A to cityB decreases Travel time  f speed is a decreasing function Graph of a function Consider the function y  f (x) . x = Independent variable and y = dependent variable. Corresponding to every value of x there is unique value of y so that we get a set of ordered pairs (x,y). These points are plotted on a graph paper and are joined by a smooth curve. It is called the graph of that function. X  axis  Independent var iable Y  axis  dependent var iable For example consider the function f (x)  x 1 Here y  f (x)  x 1  corresponding to every x, the value of y = x +1. The values of x and y 4

x -3 -2 -1 012 3 y=x+1 -2 1 0 123 4 Graph of Y = x+1 Constant function y  f (x)  k y  f (x)  0 is x  axis y  x is identify function y = -x 5

Q.2 Draw the graph of f(x) = x2; y = f(x) =x2 Y is the square of x y x square x2  x -2 -1 0 1 2 y=f(x)=x2 4 1 0 1 4 Graph f (x)  x2 This shape is called parabola The graph of f(x) = x2 is a parabola 6

Q.3 Draw graph of y=f(x) = x3 x 2 1 0 1 2 y  x3 8 1 0 1 8 Constant function f(x) = k x 2 1 0 1 2 f (x)  k k k k k k 7

Note (x-axis) f(x) = 0 or y = 0 is a constant function and it is the x-axis Linear function f(x) = ax +b or y = ax +b Intercept form of a straight line x  y 1 ab  f (x)  y  ax  b can be converted in to intercept form y = ax +b ax  y  b x  y 1  b  b   a  8

Quadratic function f (x)  ax2  bx  c a  0 concave up a  0 concave down 1) f (x)  ax2  bx  c a  0 b2  4ac  0 2) f (x)  ax2  bx  c a  0 b2  4ac  0 3) f (x)  ax2  bx  c a  0 b2  4ac  0 4) f (x)  ax2  bx  c a  0 b2  4ac  0 5) f (x)  ax2  bx  c a  0 b2  4ac  0 6) f (x)  ax2  bx  c a  0 b2  4ac  0 Before having the graphs of some other functions we may introduce  and   9

The concept plus infinity  Consider the function f (x)  2x f 0  1 f 1  2 f 2  22  4 f 3  23  8 f  0  210  1024 . As x increases f (x)  2x will increase much faster than the increase in x Now consider the function f (x)  10x x 012 3 4 5 6 9 12 y-f(x)10x 1 10 100 1000 10000 100000 1 0 6= m illio n 1 0 9 b illio n 1 0 12 T rillio n It can be seen that as x increases f (x)  10x , increases much much faster than x. Hence when x is very big number f (x)  10x tends to a very, very big number and it is denoted by  (Read as + infinity or positive  ) The concept -ve (-) infinity  consider f(x) = -10x f (0)  100  1 f (1)  101  10 f (2)  102  100 f (3)  103  1000 f (4)  104  10000 f (5)  105  100, 000 f 6  106  (million)  1000000(10 lakhs) f (9)  109  Billion f (12)  1012   Trillion f (100)  10100  Googol  verysmall number x 012 3 4 5 6 9 12 - billion - Trillion y-f(x)10x -1 -10 -100 -1000 -10000-100,000 - million As x increases -10x decreases much faster than the increase in x. When ‘x’ is a very big number -10x will be a very very small number which can not be visulized, which can not be writtern on paper and which can not be operated. This very very small number is represented by -  and is called -ive or minus infinity Is infinity a number? No, infinity  is not a real number. It is only a concept, an idea. It can not be measured. Even the far away galaxies can not comepte with infinity. 10

Since  is not a number the mathematical operations, Algebraic laws, laws of exponents etc are not valid in  Limit of a function: Consider the function y  f (x) . When the independent variable ‘x’ approaches or x tends to a constant value ‘a’ (denoted by x  a ) if the dependent variable y approaches to another constant value ‘k’ (denoted by y or f (x)  k ) we say, the limit of y = f(x) when x tends to a  x  a  is k. It is denoted by \\ Lt f (x)  k xa Here (i) The variable ‘x’ may or may not become exactly equal to ‘a’. ii) f (x) may or may not take the value k Example -1 consider the function f(x) = x2 x 1.5 1.8 1.9 1.999 2 2.1 2.2 2.5 f (x)  x2 2.25 3.24 3.61 3.996001 4 4.41 4.84 6.25 From table when x  2 from either side the value of f (x)  x2  4 and we write Lt f (x)  Lt x2  4 x2 x2 Here ‘x’ takes the value 2 and f (x)  x2 takes the value 4 Right Hand Limit (RHL) and Left Hand Limit (LHL) From the table it can be seen that when x  2 from x  2 , the value of f (x)  x2 tends to 4 and it is denoted by Lt f (x)  Lt x2  4 and is called the LHL x2 x2 Also from table when x  2 from x  2 then also f (x)  x2  4 and it is denoted by Lt f (x)  Lt x2  4 and is called RHL x2 x2 11

Note 1) Lt f (x)  Lt x2  4 RHL  LHL  4 x2 x2 In general Lt f (x)  k  Lt f (x)  Lt f (x)  k xa xa xa 2) If RHL  LHL  Lt f (x) Does not exist xa Now consider the function f (x)  x2 1 x 1 x 0.99 0.999 1 1.01 1.1 f (x)  x2 1 1.99 1.999 11  0 2.01 2.1 x 1 11 0 not defined From table when x 1 from x 0 the value of f (x)  x2 1  2 and we write x 1 LHL  Lt f (x)  Lt x2 1  2 x 1 x1 x 1 Also when x 2 from x  1, then also the value of f (x)  x2 1  2 and we write x 1 RHL  Lt f (x)  Lt x2 1  2 x2 x2 x 1 The RHL  LHL ie Lt x2 1  Lt x2 1  2 x2 x 1 x2 x 1  Lt x2 1  2 x2 x 1 Objective of limit 12

f (x)  x2 1  f (1)  11  0 x 1 11 0 0 0 is undefined (Not a finite quantity / Exact value ) The concept limit gives you the expected value (not exact value) of f (x)  x2 1 and the expected value is 2 x 1 So the objective of limit is to find the expected value (and the exact value) of a function at a point where the direct subtitution results in an undefined value f (x)  sin x  f (0)  sin 0  0 undefined x 00 x 0.2 0.05 0 0.01 0.03 sin x .993347 .999583 sin 0 .999983 .99985 x0 Lt sin x  1  Lt sin x  1 xx 0  xx  0  Lt sin x  1 x0 x Result If RHL  LHL at x a then Lt f (x) does not exist xa Reciprocal function (Rectangular Hyperbola) f (x)  1 x 13

Lt 1  1  0 x x  Lt 1  1  0 xx  v.Important Lt 1  1   xx0 0  Lt 1  1   xx0 0  x .001 .0001 .00001 0 0.00001 .0001 .001 .01 .1 f (x)  1 1000 10000 100000 100000 10.000 1 x 0 1000 100 10 undefined 105 104 14

x 10 100 1000 10,000 ......  y  1 0.1 0.01 0.001 0.0001 1/   0 x Lt 1  1  0 x x  x 10 100 1000 10000  y  1 0.1 0.01 .001 .0001 1  0 x  Lt 1  1  0 xx  Exponential function  f (x)  ax y  2x , y  3x , y  10x ....... a   a  0 2   2  0 3   3  0 15

case 2 f (x)  ax 0  a  1 a  0 a    1   0  1     2   2  Natural exponential function f (x)  ex e   e   0 1 Lt   4 x x  0 10  10 1   5   1 x 16

Lt 1 Lt 1 1 Find x0 1 x0 1  Lt 4x  5x x 32 x 32 x x Logarithms 23 =8 Then we say log2 8  3 34  81 Then we say log3 81  4  1 3 1 1  2  8     log  8  3  1  2  In general a m  k  loga k  m (Read it as logarithm of k to the base ‘a’ is m) logarithmic function f (x)  loga x where x is a +ive real no. and a > 0 and a  1 is called the logarithmic function Natural logarithmic function When base a  e  2.72 17

Properties log ab  log a  log b log  a   log a  log b  b  log a m  m log a  log a  log  1   a  Series expansion of functions 5! = 1 × 2 × 3 × 4 × 5 ex  1 x  x2  x3  ....... 1! 2! 3! ex  1 x  x2  x3  ....... 1! 2! 3! log 1 x   x  x2  x3  ......... 23 log 1 x   x  x2  x3  . x4 ........ 22 4 tan x  x  x3  2x5  ...... 3 15 18

sin x  x  x3  x5  ...... 3! 5! cos x  1 x2  x4  ........ 2! 4! ax  1 x log a   x log a 2 1! 2! 1 xx  1 nx  n(n 1) x2  n n 1n  2 x3  ....... when x  1 1 2 1 23 L s ti n x = 1  0x  x → → →→ t at n x = 1  L 0x   x ta x- 1= l o g a Prove using expansion method 0x L  lo g t 1 + x 1 x x =  0  L  x Some important limits L sint x  L tx 1 1) x 0 x 0 x sin x → → → → →→ L →→tant xLtx 1 2) x 0 x 0 x tan x L xt n  an n a n1 3) x 0 xa L 0 et x 1  1 4) x x L lt og 1 x 1 5) x 0 x L 0 at x 1  log a 6) x x 19

L 1t  cos x  1 7) x 0 x2 2 → Note : In all these limits the direct substitution is undefined. Hence Limit gives us the expected value of the fn when x  0 or a etc Questions find sin 5x Lt tan 3x 1) Lt 2) x0 tan 5x x0 x x3 8 log 1 2x 3) Lt 4) Lt x0 x  2 x0 x 5) Lt 2x 1 x0 x Limits of Rational functions a0 f m  n  b0 Lt a0xn  a1xn1  ......an   0 mn b0nm  b1nm1  .......bm  nm x    1) Lt 5x3  2x2 1  5 x 4x3  3x  7 4 2) Lt 5x2  7x 1  0 x 4x3  3x2 2 3) Lt 2x3  3x 1   x 4x3  2x  7 Sine function : f(x) = sin x 1800   radians  900   radians 2 2  2180  3600 and so on 20

x 0   90 180   270  3  360  2 22 f (x)  sin x 0 1 0 1 0 f(x) sin x is periodic with period = 2 Now cut and paste cosine function  f (x)  cos x 0   90   180 3  270 2  360 22 10 1 0 1 21

f(x) = cos x is periodic with period = 2 so cut and pase Tangent function f (x)  tan x tan 0  0 tan    2 tan        2  f (x)  tan x tan 0  0 tan    2 tan        2  Differentiation 22

Chord :Line segment joining exactly two points Secant : Line segment joining two or more points Tangent : Limiting line of a secant Derivative or Differential Coefficient  Consider the function y = f (x) ; LetA x f (x) and B x  h f x  h be two points on the graph of f(x) as shown below Slope of secant AB f (x  h)  f x xhx Slope of secant AB  f  x  h  f x  h 23

 Slope of tangent at A  Lt f xhf x h0 h This limit, if it exists, is called the derivative or differential coefficient of y  f (x) w.r.t. x and is called the ab initio derivative or the derivative from first principles. It is denoted by dy or f 1(x) dx The process of finding the Derivative is called differentiation. Result -1  In Geometrical sense dy or f 1(x) is the slope of tangent at the point x f x dx Result -2 dy In physical sense dx is the rate of change of y w.r.t.x Questions Find the at-initio Derivative of 1) f (x)  k 2) f (x)  x2 3) f (x)  x3 4) f (x)  1 5) f (x)  ex x List of standard Derivatives 1) d k0 dx 2) d x 1 dx d xn  n xn1 3) dx d 1  1 4) x x2 dx d x  1 5) dx 2x 6) d log x  1 dx x 24

7) d ex  ex dx 8) d a x  a x log a dx 9) d sin x  cos x dx 10) d cos x   sin x dx 11) d tan x  sec2 x dx 12) d sec x  sec x tan x dx 13) d cos ecx   cos ec x cot x dx d 1  n 14)) xn x n1 dx Algebra of Derivatives d k f x  k d f x 1) dx dx 2) d f x  g x  d f x   d g(x) dx dx dx 3) Product Rule d f x g x  f x g1 x  g x f 1 x  dx 4) Quotien Rule d f x g xf 1 x  f xg1 x dx gx  g  x 2 25

5) Power Rule d f xn  n f x n1 d f x dx dx 6) Reciprocal Rule d f 1   f 1 d f x dx x x2 dx Function of a function and chain Rule f g  x  , g f  x  etc are function of functions 1) df g x f g xg1 x dx 2) d g f  x   g1 f xf 1 x dx Derivative of f(x) w.r.t. another variable ‘t’ d f x  f 1 x dx dt dt 1) d x2  2x dx dt dt 2) d sin x  cos x dx dt dt 3) d sin2 x  2 sin x cos x dx dt dt 4) d sin y  cos y dy dx dt 5) d log t 1 dt du t du Parametric Differentiation x  f t and y  t 26

y = f(x) is a parametric function in parameter ‘t’  dy  dy   dt  dx  dx   dt  Ex:1 x  2t2 y  4t 2) x  a cos t y  a sin t dx  4t dy  4 dt dt dx  a sin t dy  a cos t dy  4  1 dt dt dx 4t t dy  a cos t   cot t dx a sin t Physical Application of Derivatives S  f (t)  ds  Velocity dv  Acceleration dt dt Q.1) A particle is protected vertically upwards satisfies S  60t 16t2 . What is the velocity when t =0 S  60t 16t2  dy  v  60 d V  ds  60  32t when t  0 dt Q.2) Velocity v  ks2 . Then the acceleration is  a  dv  2ks ds  2ks ks2  2k2s3 dt dt Q.3) A circular plate is heated uniformly and its area exponds 3c times as fast as its radius. What is the value of ‘c’ when r = 6 Area  A  r2 Diff.w.r.t ‘t’ 27

dA  2 dr Given dA  3c dx dt dt dt dt 3c dr  2r dr Given r  6 dt dt 3c  2 6  c  26  4 3 Geometrical Applications Q.1) What is the slope of tangent at ( 1 4) to the curve f  x  3x2  5x  6 f1 x  6x  5 slope at (1 4) = f 1 1  6  5  1 Q.2) Slope of tangent of f (x)  x2  1 at 1, 0  x2 f 1  x  2x   2   2x  2  x3  x3 f 1 1  2  2  4 Q.3) Slope of tangent at 1  3 to the curve y2 ey  9e3x2 ie y2ey  9e3x2 y2ey dy  ey 2y dy  9e3 2x put x  1 y  3 dx dx 9 e3 dy  6e3 dy  18 e3 dx dx dy 9  6  18  dy  18  6 dx dx 3 Increasing and decreasing function ↓↑ f1 x > 0 ⇒ f x i s s tri c tl y   f1 x < 0 ⇒ f x i s s tri c tl y   28

f x  x2  f1 x  2x  0 when x  0  when x  0 1)   0 f x  x2 S  when x  0 and S  when x  0 2) f  x   x3 f 1  x   3x2  0 for all x f  x   x3 is S  for all x INTEGRATION list of Integrals  xndx  xn1  c c: Integrating constant n 1  dx  x  c  k dx  kx  c; k is a constant  sin x dx   cos x  c  cos x dx  sin x  c  sec x tan x dx  sec x  c  cos ec x cot x dx   cos ec x  c  sec2 x dx  tan x  c  cos ec2x dx   cot x  c  1 dx  log x  c x 29

 exdx  ex  c Examples 1. sin 2x dx   cos 2x  c [ Divide by co-efficient of x 2 2.  cos 3x dx  sin 3x  c 3  3. 3x2  5x  8 dx  3 x3  5 x2  8x  c 32  x3  5x2  8x  c 2 4.  3 sin x  4 sin 2x  dx  3x  cos x  4   cos 2x  c 2  5. 3sin 2x  6 cos 4x  e2x dx  3  cos 2x  6 sin 4x  e2x  c 2 42  3cos 2x  3sin 4x  e2x  c 2 22 6.  x 2  x  4  dx  x3  x2  4 log x  c  x  3 2   7. x3  4x2  e3x dx  x4  4 x3  e3x  c 4 33  x4  4x3  e3x  c 43 3 30

Definite integrals -Area under the curve Area under the curve y - f(x) from x = a to x = b is bb  y dx  f xdx aa a: lawer limit b: upper limit 1. Find the area under the curve y  x2 from x = 0 to x = 1 solution Shaded portion is the required area 11 Area   y dx   x2dx 00   x3 1 No need to write integrating  3  constant on definite integrals  0 31

  1    03   Put upper limit Ist  3   3      and lower limit  1 sq.units 3 2. Find the area under the curve y  x from x  1 to x  4 4 x3/2 4 3  Required area = xdx   1  2 1 4 x1/2dx  2  x 3/ 2 4 1 3 1  x 1 1 4  2  43/2  2 13/2  1 2  33      2 8  2 1 33 1  2 1  16  2  14 sq.units 33 3 3) cos 2x dx  sin 2x   :180  2      sin 2  sin 2 22 000 32

 4)2 sin 2x dx   cos 2x 2 0 2 0   cos 2      cos 2 0   2 2   2          cos   cos 0   1  1  1  1  1 22 2 2 22 1 5)  x2  4x 1dx 1  x3  4 x2  1 3 2 x  1  1  4  1  1   1  4  1  1  3 2  3 2  1  2 1 1  2 1 33  22 3 8 3 33