Partial Derivatives. Limits. For a single variable function f (x), the limit lim

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1 Limits Prtil Derivtives For single vrible function f (x), the limit lim x f (x) exists only if the right-hnd side limit equls to the left-hnd side limit, i.e., lim f (x) = lim f (x). x x + For two vribles function f (x,y), the limit lim f (x,y) exists only if the vlue of the (x,y) (,b) limit is unique for (x,y) pproches to (,b) in ny pth. If we cn find ny two different pths which give different vlues for follows tht the limit does not exist. 1. Find the limit, if exists. () x 2 y 2 x 2 y 3 lim (b) lim (x,y) (0,0) x 2 + y 2 (x,y) (0,0) 2x 2 + y 2 lim f (x,y), then it (x,y) (,b) Continuity A function f (x) is continuous t x = if lim f (x) = f (). x For the two vribles function f (x,y), the function is continuous t (,b) if lim f (x, y) exists nd (x,y) (,b) lim f (x,y) = f (,b). (x,y) (,b) 1. Determine which of the following function is continuous t (0, 0). x 2 y 3 if (x,y) (0, 0) () f (x,y) = 2x 2 + y 2 (b) f (x,y) = 1 if (x,y) = (0, 0) x 2 y 3 if (x,y) (0, 0) 2x 2 + y 2 0 if (x,y) = (0, 0) First Principle The derivtive (rtes of chnge) of the function f (x) t x =, denoted by f (), is defined by if the limit exists. f () = lim x 0 f ( + x) f (), x If f is function of two vribles, f (x,y), its prtil derivtives t the point (,b), denoted by f x (,b) nd f y (,b), re defined by f ( + x,b) f (,b) f x (,b) = lim x 0 x if the limits exist. nd f y (,b) = lim y 0 f (,b + y) f (,b), y 1. Use the definition of prtil derivtives to find f x (,b) nd f y (,b) t the given point (,b), if exist. 1

2 () f (x,y) = x 2 + y 2 ; (0, 0) (b) f (x,y) = x ; (1, 1) y 2. Use the First Principle to find f x (x,y) nd f y (x,y). () f (x,y) = 16 4x 2 y 2 (b) f (x,y) = 2x 3y Prtil Derivtives There re mny lterntive nottions for prtil derivtives. For instnce, insted of f x we cn write f s the first prtil derivtive of f (x,y) with respect to x. If z = f (x,y), we write x f x (x,y) = f x = f x = f (x,y) = x x = z x nd f y (x,y) = f y = f y = f (x,y) = y y = z y. The rules for finding prtil derivtives of z = f (x,y): To compute f x, regrd y s constnt nd differentite f (x,y) with respect to x. To compute f y, regrd x s constnt nd differentite f (x,y) with respect to y. Apply the Product Rule, Quotient Rule, or Chin Rule if necessry. 1. Find the first prtil derivtives of the given functions. () f (x,y) = x 3 y 5 (b) f (x,y) = 2x 3y (c) f (x,y) = x3 + y 3 x 2 + y 2 (d) f (x,y) = (3xy 2 x 4 + 1) 4 (e) f (x,y) = ln (x 2 + y 2 ) (f) f (x,y) = e x tn (x y) Higher Derivtives If f is function of two vribles, then its prtil derivtives f x nd f y re lso functions of two vribles, so we cn consider their prtil derivtives (f x ) x, (f x ) y, (f y ) x, nd (f y ) y, which re clled the second prtil derivtives of f. If z = f (x,y), we use the following nottion: (f x ) x = f xx = x (f y ) x = f yx = x ( ) f = 2 f x x = 2 z 2 x 2, ( ) f = 2 f y x y = 2 z x y, nd (f x) y = f xy = y (f y) y = f yy = y ( ) f = 2 f x y x = 2 z y x, ( ) f = 2 f y y = 2 z 2 y 2. A function f is sid to be of clss C k if ll the k-th order prtil derivtives of f exist nd continuous in the domin of the function. If z = f (x,y) is C 2 -function, then f xy = f yx or 2 f y x = 2 f x y. 1. Find ll the second prtil derivtives for the following functions nd identify which one is C 2 -function. 2

3 () f (x,y) = x 5 y 4 3x 2 y 3 + 2x 2 (b) f (x,y) = x 2 y + x y (c) f (x,y) = sin (x + y) + cos (x y) (d) f (x,y) = sin 1 (xy 2 ) 2. Find the indicted prtil derivtive. () f (x,y) = e xy2 ; f xxy (b) z = x sin y; 3 z y 2 x (c) f (x,y,z) = x 5 + x 4 y 4 z 3 + yz 2 ; f xyz (d) z = ln sin (x y); 3 z y x 2 3. Verify tht the function u = e α2 k 2t sin kx is solution of the het conduction eqution u t = α 2 u xx. 4. Show tht the following functions re solutions of the wve eqution u tt = 2 u xx. () u = sin (kx) sin (kt) (b) u = (x t) 6 + (x + t) 6 Chin Rule Recll the Chin Rule for functions of single vrible gives the rule for differentiting composite function: If y = f (x) nd x = g (t), where f nd g re differentible functions, then y is indirectly differentible function of t, i.e., y = f (g (t)) nd dy dt = dy dx dx dt. Extending this ide, there re severl versions of the Chin Rule for functions of more thn one vrible: Cse I: Suppose tht z = f (x,y) is differentible function of x nd y, where x = g (t) nd y = h (t) re both differentible functions of t. Then dz dt = dx x dt + dy y dt. Cse II: Suppose tht z = f (x,y) is differentible function of x nd y, where x = g (s,t) nd y = h (s,t), nd the prtil derivtives g s, g t, h s, nd h t exist. Then s = x x s + y y s nd t = x x t + y y t. Cse III: Suppose tht z = f (x) is differentible function of x, where x = g (s,t), nd the prtil derivtives g s nd g t exist. Then s = dz x dx s nd t = dz x dx t. 1. Use the chin rule to find dz dt. () z = x 2 + y 2, x = t 3, y = 1 + t 2 (b) z = 6x 3 3xy + 2y 2, x = e t, y = cost 2. Use the chin rule to find s nd t. 3

4 () z = x 2 sin y, x = s 2 + t 2, y = 2st (b) z = sinxcos y, x = (s t) 2, y = s 2 t 2 3. If z = f (x y), show tht x + y = 0. [Hint: let u = x y] 4. If z = f (x,y), where x = r cos θ, y = r sin θ nd f is C 2 -function, () find nd r θ nd (b) show tht ( ) 2 + x ( ) 2 = y ( ) r r 2 ( ) 2 nd 2 z θ x + 2 z 2 y = 2 z 2 r z 2 r 2 θ r r. 5. Use the substitution u = x + ct nd v = x ct to reduce the wve eqution c 2 2 w x = 2 w 2 t 2 to the form 2 w u v = Suppose g (u) is C 1 -function of u nd let w = xy g differentition to evlute the expression x w x + y w y w. 7. Let ( ) x. Use the chin rule of y u = f (x,y) = x 2 + xy y 2, v = g (x,y) = x 3 y ( ) so tht (u,v) = (1, 2) when (x,y) = (1, 1). () Find the prtil derivtives f x, f y, g x, g y t (x,y) = (1, 1). (b) Let x = F (u,v), y = G (u,v) be the inverse function of ( ). Find the prtil derivtives F u, F v, G u, G v t (u,v) = (1, 2). (2007 Quiz) (2007 Test) 8. Let f be C 2 function nd w (x,y,z,t) = 1 r f (r t) where r = x 2 + y 2 + z 2. () Find ll the first order prtil derivtives. (b) Show tht 2 w x + 2 w 2 y + 2 w 2 = 2 w 2 t. 2 (2008 Exm) 4

5 Totl Differentil Recll tht for function of one vrible, y = f (x), we defined the increment of y s nd the differentil of y s y = f (x + x) f (x) dy = f (x)dx nd y dy. If z = f (x,y) is function of two vribles, then the increment of z is z = f (x + x,y + y) f (x,y), where x nd y re the increments of x nd y. The differentil dz, lso clled the totl differentil, is defined by dz = f x (x,y) dx + f y (x,y)dy = dx + dy nd z dz. x y Similrly, the reltive error nd percentge error re defined s Reltive Error = dz z nd Percentge Error = dz z 100%. If we tke dx = x = x 1 x 0 nd dy = y = y 1 y 0 for computing the increment of z for (x,y) chnges from (x 0,y 0 ) to (x 1,y 1 ), then dz = f x (x 0,y 0 ) x + f y (x 0,y 0 ) y = x + x y y. (x0,y 0 ) (x0,y 0 ) For mximum increment, dz = f x (x,y) dx + f y (x,y) dy f x (x,y) x + f y (x,y) y = x x + y y. 1. Find the totl differentil of the given functions. () z = x 2 y 3 (b) z = x 4 5x 2 y + 6xy (c) w = x sin (yz) (d) w = ln x 2 + y 2 + z 2 2. If z = 5x 2 + y 2 nd (x,y) chnges from (1, 2) to (1.05, 2.1), compute the vlue of z nd dz. 3. If z = x 2 xy + 3y 2 nd (x,y) chnges from (3, 1) to (2.96, 0.95), compute the vlue of z nd dz. 4. The dimensions of closed rectngulr box re mesured s 80cm, 60cm, nd 50cm, respectively, with possible error of 0.2cm in ech dimension. Use differentils to estimte the mximum error in clculting the surfce re of the box. 5

6 5. If R is the totl resistnce of three resistors, connected in prllel, with resistnces R 1, R 2, R 3, then 1 R = 1 R R R 3. If the resistnces re mesured in ohms s R 1 = 25Ω, R 2 = 40Ω, nd R 3 = 50Ω, with possible error of 0.5% in ech cse, i.e., dr i R i = 0.5%, where i = 1, 2, 3, estimte the mximum error in the clculted vlue of R. Liner Approximtion By using the totl differentil, we cn estimte the vlue of f (x 1,y 1 ) when f (x 0,y 0 ) is known nd (x 1,y 1 ) is close to (x 0,y 0 ). This is clled the liner pproximtion. Since z dz, then f (x 1,y 1 ) = f (x 0 + x,y 0 + y) = f (x 0,y 0 ) + z f (x 0,y 0 ) + dz. 1. Use differentils to pproximte the vlue of f t the given point. () f (x,y) = x 2 y 2, (5.01, 4.02) (b) f (x,y,z) = xy 2 sin (πz), (3.99, 4.98, 4.03) 2. Use liner pproximtion to evlute the following numbers. () (1.01) 3 (b) ( ) 4 (c) (3.02) 2 + (1.97) 2 + (5.99) 2 Reltive Mxim nd Minim A function of two vribles hs reltive mximum t (,b) if f (x,y) f (,b) for ll points (x,y). The vlue of f (,b) is clled reltive mximum vlue. If f (x,y) f (,b) for ll (x,y), then (,b) is reltive minimum point nd f (,b) is reltive minimum vlue. Procedure for finding reltive mxim nd minim: { fx (x,y) = 0 I. Find ll criticl point(s) by solving the system of equtions f y (x,y) = 0. II. For ech criticl point (x 0,y 0 ), compute the vlue of H = f xx (x 0,y 0 ) f xy (x 0,y 0 ) f yx (x 0,y 0 ) f yy (x 0,y 0 ). III. () If H > 0 nd f xx (x 0,y 0 ) > 0, then f (x 0,y 0 ) is reltive minimum. (b) If H > 0 nd f xx (x 0,y 0 ) < 0, then f (x 0,y 0 ) is reltive mximum. (c) If H < 0, then f (x 0,y 0 ) is not reltive extremum nd (x 0,y 0 ) is sddle point. (d) If H = 0, then we do not hve enough informtion bout the criticl point. 1. Find ll the reltive mximum nd minimum vlues nd sddle points of the given function. () f (x,y) = x 2 + y 2 + 4x 6y (b) f (x,y) = x 3 3xy + y 3 (c) f (x,y) = 3x 2 y + y 3 3x 2 3y (d) f (x,y) = x sin y 6

7 Tylor s Formul Recll the Tylor s Formul for one vrible function f (x), where f is C n+1 -function defined on n open intervl I contining x 0, n f (i) (x 0 ) f (x) = ( x) i + f(n+1) (ζ) n ( x) (n+1) f (i) (x 0 ) ( x) i, i! (n + 1)! i! i=0 }{{}}{{} i=0 Tylor Series Error Term where x = x x 0 nd ζ is point between x nd x 0. Suppose now z = f (x,y) is C 3 -function defined in domin D of the xy-plne, nd let (x 0,y 0 ) be point in D. For ny x = x x 0 nd y = y y 0, the Tylor s Formul for f (x,y) becomes f (x,y) =f (x 0,y 0 ) + f x (x 0,y 0 ) x + f y (x 0,y 0 ) y +R }{{} 2, Three Terms Formul f (x,y) =f (x 0,y 0 ) + f x (x 0,y 0 ) x + f y (x 0,y 0 ) y + 1 [ fxx (x 0,y 0 ) ( x) 2 + 2f xy (x 0,y 0 ) x y + f yy (x 0,y 0 ) ( y) 2] +R 3, etc..., } 2! {{} Six Terms Formul where R 2 nd R 3 re the error terms. 1. Find the first six terms of the Tylor s Formul t the given point for the following functions: () x y ; (2, 1) (b) 2x + y; (5, 1) Vectors Given the points A (x 1,y 1,z 1 ) nd B (x 2,y 2,z 2 ) in the three-dimensionl spce, the vector r (or r ) represents directed line segment AB from point A to point B nd r = [x 2 x 1,y 2 y 1,z 2 z 1 ] = (x 2 x 1 )i + (y 2 y 1 )j + (z 2 z 1 )k, where i, j nd k re the unit vectors in the x, y nd z direction, respectively. Given = [ 1, 2, 3 ] nd b = [b 1,b 2,b 3 ] re two non-zero vectors in the three-dimensionl spce, nd c is constnt, the length of vector, (or the norm), is defined by = , the vector + b is defined by + b = [ 1 + b 1, 2 + b 2, 3 + b 3 ] = ( 1 + b 1 )i + ( 2 + b 2 )j + ( 3 + b 3 )k, nd the vector c is defined by c = [c 1,c 2,c 3 ] = c 1 i + c 2 j + c 3 k. If u is unit vector of, then u is vector which hs the sme direction s with length equls to 1, i.e., u = = [ 1, 2, 3 ] If b is vector prllel to, then b = c, where c 0 nd b 1 = c 1, b 2 = c 2, b 3 = c Find vector r with representtion given by the directed line segment AB. 7

8 () A (1, 3), B (4, 4) (b) A (0, 3, 1), B (2, 3, 1) 2. Find, + b, b, 2, nd 3 + 4b () = [5, 12], b = [ 2, 8] (b) = i + j + k, b = 2i j + 3k 3. Find unit vector for the given vector. () [3, 5] (b) 2i 4j + 7k The Dot Product If = [ 1, 2, 3 ] nd b = [b 1,b 2,b 3 ], then the dot product (or the inner product) of nd b is the sclr b given by b b = 1 b b b 3. If θ, 0 θ < π, is the ngle between the vectors nd b, then b = b cosθ. nd b re orthogonl, θ = π, if nd only if b = 0. 2 The vector projection of b onto is defined by θ proj b proj b = proj b = b cos θ }{{} sclr }{{} direction (unit vector) 1. Find b. = b b b = b 2. () = [2, 5], b = [ 3, 1] (b) = 2i + 3j 4k, b = i 3j + k 2. Find the ngle between the given vectors. () = [1, 2, 2], b = [3, 4, 0] (b) = i + j + 2k, b = 2j 3k 3. Determine whether the given vectors re orthogonl, prllel, or neither orthogonl nor prllel. () = [2, 4], b = [ 1, 2] (b) = [2, 4], b = [4, 2] (c) = [2, 8, 3], b = [ 1, 2, 5] 4. Find the projection vector of b onto. () = [2, 3], b = [4, 1] (b) = 2i 3j + k, b = i + 6j 2k The Cross Product If = [ 1, 2, 3 ] nd b = [b 1,b 2,b 3 ], then the cross product of nd b is the vector b given by i j k b = b 1 b 2 b 3 = [ 2b 3 3 b 2, 3 b 1 1 b 3, 1 b 2 2 b 1 ]. 8

9 The vector b is orthogonl to both nd b. If θ, 0 θ < π, is the ngle between the vectors nd b, then b = b sin θ. Two non-zero vectors nd b re prllel, θ = 0, if nd only if b = 0. The length of the cross product b is equl to the re of the prllelogrm determined by nd b: Are = A = b. The volume of the prllelepiped determined by the vectors, b, nd c is the mgnitude of their sclr triple product: Volume = V = (b c). 1. Find the cross product b. b θ c b () = [1, 0, 1], b = [0, 1, 0] (b) = j + 4k, b = 6i 5k 2. Find two unit vectors orthogonl to both = [1, 1, 1] nd b = [0, 4, 4]. 3. Find the re of the prllelogrm which bounded by the following vertices. () A (0, 1), B (3, 0), C (5, 2), nd D (2, 1) (b) P (0, 0, 0), Q (5, 0, 0), R (2, 6, 6), nd S (7, 6, 6) 4. Find the volume of the prllelepiped determined by the vectors, b, nd c. () = [1, 0, 6], b = [2, 3, 8], nd c = [8, 5, 6] (b) = 2i + 3j 2k, b = i j, nd c = 2i + 3k Grdient Vector If f is function of two vribles x nd y, then the grdient of f is the vector function f defined by f (x,y) = [f x (x,y),f y (x,y)] = f x i + f y j. Similrly, if g is function of three vribles x, y, nd z, then the grdient of g, g is defined by g (x,y,z) = [g x (x,y,z),g y (x,y,z),g z (x,y,z)] = g x i + g y j + g k. 1. Find the grdient vector for the following multiple vribles functions. () f (x,y) = x y (b) f (x,y,z) = xe yz + xye z 2. Find the grdient vector for the following functions t the given point. 9

10 () f (x,y) = xe y + 3y; (1, 0) (b) f (x,y,z) = x y + y ; (4, 2, 1) z Tngent Plne nd Norml Line Let S: f (x,y,z) = C be level surfce for function f (x,y,z). If point P (x 0,y 0,z 0 ) is on S, then f (x 0,y 0,z 0 ) is vector norml to S t P, i.e., f (x 0,y 0,z 0 ) is the norml vector. Since the norml vector is perpendiculr to ny vector on the tngent plne, then the eqution of the tngent plne for f (x,y,z) = C t P (x 0,y 0,z 0 ) is P (x 0,y 0,z 0 ) f (x 0,y 0,z 0 ) (x,y,z) r f (x 0,y 0,z 0 ) r = f x (x 0,y 0,z 0 ) (x x 0 ) + f y (x 0,y 0,z 0 ) (y y 0 ) + f z (x 0,y 0,z 0 ) (z z 0 ) = 0, where r is vector lies on the tngent plne (or line segment from P to ny point (x,y,z) on the tngent plne.) The norml line to S t P is the line pssing through P nd perpendiculr to the tngent plne nd S or prllel to the norml vector. The eqution for the norml line is r = k f (x 0,y 0,z 0 ), where k 0 = x x 0 f x (x 0,y 0,z 0 ) = y y 0 f y (x 0,y 0,z 0 ) = z z 0 f z (x 0,y 0,z 0 ). 1. Find equtions of (i) the tngent plne nd (ii) the norml line to the given surfce t the specified point. () 4x 2 + y 2 + z 2 = 24, (2, 2, 2) (b) xyz = 6, (1, 2, 3) 2. Find the point on the ellipsoid x 2 + 2y 2 + 3z 2 = 1 where the tngent plne is prllel to the plne 3x y + 3z = Find the point on the hyperboloid x 2 y 2 + 2z 2 = 1 where the norml line is prllel to the line tht joins the points (3, 1, 0) nd (5, 3, 6). Directionl Derivtives For differentible function f (x,y), the directionl derivtive of f t P (x 0,y 0 ) in the direction of ny unit vector u = [,b], denoted by D u f (x 0,y 0 ), is D u f (x 0,y 0 ) = f (x 0,y 0 ) u = f x (x 0,y 0 ) + f y (x 0,y 0 )b. The directionl derivtive is the rte of chnge of f t P in the direction of u. Similrly, for g (x,y,z) t Q (x 0,y 0,z 0 ) in the direction of ny unit vector v = [,b,c], the directionl derivtive is Since D v g (x 0,y 0,z 0 ) = g x (x 0,y 0,z 0 ) + g y (x 0,y 0,z 0 ) b + g z (x 0,y 0,z 0 ) c. D u f (x,y) = f (x,y) u = f (x,y) u cos θ = f (x,y) cos θ, then the mximum vlue of D u f (x,y) is f (x,y) when θ = 0, i.e., u hs the sme direction s f (x,y). 10

11 1. Find the directionl derivtive of the function t the given point in the direction of the vector w. () f (x,y) = x, (6, 2), w = [ 1, 3] y ( y ) (b) g (x,y,z) = x tn 1, (1, 2, 2), w = [1, 1, 1] z (c) f (x,y) = e x cosy, ( 1, 6) π, w = i j (d) g (x,y,z) = z 3 x 2 y, (1, 6, 2), w = 3i + 4j + 12k 2. Find the direction tht the following functions increse most rpidly t the given point nd find the mximum rte of increse. () f (x,y) = x 2 + 2y, (4, 10) (b) f (x,y) = cos (3x + 2y), ( π, ) 6 π 8 Lgrnge Multipliers If we wnt to find the mximum nd minimum vlue of function f under condition(s) or constrint(s), then we cn pply the method of Lgrnge multipliers. Suppose we wnt to minimize the function f (x,y) under the constrint, function g (x,y) = k, then we define the Lgrnge function s the following: L (x,y,λ) = f (x,y) λ [g (x,y) k]. If (x 0,y 0 ) is minimum for the originl constrined problem, then there exists λ such tht (x 0,y 0,λ) is sttionry point or criticl point for the Lgrnge function. Sttionry points re those points where the prtil derivtive of L re zero, i.e., L x = 0, L y = 0, nd L λ = 0. Since nd L x = f x (x,y) λg x (x,y) L y = f y (x,y) λg y (x,y) L λ = [g (x,y) k] L = [L x, L y, L λ ] = [f x (x,y) λg x (x,y),f y (x,y) λg y (x,y),g (x,y) k] = 0, then f x (x,y) λg x (x,y) = 0 f y (x,y) λg y (x,y) = 0 g (x,y) k = 0 f x (x,y) = λg x (x,y) = f y (x,y) = λg y (x,y) g (x,y) = k = f = λ g with g (x,y) = k. The bove system of equtions cn be extended to higher dimensions nd more constrints. Procedure for the method of Lgrnge multipliers: Suppose we wnt to find the mximum nd minimum vlue of f (x,y,z) subject to the constrint g (x,y,z) = k, then f x (x,y,z) = λg x (x,y,z) f I. Solve the system of equtions y (x,y,z) = λg y (x,y,z) nd find ll sets of (x f z (x,y,z) = λg z (x,y,z) 0,y 0,z 0,λ). g (x,y,z) = k 11

12 II. Evlute f t ll the points (x 0,y 0,z 0 ) from Step I. The lrgest vlue is the mximum vlue of f; the smllest is the minimum vlue of f. If there re two constrints, g nd h, then we hve to solve the following system of equtions: f x (x,y,z) = λg x (x,y,z) + µh x (x,y,z) f y (x,y,z) = λg y (x,y,z) + µh y (x,y,z) f z (x,y,z) = λg z (x,y,z) + µh z (x,y,z). g (x,y,z) = k 1 h (x,y,z) = k 2 1. Use Lgrnge multipliers to find the mximum nd minimum vlues of the function subject to the given constrint(s). () f (x,y) = x 2 y 2 ; x 2 + y 2 = 1 (b) f (x,y) = 2x + y; x 2 + 4y 2 = 1 (c) f (x,y,z) = x + 2y; x + y + z = 1, y 2 + z 2 = 4 (d) f (x,y,z) = yz + xy; y 2 + z 2 = 1, xy = 1 2. Find the point on the cycle x 2 + y 2 = 4 tht hs the shortest distnce to the point (3, 1). 3. Find the shortest distnce from the point (2, 2, 3) to the plne 6x + 4y 3z = Find three positive numbers whose sum is 100 nd whose product is mximum. 5. A flt circulr plte hs the shpe of circulr disc x 2 + y 2 1. The plte, including the boundry where x 2 + y 2 = 1, is heted so tht the temperture t ny point (x,y), is T = x 2 x + 2y 2. Find the hottest nd coldest points on the plte nd the temperture t ech of these points. 6. Let f (x,y) = 6x 2 y 3x 2 2y 2, find the mximum nd minimum vlues of f on the region D = {(x,y) : 3x 2 + 2y 2 1}. (2008 Exm) Implicit Differentition Suppose tht z is given implicitly s function z = f (x,y) by n eqution of the form F (x,y,z) = 0. This mens tht F (x,y,f (x,y)) = 0 for ll (x,y) in the domin of f. By the Implicit Function Theorem, F x = x F = F x F z nd y = F y F = F y F z. 1. Find x nd y for xyz = cos(x + y + z), where z = f (x,y). 2. Find 2 z x 2 nd 2 z y x for x2 + y 2 z 2 = 2x (y + z), where z = f (x,y). 12

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