MSBSHSE Class 12 Maths and Statistics Part 2 Textbook PDF

 The Constitution of India Chapter IV A Fundamental Duties ARTICLE 51A Fundamental Duties- It shall be the duty of every citizen of India- (a) to abide by the Constitution and respect its ideals and institutions, the National Flag and th...

 The Constitution of India Chapter IV A Fundamental Duties ARTICLE 51A Fundamental Duties- It shall be the duty of every citizen of India- (a) to abide by the Constitution and respect its ideals and institutions, the National Flag and the National Anthem; (b) to cherish and follow the noble ideals which inspired our national struggle for freedom; (c) to uphold and protect the sovereignty, unity and integrity of India; (d) to defend the country and render national service when called upon to do so; (e) to promote harmony and the spirit of common brotherhood amongst all the people of India transcending religious, linguistic and regional or sectional diversities, to renounce practices derogatory to the dignity of women; (f) to value and preserve the rich heritage of our composite culture; (g) to protect and improve the natural environment including forests, lakes, rivers and wild life and to have compassion for living creatures; (h) to develop the scientific temper, humanism and the spirit of inquiry and reform; (i) to safeguard public property and to abjure violence; (j) to strive towards excellence in all spheres of individual and collective activity so that the nation constantly rises to higher levels of endeavour and achievement; (k) who is a parent or guardian to provide opportunities for education to his child or, as the case may be, ward between the age of six and fourteen years.  The Coordination Committee formed by GR No. Abhyas - 2116/(Pra.Kra.43/16) SD - 4 Dated 25.4.2016 has given approval to prescribe this textbook in its meeting held on 30.01.2020 and it has been decided to implement it from the educational year 2020-21. Mathematics and Statistics (Arts and Science) Part - II STANDARD - XII Maharashtra State Bureau of Textbook Production and Curriculum Research, Pune - 411 004 Download DIKSHA App on your smartphone. If you scan the Q.R.Code on this page of your textbook, you will be able to access full text and the audio-visual study material relevant to each lesson provided as teaching and learning aids. First Edition : 2020 © Maharashtra State Bureau of Textbook Production First Reprint : 2021 and Curriculum Research, Pune- 411 004. Maharashtra State Bureau of Textbook Production and Curriculum Research reserves all rights relating to the book. No part of this book should be reproduced without the written permission of the Director, Maharashtra State Bureau of Textbook Production and curriculum Research, Pune. Mathematics and Statistics Co-ordinator (Arts and Science) Ujjwala Shrikant Godbole Commitee Members I/C Special Officer for Mathematics Dr. Mangla Narlikar (Chairman) Dr. Sharad Gore (Member) Shri. Prahallad Chippalagatti (Member) Cover & Shri. Prasad Kunte (Member) Computer Drawings Shri. Sujit Shinde (Member) Shri. Sandip Koli, Artist, Mumbai Smt. Prajkti Gokhale (Member) Shri. Ramakant Sarode (Member) Smt. Ujjwala Godbole (Member-Secretary) Production Sachchitanand Aphale Chief Production Officer Mathematics and Statistics Sanjay Kamble (Arts and Science) Production Officer Study Group Members Prashant Harne Dr. Ishwar Patil Dr. Pradeep Mugale Asst. Production Officer Dr. Pradnyankumar Bhojnkar Shri. Uday Mahindrakar Shri. Prafullachandra Pawar Shri. Balkrishna Mapari Shri. Sachin Batwal Shri. Devanand Bagul Paper Shri. Swapnil Shinde Smt. Swati Powar 70 GSM Cream wove Shri. Pradipkumar Bhavsar Smt. Mahadevi Mane Shri. Amit Totade Shri. Dilip Panchamia Print Order No. Smt. 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Prabhadevi Mumbai- 400 025  The Constitution of India Preamble WE, THE PEOPLE OF INDIA, having solemnly resolved to constitute India into a SOVEREIGN SOCIALIST SECULAR DEMOCRATIC REPUBLIC and to secure to all its citizens: JUSTICE, social, economic and political; LIBERTY of thought, expression, belief, faith and worship; EQUALITY of status and    of opportunity; and to promote among them all FRATERNITY assuring the dignity of the individual and the unity and integrity of the Nation; IN OUR CONSTITUENT ASSEMBLY this twenty-sixth day of November, 1949, do HEREBY ADOPT, ENACT AND GIVE TO OURSELVES THIS CONSTITUTION. NATIONAL ANTHEM  PREFACE Dear Students, Welcome to Standard XII, an important milestone in your life. Standard XII or Higher Secondary School Certificate opens the doors of higher education. Alternatively, you can pursue other career paths like joining the workforce. Either way, you will find that mathematics education helps you considerably. Learning mathematics enables you to think logically, constistently, and rationally. The curriculum for Standard XII Mathematics and Statistics for Science and Arts students has been designed and developed keeping both of these possibilities in mind. The curriculum of Mathematics and Statistics for Standard XII for Science and Arts students is divided in two parts. Part I deals with topics like Mathematical Logic, Matrices, Vectors and Introduction to three dimensional geometry. Part II deals with Differentiation, Integration and their applications, Introduction to random variables and statistical methods. The new text books have three types of exercises for focused and comprehensive practice. First, there are exercises on every important topic. Second, there are comprehensive exercises at the end of all chapters. Third, every chapter includes activities that students must attempt after discussion with classmates and teachers. Additional information has been provided on the E-balbharati website (www.ebalbharati.in). We are living in the age of Internet. You can make use of modern technology with the help of the Q.R. code given on the title page. The Q.R. code will take you to links that provide additional useful information. Your learning will be fruitful if you balance between reading the text books and solving exercises. Solving more problems will make you more confident and efficient. The text books are prepared by a subject committee and a study group. The books (Paper I and Paper II) are reviewed by experienced teachers and eminent scholars. The Bureau would like to thank all of them for their valuable contribution in the form of creative writing, constructive and useful suggestions for making the text books valuable. The Bureau hopes and wishes that the text books are very useful and well received by students, teachers and parents. Students, you are now ready to study. All the best wishes for a happy learning experience and a well deserved success. Enjoy learning and be successful. (Vivek Gosavi) Pune Director Date : 21 February 2020 Maharashtra State Bureau of Textbook Bharatiya Saur : 2 Phalguna 1941 Production and Curriculum Research, Pune.  Mathematics and Statistics XII (Part II) Arts and Science Sr. No Area / Topic Sub Unit Competency Statement The students will be able to state and use standard formulas of derivative of standard functions use chain rule of derivatives 1. Differentiation Differentiation find derivatives of the logarithm, implicit, inverse and parametric functions find second and higher order derivatives. find equations of tangents and normal to a curve determine nature of the function- increasing or decreasing Applications Applications of 2. find approximate values of the of Derivatives Derivatives function examine function for maximum and minimum values verify mean value theorems understand the relation between derivative and integral use the method of substitution Indefinite Indefinite 3. solve integrals with the help of Integration Integration integration by parts solve the integrals by the method of partial fractions understand integral as a limit of sum Definite Definite the properties of definite integral 4. Integration Integration state the properties of definite integral and use them to solve problems  Application Application of Definite of Definite find the area under the curve, bounded 5. Integration Integration by the curves using definite integrals. form a differential equation and find its order and degree solve the first order and first degree differential equation by various Differential Differential 6. methods Equation Equation apply the differential equations to study the polpulation, growth and decay in amount of substance and physics. understand the random variable and its types. find probability mass function and its probability distribution. Probability Probability 7. find the expected value, variance and Distribution Distribution the standard deviation find the probability density function of continuous random variable find distribution function of c.r.v. understand random experiment with two or more outcomes. determine probability distribution of Binomial Binomial random experiment with parameters n 8 Distribution Distribution and p. find mean, variance, expected value and standard deviation for the binomial distribution.  INDEX Sr. No. Chapter Name Page No. 1 Differentiation 1-64 2 Applications of Derivatives 65-94 3 Indefinite Integration 95-150 4 Definite Integration 151-177 5 Application of Definite Integration 178-190 6 Differential Equations 191-218 7 Probability Distributions 219-244 8 Binomial Distribution 245-255 Answers 256-276  1. DIFFERENTIATION Let us Study Derivatives of Composite functions. Geometrical meaning of Derivative. Derivatives of Inverse functions Logarithmic Differentiation Derivatives of Implicit functions. Derivatives of Parametric functions. Higher order Derivatives. Let us Recall The derivative of f (x) with respect to x, at x = a is given by The derivative can also be defined for f (x) at any point x on the open interval as. If the function is given as y = f (x) then its derivative is written as. For a differentiable function y = f (x) if δx is a small increment in x and the corresponding increment in y is δy then. Derivatives of some standard functions. y = f (x) y = f (x) c (Constant) 0 sec x sec x tan x xn nxn−1 1 1 cosec x − cosec x cot x − 2 x x cot x − cosec2 x 1 n ex ex − xn xn+1 ax a x log a 1 log x x sin x cos x cos x − sin x 1 log a x tan x sec2 x x log a Table 1.1.1 1 Rules of Differentiation : If u and v are differentiable functions of x such that (i) y = u ± v then (ii) y = uv then u (iii) y= where v ≠ 0 then v Introduction : The history of mathematics presents the development of calculus as being accredited to Sir Isaac Newton (1642-1727) an English physicist and mathematician and Gottfried Wilhelm Leibnitz (1646- 1716) a German physicist and mathematician. The Derivative is one of the fundamental ideas of calculus. It's all about rate of change in a function. We try to find interpretations of these changes in a mathematical way. The symbol δ will be used to represent the change, for example δx represents a small change in the variable x and it is read as "change in x" or "increment in x". δy is the corresponding change in y if y is a function of x. We have already studied the basic concept, derivatives of standard functions and rules of differentiation in previous standard. This year, in this chapter we are going to study the geometrical meaning of derivative, derivatives of Composite, Inverse, Logarithmic, Implicit and Parametric functions and also higher order derivatives. We also add some more rules of differentiation. Let us Learn 1.1.1 Derivatives of Composite Functions (Function of another function) : So far we have studied the derivatives of simple functions like sin x, log x, e x etc. But how about the derivatives of sin , log sin (x2 + 5) or e tan x etc ? These are known as composite functions. In this section let us study how to differentiate composite functions. 1.1.2 Theorem : If y = f (u) is a differentiable function of u and u = g (x) is a differentiable function of x such that the composite function y = f [g (x)] is a differentiable function of x then. Proof : Given that y = f (u) and u = g (x). We assume that u is not a constant function. Let there be a small increment in the value of x say δx then δu and δy are the corresponding increments in u and y respectively. As δx, δu, δy are small increments in x, u and y respectively such that δx ≠ 0, δu ≠ 0 and δy ≠ 0. We have. Taking the limit as δx → 0 on both sides we get, 2  As δx → 0, we get, δu → 0 (⸪ u is a continuous function of x)..... (I) Since y is a differentiable function of u and u is a differentiable function of x. we have, and..... (II) From (I) and (II), we get..... (III) The R.H.S. of (III) exists and is finite, implies L.H.S.of (III) also exists and is finite. Then equation (III) becomes, Note: 1. The derivative of a composite function can also be expressed as follows. y = f (u) is a differentiable function of u and u = g (x) is a differentiable function of x such that the composite function y = f [ g (x)] is defined then. 2. If y = f (v) is a differentiable function of v and v = g (u) is a differentiable function of u and u = h (x) is a differentiable function of x then. 3. If y is a differentiable function of u1, ui is a differentiable function of ui+1 for i = 1, 2,..., n−1 and un is a differentiable function of x, then This rule is also known as Chain rule. 3 1.1.3 Derivatives of some standard Composite Functions : dy dy y y dx dx [ f (x)] n n [ f (x)] n−1 ⋅ f '(x) cot [ f (x)] − cosec2 [ f (x)]⋅ f '(x) f '(x) cosec [ f (x)] − cosec [ f (x)] ⋅ cot [ f (x)] ⋅ f '(x) √ f (x) 2√ f (x) a f (x) a f (x) ⋅ log a ⋅ f '(x) 1 n ⋅ f '(x) − e f (x) e f (x) ⋅ f '(x) [ f (x)]n [ f (x)] n+1 f '(x) sin [ f (x)] cos [ f (x)]⋅ f '(x) log [ f (x)] cos [ f (x)] − sin [ f (x)]⋅ f '(x) f (x) tan [ f (x)] sec2 [ f (x)]⋅ f '(x) f '(x) log a [ f (x)] sec [ f (x)] sec [ f (x)] ⋅ tan [ f (x)] ⋅ f '(x) f (x) log a Table 1.1.2 SOLVED EXAMPLES Ex. 1 : Differentiate the following w. r. t. x. (i) y = √ x2 + 5 (ii) y = sin (log x) (iii) y = e tan x 3 (iv) log (x5 + 4) (v) 53 cos x − 2 (vi) y= (2x2 − 7)5 Solution : (i) y = √ x2 + 5 Method 1 : Method 2 : Let u = x2 + 5 then y = √ u , where y is We have y = √ x2 + 5 a differentiable function of u and u is a Differentiate w. r. t. x differentiable function of x then..... (I) [Treat x2 + 5 as u in mind and use the formula Now, y = √ u of derivative of √ u ] Differentiate w. r. t. u and u = x2 + 5 Differentiate w. r. t. x du d 2 = (x + 5) = 2x dx dx Now, equation (I) becomes, 4 (ii) y = sin (log x) Method 1 : Method 2 : Let u = log x then y = sin u, where y is We have y = sin (log x) a differentiable function of u and u is a Differentiate w. r. t. x differentiable function of x then [sin (log x)]..... (I) [Treat log x as u in mind and use the formula Now, y = sin u of derivative of sin u] Differentiate w. r. t. u = cos u and u = log x Differentiate w. r. t. x 1 = x Now, equation (I) becomes, Note : Hence onwards let's use Method 2. (iii) y = e tan x (iv) Let y = log (x5 + 4) Differentiate w. r. t. x Differentiate w. r. t. x [e tan x] [log (x5 + 4)] 3 (v) Let y = 53 cos x − 2 (vi) Let y = (2x2 − 7)5 Differentiate w. r. t. x Differentiate w. r. t. x [53 cos x − 2] dy = 53 cos x − 2 · log 5 × (3 cos x − 2) dx dy = − 3 sin x · 53 cos x − 2 · log 5 dx 5 Ex. 2 : Differentiate the following w. r. t. x. (i) y = √ sin x3 (ii) y = cot2 (x3) (iii) y = log [cos (x5)] (iv) y = (x3 + 2x − 3)4 (x + cos x) 3 (v) y = (1 + cos2 x) 4 × √ x + √tan x Solution : (i) y= √ sin x3 (ii) y = cot2 (x3) Differentiate w. r. t. x Differentiate w. r. t. x = 2 cot (x3) [cot (x3)] = 2 cot (x3)[− cosec2 (x3)] (x3) = − 2 cot (x3)cosec2 (x3)(3x2) dy ∴ ∴ = − 6x2 cot (x3)cosec2 (x3) dx (iii) y = log [cos (x5)] Differentiate w. r. t. x dy = (log [cos (x5)]) dx dy ∴ = − tan (x5) (5x4) = − 5x4 tan (x5) dx (iv) y = (x3 + 2x − 3)4 (x + cos x) 3 Differentiate w. r. t. x = (x3 + 2x − 3)4 ⋅ (x + cos x) 3 + (x + cos x) 3⋅ (x3 + 2x − 3)4 6  = (x3 + 2x − 3)4 ⋅3 (x + cos x) 2 ⋅ (x + cos x) + (x + cos x) 3⋅ 4(x3 + 2x − 3)3 ⋅ (x3 + 2x − 3) = (x3 + 2x − 3)4 ⋅3 (x + cos x) 2 (1 − sin x) + (x + cos x) 3 ⋅ 4(x3 + 2x − 3)3 (3x2 + 2) dy ∴ = 3(x3 + 2x − 3)4 (x + cos x) 2 (1 − sin x) + 4 (3x2 + 2) (x3 + 2x − 3)3 (x + cos x)3 dx (v) y = (1 + cos2 x) 4 × √ x + √tan x Differentiate w. r. t. x Ex. 3 : Differentiate the following w. r. t. x. (i) y = log3 (log5 x) (ii) (iii) (iv) (v) (vi) 7 Solution : (i) y = log3 (log5 x) = log3 = log3 (log x) - log3 (log 5) ∴ y = - log3 (log 5) Differentiate w. r. t. x [Note that log3(log 5) is constant] ∴ (ii) ∴ [ ⸪ log e = 1] Differentiate w. r. t. x ∴ 8 (iii) ∴ Differentiate w. r. t. x ∴ (iv) ∴ Differentiate w. r. t. x 9  (v) (vi) y = a cot x [⸪ alog a f (x) = f (x)] Differentiate w. r. t. x [⸪ alog a f (x) = f (x)] (a cot x ) = sin2 x + cos2 x = a cot x log a · (cot x) ∴ y =1 Differentiate w. r. t. x =a cot x log a (− cosec2 x) − cosec2 x· a cot x log a Ex. 4 : If f (x) = √ 7g (x) − 3 , g (3) = 4 and g' (3) = 5, find f ' (3). Solution : Given that : f (x) = √ 7g (x) − 3 Differentiate w. r. t. x ∴ For x = 3, we get 35 7 = = [Since g (3) = 4 and g' (3) = 5] 2(5) 2 10 Ex. 5 : If F (x) = G {3G [5G(x)]}, G(0) = 0 and G' (0) = 3, find F' (0). Solution : Given that : F (x) = G {3G [5G(x)]} Differentiate w. r. t. x F' (x) = G {3G [5G(x)]} = G' {3G [5G(x)]}3· [G [5G(x)]] = G' {3G [5G(x)]}3·G' [5G(x)] 5· [G(x)] F' (x) = 15·G' {3G [5G(x)]}G' [5G(x)] G' (x) For x = 0, we get F' (0) = 15·G' {3G [5G(0)]}G' [5G(0)] G' (0) = 15·G' [3G (0)]G' (0)·(3) [⸪ G(0) = 0 and G' (0) = 3] = 15·G' (0)(3)(3) = 15·(3)(3)(3) = 405 Ex. 6 : Select the appropriate hint from the hint basket and fill in the blank spaces in the following paragraph. [Activity] "Let f (x) = sin x and g (x) = log x then f [g(x)] = _ _ _ _ _ _ _ _ _ _ _ _ _ and g [ f (x)] = _ _ _ _ _ _ _ _ _. Now f ' (x) = _ _ _ _ _ _ _ _ _ _ and g' (x) = _ _ _ _ _ _ _ _. The derivative of f [g(x)] w. r. t. x in terms of f and g is _ _ _ _ _ _ _ __ _ _ _ _. Therefore [ f [g(x)]] = _ _ _ _ _ _ _ _ _ and = _ _ _ _ _ _ _ _ _ _ _ _. The derivative of g [ f (x)] w. r. t. x in terms of f and g is _ _ _ _ _ _ _ __ _ _ _ _. Therefore [g [ f (x)]] = _ _ _ _ _ _ _ _ _ and = _ _ _ _ _ _ _ _ _ _ _." cos (log x) Hint basket : { f ' [g(x)]·g' (x), , 1, g' [ f (x)]·f ' (x), cot x, √ 3, x 1 sin (log x), log (sin x), cos x, } x 1 cos (log x) Solution : sin (log x), log (sin x), cos x, , f ' [g(x)]·g' (x), , 1, g' [ f (x)]·f ' (x), cot x, √ 3. x x EXERCISE 1.1 (1) Differentiate w. r. t. x. 5 (i) (x3 − 2x − 1) (ii) (v) (iii) (iv) (vi) 11 (2) Differentiate the following w.r.t. x (xv) (i) cos (x2 + a2) (ii) (iii) (iv) (xvi) cot3[log (x3)] 3x+3 (v) (vi) 5sin (vii) cosec (√cos x) (viii) log [cos (x3 − 5)] (xvii) (x) cos2 [log (x2 + 7)] 2 x − 2 cos2 x (ix) e3 sin (xi) tan [cos (sin x)] (xii) sec[tan (x4 + 4)] (xviii) (xiii) elog [(log x) − log x2 ] (xiv) sin √sin √x 2 (xix) y = (25)log5 (sec x) − (16)log4 (tan x) (xv) log[sec (e )] x2 (xvi) loge2 (log x) (xx) (xvii) [ log [log(log x)]] 2 (xviii) sin2 x2 − cos2 x2 (4) A table of values of f, g, f ' and g' is given (3) Differentiate the following w.r.t. x x f (x) g(x) f '(x) g'(x) (i) (x2 + 4x + 1)3 + (x3 − 5x − 2)4 5 8 2 1 6 −3 4 (ii) (1 + 4x) (3 + x − x2) 4 3 4 5 −6 (iii) (iv) 6 5 2 −4 7 (v) (1 + sin2 x)2 (1 + cos2 x)3 (i) If r(x) = f [g(x)] find r' (2). (vi) √cos x + √cos √x (ii) If R(x) = g[3 + f (x)] find R' (4). (vii) log (sec 3x + tan 3x) (viii) (iii) If s(x) = f [9 − f (x)] find s' (4). (iv) If S(x) = g [ g(x) ] find S' (6). (ix) (5) Assume that f ' (3) = − 1, g' (2) = 5, g (2) = 3 (x) (xi) and y = f [g(x)] then (xii) log [tan3 x·sin4 x·(x2 + 7) ] 7 (6) If , f (1) = 4, g (1) = 3, (xiii) f ' (1) = 3, g' (1) = 4 find h' (1). (7) Find the x co-ordinates of all the points on the curve y = sin 2x − 2 sin x, 0 ≤ x < 2π (xiv) where 0. 12 (8) Select the appropriate hint from the hint basket and fill up the blank spaces in the following paragraph. [Activity] "Let f (x) = x2 + 5 and g (x) = ex + 3 then Therefore [ f [g(x)]] = _ _ _ _ _ _ _ _ _ and f [ g (x)] = _ _ _ _ _ _ _ _ and = _ _ _ _ _ _ _ _ _ _ _. g [ f (x)] =_ _ _ _ _ _ _ _. The derivative of g [ f (x)] w. r. t. x in terms of Now f '(x) = _ _ _ _ _ _ _ _ and f and g is _ _ _ _ _ _ __ _ _ _ _. g' (x) = _ _ _ _ _ _ _ _. The derivative of f [g (x)] w. r. t. x in terms Therefore [g [ f (x)]] = _ _ _ _ _ _ _ _ _ of f and g is _ _ _ _ _ _ _ _. and = _ _ _ _ _ _ _ _ _ _ _." Hint basket : { f ' [g(x)]·g' (x), 2e2x + 6ex, 8, g' [ f (x)]· f ' (x), 2xe x +5, − 2e6, e2x + 6ex + 14, e x +5 + 3, 2x, ex} 2 2 1.2.1 Geometrical meaning of Derivative : Consider a point P on the curve f (x). At x = a, the coordinates of P are (a, f (a)). Let Q be another point on the curve, a little to the right of P i.e. to the right of x = a, with a value increased by a small real number h. Therefore the coordinates of Q are ((a + h), f (a + h)). Now we can calculate the slope of the secant line PQ i.e. slope of the secant line connecting the points P (a, f (a)) and Q ((a + h), f (a + h)), by using formula for slope. Slope of secant PQ Fig. 1.2.1 Suppose we make h smaller and smaller then a + h will approach a as h gets closer to zero, Q will approach P, that is as h → 0, the secant coverges to the tangent at P. ∴ (Slope of secant PQ) = = f ' (a) So we get, Slope of tangent at P = f ' (a)..... [ If limit exists] Thus the derivative of a function y = f (x) at any point P (x1, y1) is the slope of the tangent at that point on the curve. If we consider the point a − h to the left of a, h > 0, then with R = ((a − h), f (a − h)) we will find the slope of PR which will also converge to the slope of tangent at P. For Example : If y = x2 + 3x + 2 then slope of the tangent at (2,3) is given by Slope = (2x + 3)(2,3) = 2 (2) + 3) ∴ m=7 13 1.2.2 Derivatives of Inverse Functions : We know that if y = f (x) is a one-one and onto function then x = f ( y) exists. If f −1 ( y) is −1 differentiable then we can find its derivative. In this section let us discuss the derivatives of some inverse functions and the derivatives of inverse trigonometric functions. x+2 Example 1 : Consider f (x) = 2x − 2 then its inverse is f −1 (x) =. Let g(x) = f −1 (x). 2 d d 1 If we find the derivatives of these functions we see that [ f (x)] = 2 and [g (x)] =. dx dx 2 These derivatives are reciprocals of one another. Example 2 : Consider y = f (x) = x2. Let g = f −1. ∴ g ( y) = x = √ y 1 ∴ g' ( y) = also f ' (x) = 2x 2√y d d d Now [g ( f (x))] = = 1 and g [ f (x)] = x ∴ [g ( f (x))] = (x) = 1 dx dx dx 1 1 1 At a point ( x, x2 ) on the curve, f ' (x) = 2x and g' ( y) = = =. 2√y 2x f '(x) 1.2.3 Theorem : Suppose y = f (x) is a differentiable function of x on an interval I and y is One-one, onto and dy d −1 1 dx dy ≠ 0 on I. Also if f −1( y) is differentiable on f (I ) then [ f ( y)] = or = where ≠ 0. dx dy f '(x) dy dx Proof : Given that y = f (x) and x = f −1 (y) are differentiable functions. Let there be a small increment in the value of x say δx then correspondingly there will be an increment in the value of y say δy. As δx and δy are increments, δx ≠ 0 and δy ≠ 0. δx δy We have, × =1 δy δx δx δy ∴ = , where ≠0 δy δx Taking the limit as δx → 0, we get, as δx → 0, δy → 0,..... (I) Since y = f (x) is a differentiable function of x. 14  dy we have, and ≠ 0..... (II) dx From (I) and (II), we get..... (III) δx exists and is finite. ∴ lim dy dx As ≠ 0, δy→0 = exists and is finite. dx δy dy dx dy Hence, from (III) = where ≠0 dy dx An alternative proof using derivatives of composite functions rule. We know that f −1 [ f (x)] = x [Identity function] Taking derivative on bothe sides we get, d d [ f −1 [ f (x)]] = (x) dx d dx i.e. ( f −1)' [ f (x)] [ f (x)] = 1 dx i.e. ( f −1)' [ f (x)] f ' (x) = 1 1 ∴ ( f −1)' [ f (x)] =..... (I) f ' (x) So, if y = f (x) is a differentiable function of x and x = f −1 ( y) exists and is differentiable then ( f −1)' [ f (x)] = ( f −1)' ( y) = dx dy and f ' (x) = dy dx ∴ (I) becomes dx dy = where ≠0 dy dx SOLVED EXAMPLES Ex. 1 : Find the derivative of the function y = f (x) using the derivative of the inverse function x = f −1 ( y) in the following (i) (ii) (iii) y = ln x Solution : (i) We first find the inverse of the function y = f (x), i.e. x in term of y. y3 = x + 4 ∴ x = y3 − 4 ∴ x = f −1 ( y) = y3 − 4 dy = = dx for x ≠ −4 15  (ii) We first find the inverse of the function y = f (x), i.e. x in term of y. 2 y2 = 1 + √ x i.e. √ x = y2 − 1, ∴ x = f −1 ( y) = ( y2 − 1) dy = = dx (iii) y = log x We first find the inverse of the function y = f (x), i.e. x in term of y. y = log x ∴ x = f −1 ( y) = e y dy 1 1 1 = = = y = ln x =. dx e e x Ex. 2 : Find the derivative of the inverse of Ex. 3 : Let f and g be the inverse functions of function y = 2x3 − 6x and calculate its each other. The following table lists a value at x = −2. few values of f, g and f ' Solution : Given : y = 2x3 − 6x x f (x) g(x) f '(x) Diff. w. r. t. x we get, 1 dy −4 2 1 = 6x2 − 6 = 6 (x2 − 1) 3 dx dx 1 −4 −2 4 we have, = dy find g' (−4). dx 1 Solution : In order to find g' (−4), we should first ∴ = dy 6 (x − 1) 2 find an expression for g' (x) for any input at x = − 2, x. Since f and g are inverses we can use we get, y = 2(−2)3 − 6(−2) the following identify which holds for = − 16 + 12 = − 4 any two diffetentiable inverse functions. 1 g' (x) =... [check, how?] f ' [g(x)]... [Hint : f [g(x)] = x] 1 1 = ∴ g' (−4) = 6 ((−2)2 − 1) f '[g (−4)] 1 1 = 1 18 = = f '(1) 4 16 Ex. 4 : Let f (x) = x5 + 2x − 3. Find ( f −1)' (−3). Solution : Given : f (x) = x5 + 2x − 3 Diff. w. r. t. x we get, f '(x) = 5x4 + 2 Note that f (x) = −3 corresponds to x = 0. 1 ∴ ( f −1)' (−3) = f ' (0) 1 1 = = 5(0) + 2 2 1.2.4 Derivatives of Standard Inverse trigononmetric Functions : We observe that inverse trigonometric functions are multi-valued functions and because of this, their derivatives depend on which branch of the function we are dealing with. We are not restricted to use these branches all the time. While solving the problems it is customary to select the branch of the inverse trigonometric function which is applicable to the kind of problem we are solving. We have to pay more attention towards the domain and range. π π dy 1 1. If y = sin−1 x, −1 ≤ x ≤ 1, − ≤ y ≤ then prove that = , |x| < 1. 2 2 dx √1 − x2 π π Proof : Given that y = sin−1 x, −1 ≤ x ≤ 1, − ≤ y ≤ 2 2 ∴ x = sin y.... (I) Differentiate w. r. t. y dx d = (sin y) dy dy dx = cos y = ± √cos2 y = ± √1 − sin2 y dy dx ∴ = ± √1 − x2.... [⸪ sin y = x] dy π π But cos y is positive since y lies in 1st or 4th quadrant as − ≤ y ≤ 2 2 dx ∴ = √1 − x2 dy dy We have = dx dy 1 ∴ = , |x| < 1 dx √1 − x2 dy 1 2. If y = cos−1 x, −1 ≤ x ≤ 1, 0 ≤ y ≤ π then prove that =−. dx √1 − x2 [As home work for students to prove.] 17  dy 1 3. If y = cot−1 x, x ∈ R, 0 < y < π then =−. dx 1 + x2 Proof : Given that y = cot−1 x, x ∈ R, 0 < y < π ∴ x = cot y.... (I) Differentiate w. r. t. y dx d = (cot y) dy dy dx = − cosec2 y = − (1 + cot2 y) dy dx ∴ = − (1 + x2).... [⸪ cot y = x] dy dy We have = dx dy 1 dy 1 ∴ = ∴ =− dx − (1 + x2) dx 1 + x2 π π dy 1 4. If y = tan−1 x, x ∈ R, − < y < then =. [left as home work for students to prove.] 2 2 dx 1 + x2 π dy 1 5. If y = sec−1 x, such that |x| ≥ 1 and 0 ≤ y ≤ π, y ≠ then = if x > 1 2 dx x√ x2 − 1 dy 1 =− if x < − 1 dx x√ x2 − 1 π Proof : Given that y = sec−1 x, |x| ≥ 1 and 0 ≤ y ≤ π, y ≠ 2 ∴ x = sec y.... (I) Differentiate w. r. t. y dx d = (sec y) dy dy dx = sec y · tan y dy dx ∴ = ± sec y ·√tan2 y dy = ± sec y ·√sec2 y − 1 dx Fig. 1.2.2 ∴ = ± x √ x2 − 1..... [⸪ sec y = x] dy We use the sign ± because for y in 1st and 2nd quadrant. sec y · tan y > 0. Hence we choose x √ x2 − 1 if x > 1 and − x √ x2 − 1 if x < − 1 In 1st quadrant both sec y and tan y are positive. In 2nd quadrant both sec y and tan y are negative. ∴ sec y · tan y is positive in both first and second quadrant. 18  Also, for x > 0, x √ x2 − 1 > 0 and for x < 0, − x √ x2 − 1 > 0 dx = x √ x2 − 1 , when x > 0, |x| > 1 i.e. x > 1 dy = − x √ x2 − 1 , when x < 0, |x| > 1 i.e. x < − 1 dy 1 = if x > 1 dx x√ x2 − 1 dy 1 =− if x < − 1 dx x√ x2 − 1 Note 1 : A function is increasing if its derivative is positive and is decreasing if its derivative is negative. Note 2 : The derivative of sec−1 x is always positive because the graph of sec−1 x is always increasing. π π dy 1 6. If y = − cosec x, such that |x| ≥ 1 and − ≤ y ≤ , y ≠ 0 then =− if x>1 2 2 dx x√ x2 − 1 dy 1 = if x 1 √1 − x2 − ≤y≤ x√ x2 − 1 0≤y≤π 2 2 sec −1 x 1 1 π −1 ≤ x ≤ 1 − for x < − 1 y≠ cos −1 x − , |x| < 1 x√ x2 − 1 2 √1 − x2 0≤y≤π 1 x∈R tan −1 x 1 |x| ≥ 1 π π 1 + x2 − 1 π π 2 2 cosec −1 x x√ x2 − 1 − ≤y≤ 1 1 2 2 − x∈R for x < − 1 cot −1 x x√ x2 − 1 y≠0 1 + x2 0 1 tan −1 [ f (x)] cosec −1 [ f (x)] 1 + [ f (x)]2 f (x) √[ f (x)] − 1 2 Table 1.2.2 Some Important Formulae for Inverse Trigonometric Functions : (1) sin−1 (sin θ) = θ, sin(sin−1 x) = x (2) cos−1 (cos θ) = θ, cos(cos−1 x) = x (3) tan−1 (tan θ) = θ, tan(tan−1 x) = x (4) cot−1 (cot θ) = θ, cot(cot−1 x) = x (5) sec−1 (sec θ) = θ, sec(sec−1 x) = x (6) cosec−1 (cosec θ) = θ, cosec(cosec−1 x) = x π π π π (7) sin−1 (cos θ) = sin−1 sin −θ = −θ (8) cos−1 (sin θ) = cos−1 cos −θ = −θ 2 2 2 2 π π π π (9) tan−1 (cot θ) = tan−1 tan −θ = −θ (10) cot−1 (tan θ) = cot−1 cot −θ = −θ 2 2 2 2 π π (11) sec−1 (cosec θ) = sec−1 sec −θ = −θ 2 2 π π (12) cosec−1 (sec θ) = cosec−1 cosec −θ = −θ 2 2 1 1 (13) sin−1 (x) = cosec−1 (14) cosec−1 (x) = sin−1 x x 1 1 (15) cos−1 (x) = sec−1 (16) sec−1 (x) = cos−1 x x 1 1 (17) tan−1 (x) = cot−1 (18) cot−1 (x) = tan−1 x x π π π (19) sin−1 x + cos−1 x = (20) tan−1 x + cot−1 x = (21) sec−1 x + cosec−1 x = 2 2 2 x+y x−y (22) tan−1 x + tan−1 y = tan−1 (23) tan−1 x − tan−1 y = tan−1 1 − xy 1 + xy In above tables, x is a real variable with restrictions. Table 1.2.3 20 Some Important Substitutions : Expression Substitutions Expression Substitutions √1 − x2 x = sin θ or x = cos θ 2x x = tan θ 1 + x2 √1 − x2 x = tan θ or x = cot θ 1 − x2 x = tan θ √ x2 + 1 x = sec θ or x = cosec θ 1 + x2 a+x a−x 3x − 4x3 or 1 − 2x2 x = sin θ or x = a cos 2θ or x = a cos θ a−x a+x 4x3 − 3x or 2x2 − 1 x = cos θ 3x − x3 1+x 1−x x = tan θ or x = cos 2θ or x = cos θ 1 − 3x2 1−x 1+x 2 f (x) 1 − [ f (x)]2 a + x2 a − x2 or f (x) = tan θ or x2 = a cos 2θ or x2 = a cos θ a − x2 a + x2 1 + [ f (x)]2 1 + [ f (x)]2 Table 1.2.4 SOLVED EXAMPLES π Ex. 1 : Using derivative prove that sin−1 x + cos−1 x =. 2 Solution : Let f (x) = sin−1 x + cos−1 x..... (I) π We have to prove that f (x) = 2 Differentiate (I) w. r. t. x d d [ f (x)] = [sin−1 x + cos−1 x] dx dx 1 1 f ' (x) = − =0 √1 − x2 √1 − x2 f ' (x) = 0 ⇒ f (x) is a constant function. Let f (x) = c. For any value of x, f (x) must be c only. So conveniently we can choose x = 0, ∴ from (I) we get, π π π π f (0) = sin−1 (0) + cos−1 (0) = 0 + = ⇒ c = ∴ f (x) = 2 2 2 2 π Hence, sin x + cos x =. −1 −1 2 Ex. 2 : Differentiate the following w. r. t. x. (i) sin−1 (x3) (ii) cos−1 (2x2 − x) (iii) sin−1 (2x) 1 1+x (iv) cot−1 (v) cos−1 (vi) sin2 (sin−1 (x2)) x2 2 21 Solution : (ii) Let y = cos−1 (2x2 − x) (i) Let y = sin−1 (x3) Hence cos y = 2x2 − x... (I) Differentiate w. r. t. x. Differentiate w. r. t. x. dy d dy = (sin−1 (x3)) − sin y· = 4x − 1 dx dx dx 1 d 3 1 − 4x 1 − 4x = · (x ) dy = = √1 − (x3)2 dx dx sin y √1 − cos2 y 1 = (3x2) dy 1 − 4x √1 − x 6 ∴ =... from (I) 3x2 dx √1 − x2 (2x − 1)2 dy ∴ = dx √1 − x6 Alternate Method : If y = cos−1 (2x2 − x) (iii) Let y = sin−1 (2x) Differentiate w. r. t. x. Differentiate w. r. t. x. dy d dy d = (cos−1 (2x2 − x)) = (sin−1 (2x)) dx dx dx dx −1 d 1 d = · (2x2 − x) = x 2 · dx (2x) √1 − (2x − x) 2 2 dx √ 1 − (2 ) −1 1 = · (4x − 1) = (2x log 2) √1 − x2 (2x − 1)2 √1 − 22x 1 − 4x dy dy 2x log 2 ∴ = ∴ = dx √1 − x2 (2x − 1)2 dx √1 − 4x 1 1+x (iv) Let y = cot−1 = tan−1 (x2) (v) Let y = cos−1 x2 2 Differentiate w. r. t. x. Differentiate w. r. t. x. dy d = (tan−1 (x2)) dy d 1+x dx dx = cos−1 1 d 2 dx dx 2 = · (x ) 1 + (x2)2 dx 1 d 1+x = − · dy 2x 1+x 2 dx 2 ∴ = 1− 2 dx 1 + x4 1 1 d 1+x (vi) Let y = sin2 (sin−1 (x2)) = − × × 1− 1+x 2 1+x dx 2 2 2 = [sin (sin (x )] −1 2 = (x ) 2 2 2 ∴ y = x4 √2 1 1 = − × × Differentiate w. r. t. x. √1 − x √2 √1 + x 2 dy d 4 dy dy 1 = (x ) ∴ = 4x 3 ∴ =− dx dx dx dx 2 √1 − x2 22 Ex. 3 : Differentiate the following w. r. t. x. 1 − cos x (i) cos−1 (4 cos3 x − 3 cos x) (ii) cos−1 [sin (4x)] (iii) sin−1 2 1 − cos 3x cos x (iv) tan−1 (v) cot−1 sin 3x 1 + sin x Solution : (i) Let y = cos−1 (4 cos3 x − 3 cos x) (ii) Let y = cos−1 [sin (4x)] = cos−1 (cos 3x) π = cos−1 cos − 4x ∴ y = 3x 2 π Differentiate w. r. t. x. ∴ y= − 4x dy d 2 = (3x) Differentiate w. r. t. x. dx dx dy d π dy = − 4x = 0 − 4x log 4 ∴ =3 dx dx 2 dx dy ∴ = − 4x log 4 dx 1 − cos x 1 − cos 3x (iii) Let y = sin−1 (iv) Let y = tan−1 2 sin 3x 2 sin2( 2x ) 2 sin2( 3x2 ) = sin−1 = tan−1 2 2 sin ( 3x2 ) cos ( 3x2 ) x = sin−1 sin 3x 2 = tan−1 tan x 2 ∴ y= 3x 2 ∴ y= Differentiate w. r. t. x. 2 dy 1 Differentiate w. r. t. x. dy d x = ∴ = dy d 3x dy 3 dx dx 2 dx 2 = ∴ = dx dx 2 dx 2 cos x 1 + sin x (v) Let y = cot−1 = tan−1 1 + sin x cos x [cos ( 2x ) + sin ( 2x )]2 [cos ( 2x ) + sin ( 2x )]2 = tan −1 = tan −1 cos2 ( 2x ) − sin2 ( 2x ) [cos ( 2x ) − sin ( 2x )] [cos ( 2x ) + sin ( 2x )] cos ( 2x ) + sin ( 2x ) 1 + tan ( 2x ) x = tan−1 = tan−1....Divide Numerator & Denominator by cos cos ( 2 ) − sin ( 2 ) x x 1 − tan ( 2x ) 2 π x π x = tan−1 tan + + ∴y= 4 2 4 2 Differentiate w. r. t. x. dy d π x 1 dy 1 = + = 0 + ∴ = dx dx 4 2 2 dx 2 23 Ex. 4 : Differentiate the following w. r. t. x. 2 cos x + 3 sin x 3 sin x2 + 4 cos x2 a cos x − b sin x (i) sin −1 (ii) cos −1 (iii) sin−1 √13 5 √a2 + b2 Solution : 2 cos x + 3 sin x 3 sin x2 + 4 cos x2 (i) Let y = sin−1 (ii) Let y = cos−1 √13 5 2 3 3 4 = sin−1 cos x + cos x = cos−1 sin x2 + cos x2 √13 √13 5 5 2 3 3 4 Put = sin α, = cos α Put = sin α, = cos α √13 √13 5 5 4 9 9 16 Also, sin2 α + cos2 α = + =1 Also, sin2 α + cos2 α = + =1 3 13 25 25 2 2 3 3 And tan α = ∴ α = tan−1 And tan α = ∴ α = tan−1 3 3 4 4 y = sin−1 (sin α cos x + cos α sin x) y = cos−1 (sin α sin x2 + cos α cos x2) y = sin−1 (sin x cos α + cos x sin α) y = cos−1 (cos x2 cos α + sin x2 sin α) y = sin−1 [sin (x + α)] y = cos−1 [cos (x2 − α)] 2 3 ∴ y = x + tan−1 ∴ y = x2 − tan−1 3 4 Differentiate w. r. t. x. Differentiate w. r. t. x. dy d 2 dy d 2 3 = x + tan−1 = 1+0 = x − tan−1 = 2x − 0 dx dx 3 dx dx 4 dy dy ∴ =1 ∴ = 2x dx dx a cos x − b sin x a b (iii) Let y = sin−1 = sin−1 cos x − sin x √a2 + b2 √a2 + b2 √a2 + b2 a b Put = sin α, = cos α √a2 + b2 √a2 + b2 a2 b2 a a Also, sin2 α + cos2 α = + = 1 And tan α = ∴ α = tan−1 a +b 2 2 a +b 2 2 b b y = sin (sin α cos x − cos α sin x) −1 But sin (α − x) = sin α cos x − cos α sin x y = sin−1 [sin (α − x)] a ∴ y = tan−1 −x b Differentiate w. r. t. x. dy d a dy = tan−1 −x = 0−1 ∴ = −1 dx dx b dx 24 Ex. 5 : Differentiate the following w. r. t. x. 2x 1 (i) sin−1 (ii) cos−1 (2x √1− x2) (iii) cosec−1 1 + x2 3x − 4x3 2ex 1 − 9x2 2x − 2− x (iv) tan−1 (v) cos−1 (vi) cos−1 1− e2x 1 + 9x2 2x + 2− x 3−x 5√1− x2 − 12x 2x+1 (vii) tan−1 (viii) sin−1 (ix) sin−1 3+x 13 1 + 4x Solution : 2x (ii) Let y = cos−1 (2x √1− x2) (i) Let y = sin −1 1 + x2 Put x = sin θ ∴ θ = sin−1 x Put x = tan θ ∴ θ = tan−1 x ∴ y = cos−1 (2 sin θ √1− sin2 θ) 2 tan θ ∴ y = sin−1 y = cos−1 (2 sin θ √cos2 θ) 1 + tan2 θ y = sin−1 (sin 2θ) = 2θ y = cos−1 (2 sin θ cos θ) = cos−1 (sin 2θ) ∴ y = 2 tan−1 x π π y = cos−1 cos − 2θ = − 2θ Differentiate w. r. t. x. 2 2 dy d π = 2 (tan−1 x) ∴ y = − 2 sin−1 x dx dx 2 Differentiate w. r. t. x. dy 2 ∴ = dy d π dx 1 + x2 = − 2 sin−1 x dx dx 2 dy 2×1 =0− dx √1 − x2 dy 2 ∴ =− dx √1 − x2 1 2ex (iii) Let y = cosec−1 (iv) Let y = tan−1 3x − 4x3 1− e2x y = sin−1 (3x − 4x3) Put ex = tan θ ∴ θ = tan−1 (ex) Put x = sin θ ∴ θ = sin−1 x 2 tan θ y = tan−1 y = sin−1 (3 sin θ − 4 sin3 θ) 1 + tan2 θ y = sin−1 (sin 3θ) = 3θ y = tan−1 (tan 2θ) = 2θ ∴ y = 3 sin−1 x ∴ y = 2 tan−1 (ex) Differentiate w. r. t. x. Differentiate w. r. t. x. [tan−1 (ex)] dy d dy d = 3 (sin−1 x) =2 dx dx dx dx dy 3 dy 2 d x 2ex ∴ = ∴ = (e ) = dx √1 − x2 dx 1 + (ex)2 dx 1− e2x 25  1 − 9x2 2x − 2− x (v) Let y = cos−1 (vi) Let y = cos−1 1 + 9x2 2x + 2− x 1 − (3x)2 2x (2x − 2− x)... [Multiply & y = cos −1 y = cos−1 1 + (3x)2 2x (2x + 2− x) Devide by 2x] Put 3x = tan θ ∴ θ = tan−1 (3x) 22x − 1 1 − (2x)2 y = cos −1 = cos − −1 1 − tan2 θ 22x + 1 1 + (2x)2 ∴ y = cos−1 1 + tan2 θ Put 2x = tan θ ∴ θ = tan−1 (2x) y = cos−1 (cos 2θ) = 2θ 1 − tan2 θ ∴ y = cos−1 − = cos−1 [− cos 2θ] 1 + tan θ 2 y = 2 tan−1 (3x) y = cos [cos (π − 2θ)] = π − 2θ −1 Differentiate w. r. t. x. y = π − 2 tan−1 (2x) = 2 [tan−1 (3x)] dy d Differentiate w. r. t. x. dx dx [π − 2 tan−1 (2x)] dy d dy 2 d = = 2 (3x) dx dx dx 1 + (3x) dx dy 2 d x 2·2x·log 2 dy 6 =0− 2 (2 ) = − ∴ = dx 1 + (2x) dx 1 + 22x dx 1 + 9x2 dy 2x+1 log 2 ∴ =− dx 1 + 22x 3−x (vii) Let y = tan−1 3+x 1 x Put x = 3 cos 2θ ∴ θ = cos−1 2 3 3 − 3 cos 2θ 3(1 − cos 2θ) 2 sin2 θ ∴ y = tan−1 = tan−1 = tan−1 3 + 3 cos 2θ 3(1 + cos 2θ) 2 cos2 θ y = tan−1 (√tan2 θ) = tan−1 (tan θ) 1 x y = θ = cos−1 2 3 Differentiate w. r. t. x. dy 1 d x = · cos−1 dx 2 dx 3 1 1 d x 1 1 1 =− =− × × 2 1− x 2 dx 3 2 9− x2 3 3 9 1 1 1 − × 9−x × = 2 √ 3 2 3 dy 1 ∴ =− dx 2√9 − x2 26  5√1− x2 − 12x (viii) Let y = sin−1 13 Put x = sin θ ∴ θ = sin−1 x 5 cos θ − 12 sin θ y = sin−1 5√1− sin θ − 12 sin θ = sin−1 5√cos θ − 12 sin θ = sin−1 2 2 ∴ 13 13 13 5 12 ∴ y = sin−1 cos θ - sin θ 13 13 5 12 Put = sin α, = cos α 13 13 25 144 Also, sin2 α + cos2 α = + =1 169 169 5 5 And tan α = ∴ α = tan−1 12 12 y = sin (sin α cos θ − cos α sin θ) = sin−1 [sin (α − θ)] = (α − θ) −1 5 ∴ y = tan−1 − sin−1 x 12 Differentiate w. r. t. x. dy d 5 1 = tan−1 − sin−1 x = 0 − dx dx 12 √1 − x2 dy 1 ∴ =− dx √1 − x2 2x+1 2·2x (ix) Let y = sin−1 = sin −1 1 + 4x 1 + (2x)2 Put 2x = tan θ ∴ θ = tan−1 (2x) 2 tan θ ∴ y = sin−1 = sin−1 (sin 2θ) = 2θ = 2 tan−1 (2x) 1 + tan θ 2 Differentiate w. r. t. x. 2 2 = 2 [tan−1 (2x)] = dy d d x x 2 · (2 ) = (2x·log 2) dx dx 1 + (2 ) dx 1 + 22x dy 2x+1 log 2 ∴ =− dx 1 + 4x Ex. 6 : Differentiate the following w. r. t. x. 4x 7x (i) tan−1 (ii) tan−1 1 + 21x2 1 − 12x2 b sin x − a cos x 5x + 1 (iii) cot−1 (iv) tan−1 a sin x + b cos x 3 − x − 6x2 27 Solution : 4x 7x (i) Let y = tan−1 (ii) Let y = tan−1 1 + 21x2 1 − 12x2 7x − 3x 3x + 4x =tan−1 =tan−1 1 + (7x) (3x) 1 − (3x) (4x) y = tan−1 (7x) − tan−1 (3x) y = tan−1 (3x) + tan−1 (4x) Differentiate w. r. t. x. Differentiate w. r. t. x. dy d dy d = [tan−1 (7x) − tan−1 (3x)] = [tan−1 (3x) + tan−1 (4x)] dx dx dx dx d d d d = [tan−1 (7x)] − [tan−1 (3x)] = [tan−1 (3x)] + [tan−1 (4x)] dx dx dx dx 1 d 1 d 1 d 1 d = 2 · (7x) − 2 · (3x) = 2 · (3x) + 2 · (4x) 1 + (7x) dx 1 + (3x) dx 1 + (3x) dx 1 + (4x) dx dy 7 3 dy 3 4 ∴ = − ∴ = + dx 1 + 49x 1 + 9x2 2 dx 1 + 9x2 1 + 16x2 b sin x − a cos x a sin x + b cos x a b + cot x (iii) Let y = cot−1 = tan −1 = tan −1 a sin x + b cos x b sin x − a cos x 1 − ( ab ) (cot x) a a π = tan−1 + tan−1 (cot x) = tan−1 + tan−1 tan −x b b 2 a π y = tan−1 + −x b 2 Differentiate w. r. t. x. dy d a π = tan−1 + −x dx dx b 2 d a d π d = tan−1 + − (x) dx b dx 2 dx = 0 + 0 − 1 dy ∴ =−1 dx 5x + 1 5x + 1 5x + 1 (iv) Let y = tan−1 = tan−1 = tan−1 3 − x − 6x2 1 + 2 − x − 6x2 1 − (6x2 + x − 2) 5x + 1 5x + 1 = tan−1 = tan−1 1 − (6x2 + 4x − 3x − 2) 1 − [2x(3x + 2) − (3x + 2)] 5x + 1 (3x + 2) + (2x − 1) = tan−1 = tan−1 1 − (3x + 2)(2x − 1) 1 − (3x + 2)(2x − 1) 28  y = tan−1 (3x + 2) + tan−1 (2x − 1) Differentiate w. r. t. x. dy d = [tan−1 (3x + 2) + tan−1 (2x − 1)] dx dx d d = [tan−1 (3x + 2)] + [tan−1 (2x − 1)] dx dx 1 d 1 d = 2 · (3x + 2) + 2 · (2x − 1) 1 + (3x + 2) dx 1 + (2x − 1) dx dy 3 2 ∴ = 2 + dx 1 + (3x + 2) 1 + (2x − 1)2 EXERCISE 1.2 (1) Find the derivative of the function y = f (x) (6) Differentiate the following w. r. t. x. using the derivative of the inverse function (i) tan−1 (log x) (ii) cosec−1 (e−x ) x = f −1 ( y) in the following (iii) cot−1 (x3) (iv) cot−1 (4x ) (i) y = √x (ii) y = 2 − √x 1 + x2 (v) tan−1 (√x) (vi) sin−1 (iii) (iv) y = log (2x − 1) 2 3 (v) y = 2x + 3 (vi) y = ex − 3 (vii) cos−1 (1 − x2) (viii) sin−1 (x 2 ) x (vii) y = e2x − 3 (viii) y = log2 (ix) cos3 [cos−1 (x3)] (x) sin4 [sin−1 (√x)] 2 (2) Find the derivative of the inverse function of (7) Differentiate the following w. r. t. x. cot−1 [cot (ex )] 2 the following (i) (i) y = x2·ex (ii) y = x cos x 1 (ii) cosec−1 (iii) y = x·7x (iv) y = x2 + log x cos (5x) (v) y = x log x 1 + cos x (iii) cos−1 (3) Find the derivative of the inverse of the 2 following functions, and also find their value 1 − cos (x2) at the points indicated against them. (iv) cos−1 2 (i) y = x5 + 2x3 + 3x, at x = 1 1 − tan ( 2x ) (ii) y = ex + 3x + 2, at x = 0 (v) tan −1 (iii) y = 3x + 2 log x , at x = 1 2 3 1 + tan ( 2x ) (iv) y = sin (x − 2) + x2, at x = 2 1 (vi) cosec−1 (4) If f (x) = x3 + x − 2, find ( f −1)' (0). 4 cos3 2x − 3 cos 2x (5) Using derivative prove 1 + cos ( 3x ) π (vii) tan −1 (i) tan−1 x + cot−1 x = sin ( 3x ) 2 π sin 3x (ii) sec−1 x + cosec−1 x =... [for | x | ≥ 1] 2 (viii) cot−1 1 + cos 3x 29  cos 7x 1 − x2 (ix) tan −1 (iii) sin−1 (iv) sin−1 (2x √1 − x2) 1 + sin 7x 1 + x2 1 + cos x ex − e −x (x) tan −1 (v) cos−1 (3x − 4x3) (vi) cos−1 1 − cos x ex + e −x 1 1 − 9x 4x + 2 (xi) tan (cosec x + cot x) −1 (vii) cos−1 (viii) sin−1 1 + 9x 1 + 24x 1 + sin ( 4x3 ) + 1 − sin ( 4x3 ) (xii) cot−1 1 − 25x2 1 − x3 (ix) sin−1 (x) sin−1 1 + sin ( 4x3 ) − 1 − sin ( 4x3 ) 1 + 25x2 1 + x3 5 (8) Differentiate the following w. r. t. x. 2x 2 1 − √x (xi) tan−1 (xii) cot−1 4 sin x + 5 cos x 1 − x5 1 + √x (i) sin−1 √41 (10) Differentiate the following w. r. t. x. √3 co