March 13, 2009 CHAPTER 3: PARTIAL DERIVATIVES AND DIFFERENTIATION

March 13, 2009 CHAPTER 3: PARTIAL DERIVATIVES AND DIFFERENTIATION 1. Parial Derivaives and Differeniable funcions In all his chaper, D will denoe an open subse of R n. Definiion 1.1. Consider a funcion f : D R and le p D, i = 1,, n. We define he parial derivaive of f wih respec o he i-h variable a he poin p as he following limi (if i exiss) f(p + e i ) f(p) (p) i 0 where {e 1,..., e n } is he canonical basis of R n, defined as follows e i = (0,..., 0, i, 0,..., 0 ) }{{}}{{} i 1 erms n i erms For example, in R 2 he canonical basis is and in R 3 he canonical basis is e 1 = (1, 0) e 2 = (0, 1) e 1 = (1, 0, 0) e 2 = (0, 1, 0) e 3 = (0, 0, 1) Remark 1.2. When n = 2, in he above definiion we le we and use he noaion, Likewise, when n = 3, we le and use he noaion, p = (x, y), f(x, y) : R 2 R f(x +, y) f(x, y) (x, y) 0 (x, y) 0 (x, y, z) 0 (x, y, z) 0 f(x, y + ) f(x, y) p = (x, y, z) f(x +, y, z) f(x, y, z) (x, y, z) 0 f(x, y +, z) f(x, y, z) f(x, y, z + ) f(x, y, z) 1

2 CHAPTER 3: PARTIAL DERIVATIVES AND DIFFERENTIATION Example 1.3. In Economics, he parial derivaives of a uiliy funcion are called marginal uiliies, he parial derivaives of a producion funcion are called marginal producs. Consider, for example he Cobb-Douglas producion funcion f(k, L) = 5K 1/3 L 2/3 where f is he number of unis produced, K is he capial and L is labor. Tha is, he above formula means ha if we use K unis of capial and L unis of labor, hen we produce f(k, L) = 5K 1/3 L 2/3 unis of a good. The consans A = 5, α = 1/3 and β = 2/3 are echnological parameers. The marginal producs wih respec o capial and labor are The marginal produc of labor, K = 5 3 K 2/3 L 2/3 L = 10 3 K1/3 L 1/3 (K, L) L is inerpreed in Economics as an approximaion o he variaion in he producion of he good when we are using K unis of capial and L unis of labor and we swich o use an addiional uni L + 1 of labor and he same unis K of capial as before. We see ha he marginal produc of labor and capial is posiive. Tha is, if we use more labor and/or more capial, producion increases. On he oher hand, he marginal produc of labor is decreasing in labor and increasing in capial. We may inerpre his as follows. Suppose ha we keep consan he amoun of capial ha we are using K. If L > L hen f(k, L + 1) f(k, L ) < f(k, L + 1) f(k, L) Tha is, an increase in he producion when we use an addiional uni of labor is decreasing in he iniial labor ha is being used. If we keep he capial consan, using on addiional uni of labor, if we are already using a lo of labor, does no increase much he producion. We may imagine ha f(k, L) is he producion of a farm produc in a piece of land where L is he number of he workers and he size K of he land is consan. The impac in he producion when hiring and addiional person is greaer if few people are working in he land as compared wih he case in which we already have a lo of people working in he land. Suppose ha he amoun of labor L is kep consan. If K > K hen f(k, L + 1) f(k, L) > f(k, L + 1) f(k, L) Tha is, he increase in he producion when we use one addiional uni of labor is larger he more capial we use. Capial and labor are complemenary. In he previous example, hiring and addiional worker has a larger effec on he producion he larger is he size of land.

CHAPTER 3: PARTIAL DERIVATIVES AND DIFFERENTIATION 3 Definiion 1.4. Consider a funcion f : D R. Le p D and suppose all he parial derivaives (p), (p),, (p) 1 2 n exis a he poin p. We define he gradien of f a p as he following vecor ( f(p) = (p), (p),, ) (p) 1 2 n Definiion 1.5. Consider a funcion f : D R. Le p D and suppose all he parial derivaives (p), (p),, (p) 1 2 n exis a he poin p. We say ha f is differeniable a p if Noe ha he limi is aken for v R n. f(p + v) f(p) f(p) v lim = 0 v 0 v Remark 1.6. A funcion of wo variables f : D R 2 is differeniable a he poin p = (a, b) if Leing f(a + v 1, b + v 2 ) f(a, b) f(a, b) (v 1, v 2 ) lim = 0 (v 1,v 2) (0,0) (v 1, v 2 ) x = a + v 1, y = b + v 2 we see ha (v 1, v 2 ) (0, 0) is equivalen o (x, y) (a, b), so we may wrie his limi as f(x, y) f(a, b) f(a, b) (x a, y b) lim = 0 (x,y) (a,b) (x a, y b) Wriing his limi explicily we wee ha f is differeniable a he poin p = (a, b) if (1.1) lim (x,y) (a,b) f(x, y) f(a, b) (a, b) (x a) (a, b) (y b) = 0 (x a)2 + (y b) 2 Example 1.7. Consider he funcion { xy 2 f(x, y) = x 2 +y if (x, y) (0, 0), 2 0 if (x, y) = (0, 0). We will show ha f is no differeniable a he poin p = (0, 0). Firs of all, we compue f(0, 0). Noe ha f(, 0) f(0, 0) (0, 0) 0 (0, 0) 0 f(0, ) f(0, 0) 0 0 3 = 0 0 0 3 = 0

4 CHAPTER 3: PARTIAL DERIVATIVES AND DIFFERENTIATION so, f(0, 0) = (0, 0). Le us use he noaion v = (x, y). Then, f is differeniable a he poin p = (0, 0) if and only if f(p + v) f(p) f(p) v 0 v 0 v f ((0, 0) + (x, y)) f(0, 0) f(p) (x, y) x2 + y 2 f(x, y) f(0, 0) (0, 0) (x, y) x2 + y 2 f(x, y) x2 + y 2 xy 2 We prove ha he above limi does no exis. Consider he funcion xy 2 g(x, y) = Noe ha and noe ha so he limi 0 lim g(, 0) 0 0 (2 2 ) = 0 3/2 3 lim g(, ) 0 0 (2 2 ) = 1 3/2 (2) 0 3/2 lim xy 2 does no exis and we conclude ha f is no differeniable a he poin (0, 0). Example 1.8. Consider now he funcion { xy 3 f(x, y) = x 2 +y if (x, y) (0, 0), 2 0 if (x, y) = (0, 0). We will show ha f is differeniable a he poin p = (0, 0). Firs of all, we compue f(0, 0). Noe ha (0, 0) 0 f(, 0) f(0, 0) (0, 0) 0 f(0, ) f(0, 0) 0 0 3 = 0 0 0 3 = 0

CHAPTER 3: PARTIAL DERIVATIVES AND DIFFERENTIATION 5 so, f(0, 0) = (0, 0). Le us use he noaion v = (x, y). Then, f is differeniable a he poin p = (0, 0) if and only if f(p + v) f(p) f(p) v 0 v 0 v f((0, 0) + (x, y)) f(0, 0) f(p) (x, y) x2 + y 2 f(x, y) f(0, 0) (0, 0) (x, y) x2 + y 2 f(x, y) x2 + y 2 xy 3 Le ε > 0. Take δ = ε and suppose ha 0 < x 2 + y 2 < δ. Then, xy 3 = x y2 y x2 y 2 y = x2 + y ( 2 x 2 + y 2) y ( x 2 + y 2) 3/2 y = So, lim = y x 2 + y 2 < δ = ε xy 3 = 0 and he funcion is differeniable a he poin (0, 0). Proposiion 1.9. Le f : D R. If f is differeniable a some poin p D, hen f is coninuous a ha poin. Example 1.10. Consider he funcion { x 3 y f(x, y) = x 4 +y if (x, y) (0, 0), 2 0 if (x, y) = (0, 0). Is i coninuous and/or differeniable a (0, 0)? One compues easily he ieraed limi ( ) lim lim f(x, y) = 0 x 0 y 0 On he oher hand, aking he curve x() =, y() = 2

6 CHAPTER 3: PARTIAL DERIVATIVES AND DIFFERENTIATION we see ha So, lim f(, 4 0 2 ) 0 2 4 = 1 2 0 lim f(x, y) does no exis. I follows ha f is no coninuous a (0, 0). In addiion, by Proposiion 1.9, f is no differeniable a (0, 0), eiher. Theorem 1.11. Le f : D R and p D. Suppose ha here is some r > 0 such ha he parial derivaives, 1,,, 2 n exis a every poin of he open ball B(p, r) and are coninuous funcions on ha ball. Then, he funcion f is differeniable a p. Example 1.12. The previous Theorem applies o show ha he funcion is differeniable a every poin of R 2. f(x, y, z) = xe yz + y sin z Definiion 1.13. A funcion f : D R is of class C 1 in D if all he parial derivaives of f exis and are coninuous funcions on D. In his case we wrie f C 1 (D). 2. Direcional derivaives Definiion 2.1. Le f : D R n R. Fix a poin p D and a vecor v R n. If he following limi exiss D v f(p) 0 f(p + v) f(p) i is called he derivaive of f a p along (he vecor) v. If v = 1, hen D v f(p) is called he direcional derivaive of f a p in he direcion of (he vecor) v Remark 2.2. Le n = 1. Le f : R R, p R and v = 1. The above definiion coincides wih he derivaive of a one variable funcion f f(p + ) f(p) (p) 0 Example 2.3. Le f : R 2 R be defined by f(x, y) = xy and ake p = (1, 1), v = (3, 4). Then, for R we have ha so, p + v = (1 + 3, 1 + 4) f(1 + 3, 1 + 4) f(1, 1) D v f(p) 0 (1 + 3)( 1 + 4) + 1 = 1 0

CHAPTER 3: PARTIAL DERIVATIVES AND DIFFERENTIATION 7 And, since v = 3 2 + 4 2 = 5, he direcional derivaive of f a p in he direcion of v is 1 v D vf(p) = 1 5 Remark 2.4. If we ake v = e i = (0,..., 0, }{{} 1, 0,..., 0) i o be he i h vecor of he canonical basis, hen D ei f(p) = i (p) is he i h parial derivaive of f a he poin p. Proposiion 2.5. Le f : D R be differeniable a he poin p D. Then, (2.1) D v f(p) = f(p) v Example 2.6. As in Example 2.3, le f : R 2 R be defined by f(x, y) = xy and ake p = (1, 1), v = (3, 4). Then, f(p) = (y, x) x=1 y= 1 = ( 1, 1) and, since f is differeniable on all of R 2, we have ha as compued in Example 2.3. D v f(p) = f(p) v = ( 1, 1) (3, 4) = 3 + 4 = 1 Remark 2.7. One may also define he derivaive of a funcion f : D R n R m along (he vecor) v. To do so, we wrie he funcion f using is coordinae funcions f(x) = (f 1 (x), f 2 (x),, f m (x)) wih f i : D R for each i = 1,, m. And define D v f(p) = (D v f 1 (p), D v f 2 (p),, D v f m (p)) We see now ha D v f(p) is a vecor in R m. Likewise we may define he direcional derivaive of f a he poin p in he direcion of a uniary vecor u. 3. inerpreaion of he gradien The formula 2.1 may be used o give an inerpreaion of he gradien as follows. Recall ha given wo vecors u, v in R n, heir scalar produc saisfies u v = u v cos θ where θ is he angle beween he wo vecors.

8 CHAPTER 3: PARTIAL DERIVATIVES AND DIFFERENTIATION u θ Applying his observaion o formula 2.1, we see ha v D v f(p) = f(p) v = f(p) v cos θ where θ is he angle beween he vecors f(p) and v. Taking v o be uniary, we see ha he derivaive of f in he direcion of v is D v f(p) = f(p) cos θ Thus, D v f(p) aains a maximum when θ = 0, ha is, when he vecors f(p) and v poin in he same direcion. aains a minimum when θ = π, ha is, when he vecors f(p) and v poin in he opposie direcions. is zero when θ = π/2 or θ = 3π/2, ha is, when he vecors f(p) and v are perpendicular. I follows ha, The funcion f grows he fases in he direcion of f(p). The funcion f decreases he fases in he direcion opposie o f(p). The funcion f remains consan in he direcions perpendicular o f(p). D f(p) = 4. The chain rule Definiion 4.1. Given a funcion f(x) = (f 1 (x), f 2 (x),, f m (x)) : D R n R m and a poin p D, we define he Jacobian marix of f a he poin p as he following marix of order m n 1(p) 2 1(p) n 2(p) 2 1(p) 1 2(p) 1. m(p) 1.. m(p) 2 2(p) n. m(p) n Remark 4.2. If f(x) = D R n R Wha is he difference beween D f(p) and f(p)? Remark 4.3. If m = n = 1 Wha is D f(p)? Definiion 4.4. A funcion f(x) = (f 1 (x), f 2 (x),, f m (x)) : D R n R m is said o be differeniable a a poin p D if each of he funcions f 1 (x), f 2 (x),, f m (x) is differeniable a p. Theorem 4.5 (The chain rule). Le g : R n R m and f : R m R l. Suppose ha g is differeniable a p R n and ha f is differeniable a g(p) R m. Then, he funcion f g is differeniable a p and D(f g)(p) = D f(g(p)) D g(p)

CHAPTER 3: PARTIAL DERIVATIVES AND DIFFERENTIATION 9 Remark 4.6. The expression D(f g)(p) = D f(g(p)) D g(p) conains he produc of 2 marices. Example 4.7 (Special case of he chain rule). Le σ : R R 2 and f : R 2 R be differeniable. Suppose ha σ() may be wrien as Then, he chain rule says ha σ() = (x(), y()) d f(x(), y()) = D(f σ)() = D f(x, y) x=x() D σ() ( ) ( ) = x=x() x () y () = (x(), y())x () + (x(), y())y () Example 4.8 (Special case of he chain rule). Le g(s, ) : R 2 R 2 and f(x, y) : R 2 R be differeniable. Suppose ha g(s, ) may be wrien as so ha hen, he chain rule says ha g(s, ) = (x(s, ), y(s, )) (f g)(s, ) = f(g(s, )) = f (x(s, ), y(s, )) D f (x(s, ), y(s, )) = D(f g)(s, ) = D f(x, y) x=x(s,) D g(s, ) = = ( ) ( ( s + s s s ) y=y(s,) + ) Tha is, (f g) s (f g) = s + = + s Example 4.9. Consider he Cobb-Douglas producion funcion f(k, L) = 5K 1/3 L 2/3 where f are he unis produced, K is capial and L is labor. Suppose ha capial and labor change wih ime K = K(), L = L() Then he producion funcion f(k(), L())

10 CHAPTER 3: PARTIAL DERIVATIVES AND DIFFERENTIATION is also a funcion of ime. Wha is he rae of change of he producion a a given ime? We may answer his quesion using he chain rule. df(k(), L()) = dk K + dl L = 5 3 K 2/3 L 2/3 dk + 10 3 K1/3 1/3 dl L Example 4.10. Suppose an agen has he following differeniable uiliy funcion u(x, y) where x is a consumpion good and y is air polluion. Then, he uiliy of he agen is increasing in x and decreasing in y, u > 0 u < 0 Suppose ha he producion of x unis of he good generaes y = f(x) unis of polluion, Wha is he opimal level of consumpion of x? The uiliy of he agen when he consumes x unis of he good and y = f(x) unis of polluion are generaed is u(x, f(x)) The agens maximizes his uiliy funcion. The firs order condiion is du(x, f(x)) = 0 dx using he chain rule we obain ha he equaion 0 = u u (x, f(x)) + (x, f(x))f (x) deermines he opimal level of producion of he good. 5. Derivaive along a curve and level surfaces Remark 5.1 (A special case of he chain rule). Le σ : R R n be a differeniable curve and le f : D R be differeniable, where D is an open subse of R n. Suppose, σ() can be wrien as where each σ() = (σ 1 (), σ 2 (),..., σ n ()) dσ i () is differeniable for every i = 1,..., n and every R. Hence, we may wrie dσ = Then, f(σ()) is differeniable and ( dσ1, dσ 2,..., dσ n d dσ f(σ()) = f(σ()) )

CHAPTER 3: PARTIAL DERIVATIVES AND DIFFERENTIATION 11 Example 5.2. If f : R 2 R is a funcion of wo variables and x = x(), y = y(), he chain rule is d (x, y) f(x(), y()) = dx() (x, y) + dy() x=x() x=x() Example 5.3. If f : R 3 R is a funcion of hree variables and x = x(), y = y(), z = z(), he chain rule is d (x, y, z) f(x(), y(), z()) = dx() (x, y, z) + dy() + x=x() z=z() x=x() z=z() Remark 5.1 provides anoher inerpreaion of he gradien. Le p D and le f : D R be differeniable, where D is some open subse of R n. Le C R and suppose he level surface S C = {x D : f(x) = C} is no empy. Le σ : R R n be a differeniable curve and suppose ha σ() S C for all R. Tha is f(σ()) = c for every R. Differeniaing and using he above chain rule we have ha 0 = d dσ f(σ()) = f(σ()) Tha is f(σ()) and dσ()/ are perpendicular for every R. (x, y, z) z x=x() z=z() dz() f(p) σ () p {x: f (x) = C} The above argumen shows ha a any poin p S C, he gradien f(p) is perpendicular o he surface level S C. Remark 5.4. Le us compue he plane angen o he graph of a funcion of wo variables. To do his, consider a differeniable funcion f : R 2 R. The graph of f is he se G = {(x, y, f(x, y)) : (x, y) R 2 }

12 CHAPTER 3: PARTIAL DERIVATIVES AND DIFFERENTIATION Fix a poin p = (a, b) R 2. Consider he following funcion of hree variables Then,he graph of f may be wrien as g(x, y, z) = f(x, y) z G = {(x, y, z) R 3 : g(x, y, z) = 0} Then, he plane T angen o G a he poin (a, b, f(a, b)) saisfies he wo following properies T conains he poin (a, b, f(a, b)). T is perpendicular o he gradien g(a, b, f(a, b)). This informaion permis us o compue he equaions for T as follows. Firs of all, i follows ha an equaion for T is g(a, b, f(a, b)) ((x, y, z) (a, b, f(a, b))) = 0 and noe ha ( ) g(a, b, f(a, b)) = (a, b), (a, b), 1 Hence, an equaion for T is he following (5.1) f(a, b) + (a, b) (x a) + (a, b) (y b) = z We may use he above o provide anoher inerpreaion of he definiion of differeniabiliy 1.5. Le P 1 (x, y) = f(a, b) + (a, b) (x a) + (a, b) (y b) For he case of a funcion of wo variables, Equaion 1.1 says ha he funcion f is differeniable a he poin (a, b) if f(x, y) P 1 (x, y) lim = 0 (x,y) (a,b) (x a, y b) In view of Equaion 5.1, he funcion f is differeniable a he poin (a, b) if he angen plane is a good approximaion o he value of he funcion f(x, y) f(a, b) + (a, b) (x a) + (a, b) (y b)