US01CMTH02 UNIT-3. exists, then it is called the partial derivative of f with respect to y at (a, b) and is denoted by f. f(a, b + b) f(a, b) lim

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1 Study material of BSc(Semester - I US01CMTH02 (Partial Derivatives Prepared by Nilesh Y Patel Head,Mathematics Department,VPand RPTPScience College 1 Partial derivatives US01CMTH02 UNIT-3 The concept of partial derivative plays a vital role in differential calculus The different ways of limit discussed in the previous section, yields different type of partial derivatives of a function 11 Definitions Consider a real valued function z = f(, y defined on E R 2 such that E contains a neighbourhood of (a, b R 2 Let a be a change in a If the limit, f(a + a, b f(a, b lim a 0 a eists, then it is called the partial derivative of f with respect to at (a, b and is denoted by f or f (a, b or z (a, b Similarly, let b be a change in b If the limit, (a,b f(a, b + b f(a, b lim b 0 b eists, then it is called the partial derivative of f with respect to y at (a, b and is denoted by f or f y (a, b or z y (a, b (a,b Notations If the partial derivatives f and f y eist at each point of E, then they are also the real valued functions on E Further, we can obtain the partial derivatives of these functions, if they are differentiable In these cases, we fi up the following notations f = 2 f 2 = f y = 2 f = ( f ( f, f yy = 2 f = ( f 2 ; and f y = 2 f =, ( f The notations of derivatives of order greater than two should be clear from the above pattern 12 Remark As we have seen in the above eample, in general, f y and f y need not be equal, even if they eist The following proposition gives a sufficient condition for them to be equal We accept it without proof However, we shall be dealing only with the functions f for which these two are equal 1

2 2 13 Proposition Consider a real valued function z = f(, y defined on E R 2 such that E contains a neighbourhood of (a, b R 2 If f y and f y eist and are continuous, then f y = f y Throughout this chapter our blanket assumption will be that the operation of taking partial derivatives is commutative That is, for our function f of two variables, f y = f y In general, we may assume that the second derivatives of functions eists and are continuous, so that, the Proposition 13 ensures our requirement 14 Eample For u = 3 3y 2, prove that 2 u + 2 u 2 2 Solution Here u = 3 3y 2 u = 32 3y 2 ; u = 6y; 2 u 2 = 6; = 0 Also prove that 2 u = 2 u 2 u = 6y = 2 u 2 u 2 = 6 2 u + 2 u 2 = 6 6 = 0 and 2 u 2 = 2 u 2 Homogeneous functions Let us observe the following epressions carefully (1 f 1 (, y = 2 y 4 3 y 3 + y 5 (2 f 2 (, y = 4 y 4 5 y y 2 The combined degree of and y in each term of the first epression is 6 and that in the second epression is 8 Can we determine whether the combined degree of and y in each term of the epression is same or not? It seems difficult to determine Let us develop 4 +y 4 the following tests Test 1: Let us take t = y Then and 2 y 4 3 y 3 + y 5 = 6 (t 4 t 3 + t 5 = 6 f(t 4 y 4 5 y y 2 = 8 (t 4 t 3 + t 2 = 8 g(t, where f and g are functions of one variable t Test 2: Now, let us replace by t and y by ty Then and f 1 (t, ty = (t 2 (ty 4 (t 3 (ty 3 + (t(ty 5 = t 6 f 1 (, y f 2 (t, ty = (t 4 (ty 4 (t 5 (ty 3 + (t 6 (ty 2 = t 8 f 2 (, y 21 Definitions A function z = f(, y is said to be a homogeneous function of degree r, if f(t, ty = t r f(, y for some real number r Otherwise, f is said to be a nonhomogeneous function

3 2 Homogeneous functions 3 22 Eample Let f : R 2 \ {(, y : y = } R defined by f(, y = y Then prove +y that f is a homogeneous function of degree 0 and f and f y eist at each point of the domain Solution Clearly, f(t, ty = f(, y = t 0 f(, y Thus f is a homogeneous function of degree 0 Now for any (, y R 2 with + y 0, we have, and f (, y = f y (, y = ( + y(1 ( y(1 ( + y 2 = 2y ( + y 2 ( + y( 1 ( y(1 ( + y 2 = 2 ( + y 2 23 Eample f : R 2 \ {(0, 0} R defined by f(, y = 5 5 y 3 +y 3 function of degree 14 5 is a homogeneous 24 Theorem (Euler s Theorem State and prove Euler s Theorem Statement : Let z = f(, y be a real valued function defined on E R 2 Suppose that f is a homogeneous function of degree n If f and f y eist on E, then + y = nz (241 Proof Since z = f(, y is a homogeneous function of, y of degree n, we can write ( y z = f(, y = n g (242 Differentiating (242 partially with respect to, we get, ( y ( y = nn 1 g + n g ( y 2 ( y ( y = nn g n 1 yg (243 Similarly, differentiating (242 partially with respect to y, we get, ( y 1 ( y = n g = n 1 g y ( y = yn 1 g (244 Adding (243 and (244 we get, + y ( y = nn g = nz This completes the proof We note that the converse of Euler s Theorem also holds That is, if a function z = f(, y satisfies (241, on a certain domain, then it must be homogeneous on that domain

4 4 25 Remark Now onwards we shall not mention the domain of the functions under discussion Also, whenever we use the derivatives of functions under discussion, we assume them to be sufficiently many times differentiable 26 Corollary Let z = f(, y be a real valued function defined on E R 2 Suppose that f is a homogeneous function of degree n and that all the second order partial derivatives of f eist and are continuous Then prove that 2 2 z + 2y 2 z z y2 = n(n 1z 2 Proof Since z = f(, y is a homogeneous function of, y of degree n, by Euler s Theorem, + y = nz (261 Differentiating (261 partially with respect to, we have, which, on multiplication by, gives 2 z y 2 z = n, 2 2 z y 2 z 2 2 z 2 + y 2 z = n = (n 1 (262 Similarly, differentiating (261 partially with respect to y and then multiplying the result by y, we get, Since 2 z = 2 z, we get, y 2 2 z 2 + y 2 z y 2 2 z 2 + y 2 z By adding (262 and (263 we have, = (n 1y = (n 1y ( z + 2y 2 z z y2 = (n 1( 2 + y = n(n 1z This completes the proof 27 Corollary Let u = u(, y be a nonhomogeneous real valued function defined on E R 2 and z = φ(u be homogeneous function of degree n Then prove that provided φ (u 0 for any (, y E u + y u = n φ(u φ (u,

5 2 Homogeneous functions 5 Proof Since z = φ(u is a homogeneous function of, y of degree n, by Euler s Theorem we have, + y = nz = nφ(u ( φ (u u ( + y φ (u u = nφ(u ( u + y ( u = n φ(u φ (u 28 Corollary (Only statementlet u = u(, y be a nonhomogeneous real valued function defined on E R 2 and z = φ(u be homogeneous function of degree n Then prove that 2 2 u 2 + 2y 2 u + y2 2 u 2 = ψ(u[ψ (u 1], where ψ(u = n φ(u φ (u, provided φ (u 0 for any (, y E 29 Eample For the following functions, verify Euler s Theorem and find 2 2 z + 2y 2 z + 2 y2 2 z 2 (1 z = n log ( y (2 z = sin 1 ( + y tan 1 ( y Solution (1 Clearly, z is a homogeneous function of degree n ( y = nn 1 log + n ( y ( y = n n 1 log y n 1 2 ( y = nn log n Also, ( ( 1 = n = n y y y = n + y ( y = nn log = nz Thus Euler s Theorem is verified By the Corollary 26, 2 2 z + 2y 2 z z y2 = n(n 1z 2 (2 Replacing by t and y by ty, f(t, ty = sin 1 ( + y tan 1 ( y = t0 f(, y Thus z = f(, y is a homogeneous function of degree 0 Now, ( = ( y 1 = 1 y y2 2 y 2 2 y2 y y 2 = y2 2 y 2 + y 2

6 6 Also, ( = 1 1 y 2 2 y 2 y = y2 2 + Thus Euler s Theorem is verified By the Corollary 26, as n = y2 2 y 2 + y 2 ( 1 = + y = 0 y y z + 2y 2 z z y2 = n(n 1z = 0, Eample If u = sin 1 ( 2 y 2, then prove the following +y (1 u + y u = 3 tan u (2 2 2 u 2 + 2y 2 u + y2 2 u 2 = 3 tan u(3 sec 2 u y 2 Solution Here u = sin 1 ( 2 y 2 is not a homogeneous function of, y Writing the given +y equation differently, we have sin u = 2 y 2, which is +y Let z = φ(u = sin u Then z = 2 y 2 +y homogeneous of degree 3 Hence by Corollary 27, u + y u = 3 φ(u = 3 sin u φ (u cos u which proves (1 Also, by Corollary 28, we have, 2 2 u 2 + 2y 2 u + y2 2 u 2 = 3 tan u[3 sec2 u 1] 3 Theorem on total differentials = 3 tan u, Throughout this section we consider only those functions of two variables that admit continuous partial derivatives on their domain of definition That is, if we are discussing about a function z = f(, y, then f, f y eist and are continuous on the domain of f 31 Theorem ( Only statement Let z = f(, y be defined on E Then dz = d + dy 4 Differentiation of composite functions In this section we shall study the differentiation of composite functions Let z = f(, y be function defined on E R 2 In turn one can have = ϕ(t and y = ψ(t, t F R This makes f a function of one independent variable t That is, t F (ϕ(t, ψ(t E f(ϕ(t, ψ(t The following theorem describes the differentiation of f with respect to t in this situation

7 5 Change of variables 7 41 Theorem ( Only statement Let z = f(, y be function defined on E R 2 and = ϕ(t, y = ψ(t, t F R Then prove that df = f d + f dy dt dt dt To etend Theorem 41 for functions of three variables, let u = f(, y, z be a function of three variables with = (t, y = y(t and z = z(t Then 5 Change of variables du dt = u d dt + u dy dt + u dz dt Like the composite functions we can also consider the following situation Let z = f(, y be function defined on E R 2 and let there be another domain F R 2 such that for each (, y E, = ϕ(u, v, y = ψ(u, v, (u, v F R 2 This is nothing but the change of variable In this case, the following theorem describes the partial derivatives of f with respect to u and v Now we prove Euler s Theorem for three variables The homogeneous functions of more than two variables are defined as in Definition 21 More eplicitly, a function H = f( 1, 2,, n of n variables is called homogeneous if there eists r R such that for f(t 1, t 2,, t n = t r f( 1, 2,, n for all t R In this case, the degree of homogeneity of H is r 51 Theorem (Euler s Theorem for Three variables Let H = f(, y, z be a real valued homogeneous function of three variables, y, z of degree n defined on E R 3 If f, f y, f z eist on E, then prove that H + y H + z H Proof Since H = f(, y, z is homogeneous function of degree n, ( y H = n φ, z = n φ(u, v, = nh (511 where u = y and v = z [ H φ = nn 1 φ(u, v + n u u + φ ] v v [ = n n 1 φ(u, v + n y φ 2 u z ] φ 2 v = n n 1 φ(u, v n 2 y φ u n 2 z φ v H = nn φ(u, v n 1 y φ u n 1 z φ v (512 Now, H = n [ φ u u + φ ] v v [ ] 1 = n φ u + 0 φ n 1 φ = v u y H = n 1 y φ u (513

8 8 Similarly, z H = n 1 z φ v (514 Adding (512, (513 and (514 we have, H + y H + z H = nn φ(u, v = nh This completes the proof As noted in case of the functions of two variables, here also we recall that the converse of Euler s Theorem also holds That is, if a function z = f(, y satisfies (511, on a certain domain, then it must be homogeneous on that domain 52 Eample Find dz dt when z = sin 1 ( y, = 3t, y = 4t 3 Also verify by the direct substitution Solution dz dt = d dt + dy dt 1 = ( y 2 1 ( y 2 12t2 = 3(1 4t2 1 ( y 2 = = = 3(1 4t 2 1 (3t 4t3 2 3(1 4t 2 (1 3t + 4t3 (1 + 3t 4t 3 3(1 4t 2 (1 t2 (1 4t 2 = t 2 On the other hand, verifying directly by putting the values of and y in z, we have z = sin 1 (3t 4t 3 dz dt = (3 12t 2 1 (3t 4t3 2 = 3(1 4t 2 1 (3t 4t3 2 = 3 1 t 2 53 Eample If z = f(, y, = r cos θ, y = r sin θ, then prove that [ ] 2 [ ] 2 [ ] 2 + = + 1 [ ] 2 r r 2 θ Solution Here, y are functions of r, θ Hence z is a composite function of r, θ Thus, r = r + = cos θ + sin θ r [ ] 2 [ ] 2 = cos 2 θ + 2 sin θ cos θ [ ] 2 r + sin2 θ (531

9 5 Change of variables 9 Also, θ = θ + = r sin θ + r cos θ θ [ ] 2 [ ] 2 = r 2 sin 2 θ 2r 2 sin θ cos θ [ ] 2 θ + r2 cos 2 θ 1 [ ] 2 [ ] 2 = sin 2 θ 2 sin θ cos θ [ ] 2 r 2 θ + cos2 θ (532 Adding (531 and (532 we get, [ ] r r 2 [ ] 2 = θ [ ] 2 + [ ] 2 54 Eample If H = f(2 3y, 3y 4z, 4z 2, then prove that 1 H H H 4 = 0 Solution Let u = 2 3y, v = 3y 4z, w = 4z 2 Then H = f(u, v, w Hence H is a composite function of, y, z Therefore, H = H u u + H v v + H w w = 2 H u + 0 H v 2 H w = 2 H u 2 H w (541 Also, H = H u u + H v v + H w w = 3 H u + 3 H v + 0 H w = 3 H u + 3 H v (542 Finally, H = H u u + H v v + H w w = 0 H u 4 H v + 4 H w = 4 H v + 4 H w (543 1 H H H 4 = H u H w H u + H v H v + H w = 0 55 Eample If z = f(, y and u = e cos y, v = e sin y Then prove that f u f + v f u v Solution u = e cos y, v = e sin y Also, u 2 + v 2 = e 2 e = u 2 + v 2 = 1 2 log (u2 + v 2 v u = tan y y = tan 1 ( v u Thus, y are functions of u, v, and so, z is a composite function of u, v Now, f u = f u + f u = f [ u u 2 + v 2 ] + f [ v u 2 + v 2 ] =

10 10 Similarly, u f [ u = u 2 u 2 + v 2 v f [ v = ] [ f v 2 u 2 + v 2 uv u 2 + v 2 ] [ f + Adding (551 and (552 we get, u f u + v f v = f 6 Differentiation of implicit functions ] f (551 uv u 2 + v 2 ] f (552 Many a times we are given an epression f(, y = c, where c R is a constant Note here that, and y are associated by a rule however we may not be able to write y as a function of In this case, we say that y is a function of, implicitly described by f(, y = c or y is an implicit function of We obtain the method of calculating dy and d d 2 y d 2 using the tools of partial derivatives 61 Theorem Let a function y of be implicitly described by f(, y = c Then prove that (1 dy d = f f y (2 d2 y d 2 = f (f y 2 2f y f f y + f yy (f 2 (f y 3 Proof We know that f is a function of and y Also, y is an implicit function of So, f is a composite function of differentiating the equation f(, y = c with respect to, we get, This proves (1 Now we prove (2 f d d + f dy d = 0 f + f dy d = 0 dy d 2 y d = d 2 d = d d ( dy d ( f f y = f y d (f d d f (f d y (f y 2 ( f y (f + (f dy d = f ( d = f f = f f y (f y + (f y dy d (f y 2 ( ( f y f + f y ( f f y f f y + f yy ( f f y = (f y 2 = f (f y 2 f y f f y f f y f y + f yy (f 2 (f y 3

11 6 Implicit functions 11 = f (f y 2 2f f y f y + f yy (f 2 (f y 3 62 Eample Find dy d when (1 sin( y ( + y = 0 (2 y = y Proof (1 Let f(, y = sin( y (+y Since f(, y = 0, by the previous theorem, we have, dy d = f cos( y + sin( y 1 = f y cos( y( 1 1 cos( y + sin( y 1 = cos( y + 1 (2 Let f(, y = y y Since f(, y = 0, by the previous theorem, we have, dy d = f f y = yy 1 y log y y log y 1 = y log y y y 1 y log y 1 63 Eample If z = yf( y and z is constant, then show that f ( y dy [y + = y[y dy f( y Solution Let F (, y = yf( y Then F (, y = z, z is constant Thus y is an implicit function of So, F + F dy = 0 (631 d Now differentiating F (, y with respect to, we get, F ( = yf(y + yf ( y y = yf( y y2 2 f ( y = y [ f( y yf ( y ] Similarly, d ] d ] F = f(y + yf ( y ( 1 = f(y + yf ( y Putting these values in (631, we have, y [ f( y yf ( y ] + f( y + yf ( y dy d = 0 [ y + dy ] f( y d = y [ y dy ] f ( y d f ( y [ ] f( y = y + dy d y y dy d

12 12 64 Eample If A, B and C are angles of a ABC such that sin 2 A + sin 2 B + sin 2 C = K, a constant, then prove that db tan C tan A = dc tan A tan B Solution Clearly, A + B + C = π So, A = π (B + C Therefore, sin A = sin(b + C Let f(b, C = sin 2 (B + C + sin 2 B + sin 2 C K Hence f(b, C = 0, ie, B is an implicit function of C So, db = f C dc f B Also, Similarly, we get, f B = f B = 2 sin(b + C cos(b + C + 2 sin B cos B db dc = sin 2(B + C + sin 2B = sin(2π 2A + sin 2B = sin 2A + sin 2B = 2 cos(b + A sin(b A = 2 cos(π C sin(b A = 2 cos C sin(b A = 2 cos C sin(a B Dividing by cos A cos B cos C, we get, db tan A tan C tan C tan A = = dc tan A tan B tan A tan B f C = 2 cos B sin(a C B sin(a C = cos cos C sin(a B cos B(sin A cos C cos A sin C = cos C(sin A cos B cos A sin B sin A cos B cos C cos A cos B sin C = sin A cos B cos C cos A sin B cos C