CONTINUITY DIFFRENTIABLITY AND DERIVATIVES - Lecture Notes

BBrilliant STUDY CENTRE MATHEMATICS (ONLINE) -2022 CHAPTER - 00 CONTINUITY, DIFFERENTIABILITY AND DERIVATIVES Revision Lt f x  k  Graph of f(x) meets line x = a at the point (a, k) where limiting point (a k) need not be xa on the graph. x= 0 x = a f x x2 x  0 f x  x2 x 1 x 0 1  a   , a  1 e   0  e  0 1   a   0, a  1 log 0   log e  1 0 log1  0 log10  2.303 1 0 a   0  0  a  1  a      log    1 0    Revise Important Methods of evaluating limits. 1

BBrilliant STUDY CENTRE MATHEMATICS (ONLINE) -2022 Continuity at a point The function y = f(x) is continuous at x = a if i) f(x) is defined at x = a ii) Lt f (x)  Lt f x  f a xa xa Lt ie n  a f(x) exists and is finite If these two conditions are satisfied at every point in an intervel [a b], then f(x) is continuous in the interval [a b] eg : 1) f(x) = x2    f(x) is defined at x=0 and f(0) = 0 is finit f  0  f (x)  x2 Lt f (x)  Lt f x 0 is x0 x0 continuous at x = 0. 2) f (x)  1 x f (0)  1 is not finite (not defined) 0 f (x)  1 is not continuous at n= 0 x   ex x  0  f  x    x 2 0 x 1   1 x 1  x i) f 0  eo  1finite    L t x  0 f x  Lt f x  Not continuous at x = 0 x0    ii) Lt f x  Lt f x  1  continuous at x = 1 x 1 x 1 Note 1: If f(x) is continuous at x = a then graph can be drawn through (around) ‘a’ without lifting the pen from the plane of the paper. Note 2: If f(x) is discontinuous at x = a there is a break at x = a so that the graph can not be drawn around ‘a’ without lifting the pen from the plane of the paper. 2

BBrilliant STUDY CENTRE MATHEMATICS (ONLINE) -2022 Questions 1. x  x4x  x x0 2x2 f    k x  0 2 is continuous at x = 0. Find k Types of Discontinuities i) Removable Discontinuity (RD) y = f(x) has a removable discontinuity at x = a if Lt f x   Lt f x   f x  xa xa ie; Lt f x exists  f x xa Removable discontinuity fx  x21 x  1  x 1  1 x 1 fx x sin 1 x  0 x 1) Ex:  1 x 0 Given f(0) = 1 3

BBrilliant STUDY CENTRE MATHEMATICS (ONLINE) -2022 Lt f x  Lt x sin 1  0 sand wich Theorem xx0 x0 Lt f xexists  f 0  RD at x0 x0 2) f  x   x2 x  0   1 x0 Lt n  0f x  0  f 0 At RD the limiting point is not on the graph. It is hole. Dirchlet function f x  1, x  rational    0, x  irrational Defined at every real number and discontinuous at every real number. 1) f x x x  rational  x  Irrational sin gle po int continuous function   0   2) Single point function  1 x  x 1 f x  sin gle po int function 3) f x  x1x1 Pointfunction Non Removable Discontinuity (NRD)    y = f(x) has a Non- Removable Discontinuity at x = a. If Lt f x  Lt f x xa xa      ie ; Lt f x xa x does not exist. If both Lt f and Lt f x are finite, but unequal, then the xa xa NRD is called Jump Discontinuity (JD). 4

BBrilliant STUDY CENTRE MATHEMATICS (ONLINE) -2022 Ex : f x  sigx  1 x  0    0 x0   1 x  0  RHL at x = 0 is 1 LHL at x = 0 is -1 RHL and LHL are finite, but not equal f  x  = sigx has jump discontinuity at x = 0 Special causes 1)DIRICHLET Function : f x   1, x  rational  0, x  Irrational   2)Single point function : 1 x  x 1  Continuous 3)Single po int continuous fx f x x , x  rational :  x  Irrational   0, (continuous at x  0 only) 4) f x  x 1 Defined only at int egers x 1 1) f x 1  ex 1 x0   1 ex 1 1 x  0 check whether continuous at x = 0 2) f  x    x  1 cot x when x  0 is continuous at x = 0. Find f(0) 3) f  x    ex 1 2 x  0 f 0  12. Find ‘a’ if(x) is continuous at x = 0 sin  x  log 1  x   a  a  5

BBrilliant STUDY CENTRE MATHEMATICS (ONLINE) -2022 4) f  x  ex 0  x  1    2  e x 1 1 x  2 check continuity at x  1 and x  2   xe 2x3  5) f x   5 x 1    a  bx 1 x 3   b 5x 3 x  5   30 x 5 For what value of ‘a’ and ‘b’ f(x) is continuous 6) f  x   x x 1  x  2 Find the points of discontinuities of f(x) A lg ebra of Continuous function Let f(x) and g(x) be continuous at x=a i) k f(x) is continouus at x = a ii) f x  g x is continuous at x = a iii) f x g x is continuous at x =a f x iv) g  x  is continuous at x = a v) f(x) and g(x) s.t. f [g(x)] is defined at x = a Let g(x) is continuous at x = a and f(x) is continuous at g(a) then f[g(x)] is continuous at x =a. Right hand derivatives at x =a (RHD) Rf ' a   Lt f a  h   f a  where h  0 x0 h Left hand derivative (LHD) at x = a Lf ' a   Lt f a  h   f a  where h  0 x0 h Ex: i) f x  x f 0  0  0, f h  h  h, f h  h  h 6

BBrilliant STUDY CENTRE MATHEMATICS (ONLINE) -2022 Rf ' 0  Lt f h   f 0  Lt h  0  1 x0 h x0 h Lf ' 0  Lt f h   f 0  Lt h  0  1 x0 h x0 h f x  x  Rf ' 0  1, Lf 1 0  1 Ex 2: Let f x  sign  1 x  0   0 x0    1 x0 Rf ' 0  Lt f h   f 0  Lt 1 0  Lt   h0 h h0 h h0 Lf ' 0  Lt f h  f 0  Lt 1 0  Lt 1   h0 h h0 h h0 h f  x   si gx Rf ' 0   and Lf ' 0   Rf ' 0   Lf ' 0   Result: If f(x) is continuous at x = a then Rf 'a  and Lf 'a  are respectively the derivatives of the Right and Left branches of f(x) at x = a Ex: f x x2 x  0 Rf '0  0  Lf '0  1   sin x x0 7

BBrilliant STUDY CENTRE MATHEMATICS (ONLINE) -2022 Geometrical Meaning of Rf 'a  and L f 'a  C (a-h f(a-h)) (a+h f(a+h)) B A (a, f(a)) a-h a a+ h Slope of secant AB = f a  h)  f (a)  f a  hf a aha h Slope of tangent at x  a  h Lt 0 f a  h  f a  Rf 'a to the right of x a    h Slope of secant AC = f a hf a  f a hf a aha h Slope of tan gent at x  a  h Lt 0 f a hf a  Lf 'a to the left of x  a    h Differentiability at x = a The functions y = f(x) is differentiable at x =a if the following conditions are satisfied. i) f(x) is continuous at x = a ii) Rf 'a  Lf 'a In geometrical sense if f(x) is differentiable at x = a then these exist a unique tangent at x = a. Relation between continuity and Differentiability All differentiable functions are continuous, but all continuous functions need not be differentiable Differentiable  Continuous Continuous  Differentiable f(x) =|x| is continuous, but not differentiable at x = 0 SHARP POINT A point of which a function is continuous, but not differentiable having finite RHD and LHD is called a sharp point. For ex : f(x) = |x| has a sharp point at x = 0. Ex:(2) f(x) = Min: (x x2) Sharp points at x = 0 and x = 1  Not differentiable at x = 0 and x = 1 8

BBrilliant STUDY CENTRE MATHEMATICS (ONLINE) -2022 f x  x 1  x 1 .Find RHD and LHD at sharp po int s f x is continuous at x  a  Is the function  and  aq Need not be unique tangent at x  a  Different at x   Ex : f  x   x 1 3 Lt h 1 3 0 Lt 1 1  h  Rf '0  h  0 h0 2    0 Lt h3  y axis is the unique tan gent Lf '0  h  0 1  h 3  0 Lt 1 1  h  h 0  0    2 h3  1 is continuous at x = 0 and there exist a unique tangent at x = 0 without being differentiable f (x)  x 3 at x = 0 Results 1) Rf 'a  and Lf 'a  are finite and equal  f(x) is continuous and different at x = 0 Ex: f (x)  x2 sin 1 x  0 and f x  0, x  0 x h2 sin 1  0 x  0.AlsoLf '0  0 Rf '(0)  Lt h0 h Continuous and different at x = 0 9

BBrilliant STUDY CENTRE MATHEMATICS (ONLINE) -2022 2) Rf '(0) and Lf '0 are finite and unequal  continuous, but  f  x   x   1  1  , x  0 not different at x  0  x x   e      0 , x0  3)   1 If either Rf 'a  or Lf 'a  or both are infinite then the function may be continuous f x  x 3 or may not be continuous. (Ex: f(x) = sigx) 4) If the value of a derivative at a point is finite then function is continuous and differentiable at that point  Ex: f x  x x  x 1 f x  x  1  2 1 1   x x 1  x x x x 1  2x x  22 x 1 f '0  0  0  0 1  1(Finite)  f (x) is continuous and differentiable at x = 0 Question: f x  x  x2 0  x 1 g x  Maxi f t  0  t  x, 0  x  1    sin x x 1 Find Non Differentiable points of f(x) f t  t  t2  downward parabola f 't 1 2t 0 t  1 at maximum 2  Also f 1  1 2 4 Graph of f(t) 0t  1  2 Maxi f t  f x 1  x 1 2 Maxi f t  1 4 10

BBrilliant STUDY CENTRE MATHEMATICS (ONLINE) -2022 g x f x  x  x2 0 x 1 2   1 , 1 x1  4 2    sin  x x 1 y = x-x2 y = 1 4 1 2 y  x  x2  dy  1 2x dx dy  1  Differentiable at x  1  4 2 dx x12 Methods of Differentiation Derivative or Differential coefficient at x = a Let y = f(x) be a differentiable function. The derivative or differential coefficient at x = a is defined as f 'a   Lt f a  h   f a  h  0 h0 h In general the derivative of y = f(x) w.r.t x is given by f ' x   dy  Lt f  x  h   f  x  h  0 dx h0 h When h  n f ' x   dy  Lt f  x  x   f  x    1 dx x0 x Let y  f x y  y  f x  x y   y  y  y  f x  x  f x From (1) dy  Lt y x  0 dx x 11

BBrilliant STUDY CENTRE MATHEMATICS (ONLINE) -2022 dy dy In Geometrical sense is the slope of tangent and inphysical sense is the rate of change of y dx dx w.r.t x. Process of finding the derivative is called Differentiation. Have a quick revision of i) Product Rule   ii) Quotient Rule  Re fer 1st year notes on iii) Power Rule  L im its and Derivatives iv) Re ciprocal Rule   v) Chain Rule  Extension of Chain Rule d f g  h  x   f ' g  h  x  g ' h  x  h '  x  dx Ex: 1) d sin log x  cos log x  1  1 dx x 2 x  2) y = sec tan ex ; dy sec e tan ex tan tan ex sec2 ex ex dx 3) sin log cos x = y dy cos log cos x  1  sin x  dx cos x Exercise 5.2  NCERT Text Page 166  Implicit and Explict Functions Functions of the form y  f (x) or x   y are called explicit functions Ex : y sin x or x  2y 1 2x  y cos y  ey Functions which are not explicit are implicit functons Ex : x2  y2  sin xy  k 12

BBrilliant STUDY CENTRE MATHEMATICS (ONLINE) -2022 Derivative of Explicit Functions a) y  f  x   standard results can be used directly. b) x   y . The procedure is given below dx Step 1 : Differentiable with respect to y and get dy dy  1 Step 2 : dx  dx     dy  Ex: 1) x  sin y  ey Different w.r.t x dx  cos y  ey  dy  cos 1 ey dy dx y 2) x  2sin y y  log y x  2sin y Different w.r.t y y  log y dx y  log y 2 cos y  2sin y 1 1  dy  y  log y2  y    dx  2y cos y  y  log y  2sin y 1 y dy y  y  log y2 dy  1 Now dx  dx   dy  Derivative of Implicit functions Method 1 : If possible convert the impicit function in to explicit and differentiate . Ex : xy = x+ y - implicit xy  y  x  yx 1  x  y  x ;  dy  x 1  x  1 x 1 dx x 12 x 12 13

BBrilliant STUDY CENTRE MATHEMATICS (ONLINE) -2022 (Answer in x only) x 1y y 1x 0 2) 2x2  3y  7  Im plicit s.t dy  1 2x2  7  3y ; y  1 2x2  7 3 dx 1x2 dy  1 4x  4 x dx 3 3 (Answer in x only) 3) dy If sin y  x sin a  y Find dx sin y  x sin a  y  Im plicit x  sin y y   exp licit  sin a Different w.r.t x dx  sin(a  y) cos y  sin y cos a  y dy sin 2 a  y dx  sin a  y  y dy sin2 a  y  dx  sin a dy 1  sin2 a  y dy dx  dx  sin2 a  y ;   sin a   dy (Answer in y only) 14

BBrilliant STUDY CENTRE MATHEMATICS (ONLINE) -2022 Questions:  1) x  1 If  2x 2y  4e2x2y then 1 log1x 2 dy  dx 1) log2 x 2) x log2 x  log2 3)x log2 x x x log2 x  log2 4) x [Option in x only convert in to y  f x ] 2) x 1 y  y 1 x  0 show that dy  1 dx 1 x2 x 1 y  y 1 x ; x2 1 y  y2 1 x  x2  x2y  y2  y2x ; x2  y2  y2x  x2y x  yx  y  xy y  x ; x  y  xy ; x  xy  y x  y1 x ; y  x exp licit  1 x dy   11xxx2    1 dx   1 x2 Derivative of Implicit Functions dy In case of implicit function we differentiate term by term w.r.t x and arrange the terms of dx 1) x2  y2  sin xy  k Different w.r.t x 2x  2y dy  cos xy  x dy  y   0 ; dy 2y  x cos xy  2x  y cos xy dx  dx  dx dy    2x  y cos xy  dx  2y  x cos xy  15

BBrilliant STUDY CENTRE MATHEMATICS (ONLINE) -2022 2) Find dy If x3  x2y  xy2  y3  81 dx 3x2  x2 dy  y2x  x 2y dy  y2  3y2 dy  0 dx dx dx dy x 2  2xy  3y2    3x 2  2xy  y2  dx  dy  3x2  2xy  y2   dx x2  2xy  3y2 Exercise 5.3 Page 169 NCERT TEXT Derivative of inverse Trignometric functions 1) f x  sin1 x ; y  sin1 x  x  sin y dx  cos y  dy  1  1 dy dx cos y x  sin y  x2  sin2 y  1 x2  1 sin2 y 1 x2  cos2 y  cos y  1 x2 ; dy  1  1 dx cos y 1 x2 d sin1 x  1 where 1 x2  0 dx 1 x2  x2  1  1  x  1 2) f x  cos1 x y  cos1 x  x  cos y  dx   sin y dy  dy   1 y ; x2  cos2 y  1 x2  1 cos2 y  sin2 y ;  sin y  1 x2 dx sin 16

BBrilliant STUDY CENTRE MATHEMATICS (ONLINE) -2022 dy   1  1 1  x  1 dx sin y 1 x2 3) Derivative of f  x   tan1 x y  tan1 x  x  tan y ; dx  sec2 y  dy  1  1 dy dx sec2 y x  tan y  1 tan2 y  sec2 y  1 x2  sec2 x dy  1 d tan1 x  1 x  R dx 1 x2 dx 1 x2 4) Derivative of f(x) = sec x y  sec x  x  sec y  dx  sec y tan y dy  dy  1  1 1 tan2 x  sec2x dx sec y tan y x  sec y  1 tan2 y  sec2 y ; tan2 y  sec2 y 1  1 tan2 y  x2 1  tan y  x2 1 ; dy  1  1 x 1 dx sec y tan y x2 1 x y  sec x is a S  fx  dy  0 f x   sec x is s  dx d sec x  1 dx x x2 1 5) Derivative of f(x) = cosec-1x y  cos ec1x  x  cos ec y dx   cos ec y cot y  dy  1  1 dy dx cos ec y cot y x  cos ec y  1 cot2 y  cos ec2y ; 1 cot2 y  x2  cot2 y  x2 1  cot y  x2 1 17

BBrilliant STUDY CENTRE MATHEMATICS (ONLINE) -2022 dy  1 But f  x  cos ec1x is S  dx x x2 1  dy  0 dy  1 dx dx x x2 1 6) Derivative of f(x) = cot x y  cot1 x  x  cot y  dx   cos ec2y dy dy  1 dx cos ec2y 1 cot2 y  cos ec2y  2  x2  cos ec2y dy  1 xR ; d cot1 x  1 xR dx 1 x2 dx 1 x2 d sin1 x  1 1  x  1 dx 1 x2 d cos1 x  1 1  x  1 dx 1 x2 d tan1 x 1 xR ; d sec1 x  1 x  1 or x  1 dx 1 x2 dx x x2 1 d cos ec1x  1 x  1 or x  1 ; d cot 1 x  1 xR dx x x2 1 dx x 1 2 1) y  sin 1 2x Find dy ; Put x  tan  1 x2 dx y  sin 1  2 tan    sin 1 sin 2  2 ; y    2 tan1 x  dy  2   tan 2a  dx 1 x2 1 2) f  x   sin1 2x 1 x2 Find dy ; Put x  sin  1 x2  cos  ; dx f x  sin1 2sin cos   sin1 sin 2  2 ; f  x   2sin1 x  f 'x   2 1 x2 18

BBrilliant STUDY CENTRE MATHEMATICS (ONLINE) -2022 3) Find the Derivative of tan1 sin x 4) d sin1 2  3cos x   1 (3sin x) dx 1 2  3cos x 2 Exponential and logarithmic functions f x  ax , x  R, a  base  0, a  1 y  10x : Common exp onential function e   y  ex : Natural exp onential function e  0 Logarithmic Function A function defined by f x  log ax where x  0 and a  0 and a  1.It is the inverse of exponential f u n c t io n f( x ) = x. a a>1 O<a<1 log ax y=x y = ax 1 y = log a 2 log a x y= log ax a  10  common log and a  e Natural log 19

BBrilliant STUDY CENTRE MATHEMATICS (ONLINE) -2022 d Results ( dx of exponential and logarithmic function) i) d ax  ax log a ii) d ex  ex dx dx iii) d amx  mamx log a iv) d a m x  m e m x dx dx v) d log x  1 v) d log10x  1 dx x dx x log10 d af x  af x log a d f  x  dx dx d  ef x  ef x f ' x dx d log f  x   f 1  f '  x  dx x 1) Find derivative of y  sin1 ex ; dy  1 xex Exercise 5.4 dx 1 e2x Page 174  2) dy 2sin xex  2sin xex cos x  ex dx dy i) y  e x 2  log x  2 y 3) Find d x 4) d e x  1 d e x  1 e x 1 Q uestion dx 2 e x dx 2x 2 ex e xy  log xy  cos xy  5  0 x0 y0 Find dy dx Logarithmic Differentiation When the functions are in the form afx ,f xgx ,f xg x and f x g  x   0, before finding the g x Derivative we take logarithms and it is called logarithmic differentiation. 20

BBrilliant STUDY CENTRE MATHEMATICS (ONLINE) -2022 Find the Derivative of 1) y  ax 2) y  3sin x 3) y  xx 4) y  xx 5) 1 6) y  x 1x 7) y  xxx 8) y  log x x  xlogx yx x 9) y  log x x  xlogx 10) y  sin x x  xsin x 11) y  ex cos3 x sin2 x 12) y   x x 1x  2 5 x  1, 2, 3, 4  3x  4x  13) f  x   1  x  1  x2  1  x 4  1  x8  find f '1  Ans : 120 Find dy if yx  xy  xx  ab ; u  yx v  xy z  xx  u  v z  ab dx du  dv  dz 0; du  u  x dy  log  , dv  v  y  log x dy  , dz  xx 1 log x dx dx dx dx  y dx y dx  x dx  dx   dy   yx log y  xy y  xx log 1 x   x  dx   x    y x  y    x y log x     Important Results d f  x gx  f  x gx  f 'x g x  g ' x log f x  dx  f x    d af xgx d dx dx  af xgx log a f x gx 1) y  sin x ex 2) log x sin x  3) y  2x3 1  xex 1  4) y  2x 1 dy at x  1 3x2  2 log x dx find and 21

BBrilliant STUDY CENTRE MATHEMATICS (ONLINE) -2022 Find derivative of 2) y   10 x2 1 sin x 3) y  exex dy at x  1 dx 21) sin xx1 4) y  x xxx 5) y  10x10x Find dy at x  1 dx 6) y  xxx  dy  xxx  1 x x  xx 1 log x 1 dx  x Derivative of Parametric functions If the relation between two variables x and y is expressed via a third variable ‘t’ then the function y = f(x) is called a parametric function.’t’ is called the parameter. ie; If x  f  t  and y  g  t  Then y  f x  is called a parametric function in parameter ‘t’ Ex : x  a cos  y  a sin  i) x2  y2  a2 . Hence x  a cos  y  a sin  is the parametric form (equation) of the circle x2  y2  a2 2) x  a cos  y  b sin a  x 2   y 2 1 is the parametric equation of the ellipse x2  y2 1  a   b  a2 b2 3) x  at2 y  2at ; y2  4a2t2  4a2 x  4ax a x  at2 y  2at is the parametric equation of parabola y2  4ax Parametric differentiation Method 1 : In case of a parameteric function if possible eliminate the parameter and then differentiate Ex : 1) x  a cos  y  a sin  ; x2  y2  a2  dy  x dx y 2) Find dy if x  a xsin1 and y  a xcos1 dx xy  asin1 x  cos1 x  xy  cons tan t 22

BBrilliant STUDY CENTRE MATHEMATICS (ONLINE) -2022  dy  y x 1 t2 y 1 t2  1 t2 dx x 1 t2 1 t2  1 t2 y  1 x 1 x 3) x 1 y   1 y2  x2     1 2     1 2  4  ;       y2  x2  4 ; 2y dy  2x  0  dy  x dx dx y 4) x  at2 y  2at ; y2  4a2t2  4a2 x  4ax a y2  4ax  dy  2y  4a  dy  4a dx dx 2y Parametric Differentiation : Method 2 In case of parametric function the derivative can also be obtained by  dy  dy   dt  dx  dx   dt  Ex : 1 x  at2 y  2at dy   dy   2a 1 dx  2at dy  2a dx  dt  2at t  dx  dt dt  dt  Ex : 2 x 1 y    1 ; dx 1 1 dy  1 1   d 2 d 2 dy  dy  2 1     1  x  d         dx  dx  2 1  1  y  d        23

BBrilliant STUDY CENTRE MATHEMATICS (ONLINE) -2022 Ex: 3 x  a  cos t  log tan t  y  a sin t  s.t dy  tan t  2  dx        dx  a   sin t  1 t sec2 t  1   a   sin t  1 t   a   sin t  1 t  dt  tan 2 2   t cos   sin  2 sin  2  2 2  1 sin2 t  a cos2 t dy  a cos t  sin t cos t  tan t ; dx cos2 t a  cos2 t  sin t sin t a    sin t x  a t  sin t y  a 1 cos t. dy   dy  a 1 cos t  dy  a sin t dx t dx dt 2 dy  a a sin t t   sin t t dx 1 cos 1 cos x  a cos2  y  a sin3  dy   1 1 Find 1  dy 2 1  3a sin2  cos  2 dx t 1  dx   3a cos2   sin a    2  1  tan 2  1 tan2   sec2  Derivative of special type Functions Containing an infinite expression 1) y  x  x  x  .... Find dy dx y xy  y2  x  y ; 2y dy  1 dy  dy 2y 1  1 dx dx dx dy  1 dx 2y 1 2) y  sin x  sin x  ...... 24

BBrilliant STUDY CENTRE MATHEMATICS (ONLINE) -2022 y  sin x  y  y2  sin x  y  2y dy  cos x  dy dx dx dy   cos x dx 2y 1 3) y  ax  ax  ax  ..... y ax  y  y2  ax  y ; 2y dt  ax log a  dy ;  dy  ax log a dx dx dx 2y 1 Result y  f x  f x  f x  ..... dy  f 'x dx 2y 1 y x2 1 x2 1   x2 1  ..... ; dy  2x dx 2y 1  y  sin x sin xsin x Find dy dx y  sin xy  log y  y log sin x ; 1 dy  y cos x  log sin x dy y dx sin x dx dy  1  log sin x   y cot x ; dy  y2 cot x  y2 cot x dx  y  dx 1 y log sin x 1 log y  y  cos x cos x... Find dy  y   cos x y dx log y  y log cos x  1 dy  y  sin x  log cos x dy y dx cos x dx dy  t1  log cos x    y tan x ; dy  y  tan x  y2 tan x dx dx 1 y log cos x log y 1 25

BBrilliant STUDY CENTRE MATHEMATICS (ONLINE) -2022 In General   y  f x f xfx... dy y2 f 'x f x  ; where log y = y log f(x) dx 1 log y  y  ax axax... ; dy  y2 ax log a  y2 log a  y2 log a dx ax 1 y log ax 1 xy log a 1 log y  y xx xx xx ; dy  dx 1) y  y  y....  x  x  x  .... Find dy dx Let y  y  y  .....  x  x  x  ....  t yt  xt t y  t2  t and x  t2  t y  t  t2 and x  t  t2 ; dy  2t 1  y  x 1 y  t2  t dt 2t 1 y  x 1 x  t2  t y  x  2t  2) yyyy...  log x  log  x  .... Find dy at x  e2  2 y  2 dx    3) xm    xm xm ... yn   yn yn ... Find dy dx Put xm  u and yn  u  uuuu...  vvv...  t 26

BBrilliant STUDY CENTRE MATHEMATICS (ONLINE) -2022 u t  vt  tu  1 and 1 tt v t t u  v nm  yn  mnm1  nyn1 dy dx dy  my sin ce xm  yn dx nx Differentiation by substitution 1) tan y  2t sin x  2t put t  tan  1 t2 1 t2 R e sult xy  K  dy  y dx x x  K  dy  y y dx x 2) y  sin2 cot 1 1 x put x  cos2  1 x 3) y  tan 1  4x   tan 1  2  3x   1 5x2   3 2x  Hint: 4x  5x  x and 5x2  5x.x 2  3x  2x 3 3 2x 1 2 x 3 Put 5x  tan A x  tan B and 2  tan C 3 Result When ; sinf x  Kcosf x  K etc Take inverse Diff. Ex : sec x2  y2  ea.Find dy x2  y2 dx 27

BBrilliant STUDY CENTRE MATHEMATICS (ONLINE) -2022 4) 1 x2  1 y2  a x  y x  sin A y  sin B cos A  cos B  2 cos A  B cos A  B 22 sin A  sin B  2 cos A  B sin A  B 22 5. Y  tan1  5ax   tan1  5x   tan1 3x  2x   6x2   a  aa   a2  16 x2  1 3x  2x  ax a  a  6.  x2  y2   ea find dy sec  x2  y2  dx   7. sin1 x  sin1 y   find dy 2 dx sin1 x  cos1 x   2    sin1 y  cos1 x  x  cos sin1 y , y  sin cos1 x  1 y2  1 sin2 cos1 x  cos2 cos1 x  x2 . Now differentiate 8. 3sin xy  4 cos xy  5 3, 4,5  (3, 4,5) is a pythagorian triple 32  42  52 Put 3  sin A and 4  cos A 55 9. y  tan 1  6x  8x3   1 12x 2    2x  tan  28

BBrilliant STUDY CENTRE MATHEMATICS (ONLINE) -2022 y  cos1  3x 4 1 x2  dy  5  find dx 10. x  cos   y  cos1  3 cos   4 sin    5 5  Put 3  cos A and 4  sin A 5 5 y  cos1 cos A    A   ex 1 ex 1  1  ex 1 ex  y  tan1  tan 1 11. 1  1  ex ex  Put ex  1  1 1  1 cos   2 sin2   tan 2  cos  cos  1 cos  2 cos2 2 2 1 1  cos  2 y  tan1 tan2   x tan1 tan    2 22 cos   ex    cos1 ex  1 1 x  ex 1  1  2 1 e2x 2  1 e2x   y  1 cos1 ex  dy    ex 2 dx dy  1 ex  1 exex  1 dx 2 e2x 1 2 e2x 1 2 e2x 1 e2x    12. y  x  a x2  a2 x4  a4 x8  a8 , x  a find dy dx 29

BBrilliant STUDY CENTRE MATHEMATICS (ONLINE) -2022 Derivative at a particular point Differentiate directly and substitute the point  1) f x  tan1 1 x2  x find f 0 2) tan 1  5ax   y then dy at x  0  a2  6x2  dx 3) y  cot1 cos2 x dy at x   dx 6 f x  cot 1  xx  xx  find f 'x  2  4)   In f ' x direct differentiation is used. Otherwise    Put t1 xx  t  xx 1 t   1 t2  1 tan2  Put t  tan   t2 2t 2 tan  f  x   cot 1   1 tan2    tan 1   1 2 tan    2 tan     tan2      tan1  tan 2  2  xx  2tan1 t  2tan1 xx Successive Differentiation (Higher order Derivation) d  dy   d2y or f \"x or y2 is the second order derivative. dx  dx  dx 2 d  d2y   d3y or f '''  x  is the third order derivative dx  dx 2  dx3   dn y  nth order Derivative dx n  dy n  nth Degree Derivative  dx  1) f x  x4 f 'x  4x3 f II x  12x2 f III x  24x f IV x  24 30

BBrilliant STUDY CENTRE MATHEMATICS (ONLINE) -2022 2) y  2x  32 d2y Find dx2 dy  2x  32  42x  3 ; d2y  42 8 dx dx 2 nth Derivatives of some Functions 1) dn xm  m! x mn dxn mn  ! 2) dn xn  n! dx n 3) dn ax  bm  m! a n ax  b mn dx n mn  ! 4) dn ax  bn  n !a n dx n 5) d emn  mnemn dx 6) dn sin ax  b  an sin  ax  b  n   dx n  2  7) dn sin x  sin  x  n   dx n  2  8) dn cos ax  b  an cos  ax  b  n   dx n  2  9) dn cos x  cos  x  n   dx n  2  10) dn log ax  b  1n1 n 1!an dx n ax  bn 11) dn log x  1n1 n 1! dx n xn 31

BBrilliant STUDY CENTRE MATHEMATICS (ONLINE) -2022 12) dn 1  1n n!an dx n ax  b ax  bn1 13) dn 1  1n n! dx n n x n1 14) dn xex  ex x  n dx n 1) y  aemx  bemx Find y10 Find y10  a  m10 emx  b  m10 emx y10  m10 aemx  bemx   m10y 2) d 20 2 cos x cos 3x  d 20 cos 4x  cos 2x 2 cos AcosB  cos(A  B)  cos(A  B) dx 20 dx 2  d20 cos 4x  d20 cos 2x dx 20 dx 20 Now use nth derivatives 3) d5 log 2x  3 dx log ax  b) dx5 use dxn 4) f  x   tan1 x Find d5 atx  0 dx5 f 'x  1  1  A B 1 x2 1 ix 1 ix 1 ix1 ix 1  A 1 ix  B1 ix  A  B 1 A1 2 A  B  0 B  1 2 d5  d4 1 1  1 1 dx5 dx4  2 1 ix 2 1 ix  32

BBrilliant STUDY CENTRE MATHEMATICS (ONLINE) -2022 dn 1 d5  1   14 i4 4!   14 i4 4!  1  24  24  24 dx n dx5 2  1 ix5 1 ix 5  2 Now use ax  b ;   Relation between y, y1 and y2 1) Find y1 2) Square and cross multiply y1 3) Get back function y 4) Differentiate once again w.r.t x 5) Divide by 2y1    1) y  cos m sin1 x  cos m sin1 x  s.t 1 x2 2  xy1  x2y  0  1)  m y1   sin m sin1 x    1 x2     2) y12 1 x2  m2 sin2 m sin1 x    3) y12 1 x2  m2 1 y2  4) y12 2x   1 x2 2y1y2  m2 2y y1    5) xy1  1 x2 y2  m2y ; 1 x2 y2  xy1  m2y  0 2) y  ea sin 1x a 1  x 2 ; y12 1 x2  a 2y2  y1  ea sin 1x   y12 2x   1 x2 2 y1y2  a2 2yy1  xy1  1 x2 y2  a2y  0 y  x  2  m  3) 1  x  n y  x  1 x2 Find relation y, y1 and y2 y1  n x  1 x2  n 1  2x  y1  n x  1 x2  n 1  x  1 x2   1  1 x2 ;   1 x2   2    33

BBrilliant STUDY CENTRE MATHEMATICS (ONLINE) -2022    y12 1 x2  x2y2 ; y12 2x1  1 x2 2y1y2  x2yy1  xy1  1 x2 y2  x2y 4) y  sin log x Find relation y y1 and y2 y1  cos log x 1  y12x 2  cos log x 2 x y12x2  1 sin2 log x  y12x2  1 y2 y12 2x  x2 2y1y2  2yy1 ; xy1  x2  y2  y  x2y2  xy1  y  0      d2yx dyx   y    x  f cos 3cos1 x . Find 1 x2 1 dx2 x dx      yx   TheQn is 1  x2 1 y2  xy1  Ans.9  y  3 1 x2 ; y12 1 x2      y1   sin 3cos1 x x  9 1 y2      y12 1 x2  9 1 y2 ; y12 2x   1 x2 2y1y2  9 2yy1     1 x2 1  x2 y2  xy1   9 y2  xy1  9y  y 1 Partial Differentiation If a dependent variable u depends on two independent variable x and y, it is denoted by u  f  xy and is called a Bivariate function. Let u  f  x y) be a bivariate funtion. The derivative u w.r.t x when y remains a constant is called the y partial derivative of u w.r.t x and is denoted by x . Thus u  Lt f x, yyf x, y f x, y x x0 y u y is the rate of change of u w.r.t. y when x remains constant. 34

BBrilliant STUDY CENTRE MATHEMATICS (ONLINE) -2022 u  3x2  3x2y  4xy2  3y3 u  6x  6xy  4y2 x u  3x2  8xy  9y2 y u  4x7  3x5y2  5y7    x u  y u  x 28x6 15x4y2  y 6x5y  35y6 x y   28x7 15x5y2  6x5y2  35y7  28x7  21x5y2  35y7  7 4x7  3x5y2  5y7  7u Euler’ Theorem u  f x, y is a bivariate homogenous function in degree n . Then x u  y u  nu x y Derivative of implicit function Consider the implicit function u  f  x, y  0  u    dy   x Then dx u y Ex:1) Let x2  x2y3  y2  0 dy   y    2x  2xy3  dx x  3x 2 y2  2y  u   y 35

BBrilliant STUDY CENTRE MATHEMATICS (ONLINE) -2022 2) Sin x  y  log x  y Find dy dx sin x  y  log x  y  0 y dy  y    cos  x  y  x 1 y   1  dx  cos  x  y  x  y  x  1   x      y    3) x2  y2  2  sin xy ;  x2  y2  sin xy  2  0 dy   u    2x  cos xy  y  dx x  2y  cos xy  x  x   y 4) exy  log xy  cos xy  5  0 Find dy dx dy  yexy  y 1  sin xy.y  y  exy  1  sin xy  y  xy   xy           dx 1 x 1 x  xexy  x xy  sin xy  x   exy  xy  sin xy      5) ax2  2hxy  by2  2gx  2fy  c  0 dy   y    2ax  2by  2g  dx x  2bx  2by  2f     x   y  Leibnitz Theorem Let u  f  x  be a bivariable function. d gx x t dt  gx  f x, t dt  g 'xf x g x  'xf x x dx x f   x   x  36

BBrilliant STUDY CENTRE MATHEMATICS (ONLINE) -2022 d ex ex sin x  log t  dt  cos xdt  ex sin x  x   ex sin x  2 log x   Ex: dx x2 x2   cos x xx  x2  ex  x  sin x   ex sin x  2 log x  Particular cases 1) d gx f  t  dt  g '(x)f g(x)   '  x  f   x  dx   x  2) d  x  f  t  dt   '  x  f    x   dx  k 3) d k f  t  dt   '  x  f   x  dx    x Questions gx  d x3 1) log t dt Find g ' e  dx x2 g x  3x2 3log x  2x 2log x g'x   x2  2x log x   4  x  log x  9 x  x    g 'e  9e  2e  42  27e  8 2) y 1 dt find d2y 1 4t2 dx 2 x 0  3) 1 x x x2 4 4t2  2f 't  thenf '4  f  dt x 4) f x is a continuous different function such that f  t dt  f  x  . Find log f(5) 0 37

BBrilliant STUDY CENTRE MATHEMATICS (ONLINE) -2022 5) f  x  is a non- negative fx defined in [0 1] s.t xx  1 f ' t 2 dt   f t  dt . Given f(0) = 0 . Then 00      A) f11 1  1 1  1  D) f 1 2 2 B) f 2 2 C) f 3 3 2 Inportant Result When f 'x  f x f 'x  1 f x f 'x dx  1dx f x log f x  x  cf x  ecex ; f 0  ec  f 0  0  f x  0 f 0 1 f x  ex f 0  2  f x  2ex f 0  k  f x  kex Derivative of a function w.r.t another function Derivative of f  x   d f x  f 'x gx g'x w.r.t g x  dx  d dx 1) Derivative of sin x  d sin x   cot x w.r.t cos x  dx  d cos x dx 2) Derivative of ax  ax log a w.r.t xa  ax a 1   38

BBrilliant STUDY CENTRE MATHEMATICS (ONLINE) -2022 3) Derivative of xx   xx 1 log x   x2x w.r.t cos xx  xx 1 log x   4) Derivative of sin x3 w.r.t x3 5) Derivative of asin1 x w.r.t sin1 x 6) Derivative of sin2 x w.r.t log x 2 Derivative of a Determinant d f1  x  f2 x  f1 ' x  f2 'x  f1 x  f2 x dx g1  x  g2 x g1 x  g2 x g11  x  g 2  x  2 xbb x b S.T d 1) 1  a x b 2  a x dx 1  32 aax 100 xbb xbb d 1 0      dx a x b 1 0a x b  x2  ab  x2  ab  x2  ab  32 aax aax 001 cos x   cos x   cos x   2) f x sin x   sin x   sin x   si    sin    sin    20 f x  5 Find  f x r 1 39