2s4dLg-04

2s4dLg-04 - MAT 473 9 4. Chain Rule. Partial Derivatives....

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Unformatted text preview: MAT 473 9 4. Chain Rule. Partial Derivatives. Proposition 4.1 (Chain Rule). Suppose f : U Rn Rm is differentiable at x U and g : V Rm R is differentiable at f (x) V . Then g f is differentiable at x, with (g f ) (x) = g (f (x)) f (x). Proof. Put y = f (x) and choose s, Ws and t, Wt as in Lemma 3.6 such that f (x + h) = f (x) + f (x)(h) + s(h)h for all h Ws , g(y + k) = g(y) + g (y)(k) + t(k)k for all k Wt , limh0 s(h) = 0, and limk0 t(k) = 0. Now let k(h) = f (x)(h) + s(h)h; certainly limh0 k(h) = 0, so we can choose > 0 such that k(h) Wt whenever h < . Thus for all h W = Ws B (0), we have (g f )(x + h) = = = = g(f (x) + f (x)(h) + s(h)h) g(y + k(h)) g(y) + g (y)(k(h)) + t(k(h))k(h) g(f (x)) + g (f (x))(f (x)(h) + s(h)h) + t(k(h))k(h) = (g f )(x) + g (f (x))(f (x)(h)) + g (f (x))(s(h))h + t(k(h)) k(h) = 0. Since h g (f (x))(f (x)(h)) = (g (f (x)) f (x)) (h) h0 h is linear, another application of Lemma 3.6 completes the proof. as h 0, so lim t(k(h)) Definition 4.2. For f : U Rn Rm , and for 1 i m, define fi : U R by for all x Rn . We call fi the ith component function of f , and we write f = (f1 , . . . , fm ) because by definition, f (x) = (f1 (x), . . . , fm (x)) for each x. Since a sequence in Rm converges if and only if its component sequences converge in R, a function f : U Rm is continuous if and only if each fi is. Similarly: fi (x) = ei , f (x) = f (x)i Now clearly limh0 g (f (x))(s(h)) = 0. Moreover, t(k(h)) k(h) = t(k(h)) f (x)(h) + s(h)h 0 h h f (x)h + s(h)h t(k(h)) = t(k(h)) (f (x) + s(h)) 0 h k(h) h. h Lemma 4.3. f : U Rn Rm is differentiable at x U if and only if each fi is, in which case f (x)i = fi (x). (That is, the ith component function of f (x) is the derivative at x of the ith component function of f .) Date: January 18, 2012 S. Kaliszewski, Department of Mathematics and Statistics, Arizona State University. 10 SPRING 2012 Proof. Observe that for any T L(Rn , Rm ) and any x U , f (x + h) - f (x) - T (h) h0 h f1 (x + h) - f1 (x) - T1 (h) fm (x + h) - fm (x) - Tm (h) = lim ,..., h0 h h f1 (x + h) - f1 (x) - T1 (h) fm (x + h) - fm (x) - Tm (h) = lim , . . . , lim , h0 h0 h h and apply the definition of derivative. lim Note that now we can write f (x)(h)i = f (x)i (h) = fi (x)(h), where in each expression the subscript means a different thing. Definition 4.4. For f : U Rn Rm and x U , and for each 1 i m and 1 j n, the partial derivative of fi at x with respect to the j th variable is the real number fi (x + tej ) - fi (x) Dj fi (x) = lim , t0 t provided the limit exists. More generally, for any v Rn , we define the directional derivative of f at x in the direction of v to be the element of Rm defined by f (x + tv) - f (x) Dv f (x) = lim , t0 t provided the limit exists. Note that Dv f (x) exists if and only if g (0) exists (in which case the two are equal), where g is the Rm -valued function defined on a neighborhood of 0 in R by g(t) = f (x + tv). (This explains the description of Dv f (x) as a "directional" derivative.) Further, by Lemma 4.3, for fixed j, the partial derivatives Dj fi (x) exist for all i if and only if the directional derivative Dej f (x) exists, in which case Proposition 4.5. If f : U Rn Rm is differentiable at x U , then Dv f (x) exists for each v Rn , with Dv f (x) = f (x)(v). In particular, Dj fi (x) exists for each 1 i m and 1 j n, and [f (x)] = Dj fi (x) The matrix [f (x)] is referred to as the Jacobian of f at x. Proof. Choose r for f as in Lemma 3.6, and fix v Rn . Then f (x + tv) - f (x) f (x)(tv) + r(tv)tv lim = lim t0 t0 t t tv = f (x)(v), = f (x)(v) + lim r(tv) t0 t Date: January 18, 2012 S. Kaliszewski, Department of Mathematics and Statistics, Arizona State University. Dj fi (x) = ei , Dej f (x). MAT 473 11 since r(tv) 0 as t 0 and tv/t is bounded. Now setting v = ej and taking ith coordinates gives which shows that the (i, j)-entry of [f (x)] is Dj fi (x). Dj fi (x) = ei , Dej f (x) = ei , f (x)(ej ), Exercise 4.1. Suppose f : R2 R is defined by xy if (x, y) = (0, 0) f (x, y) = x2 + y 2 0 if (x, y) = (0, 0). Exercise 4.2. Suppose f : R2 R is defined by 2 xy if (x, y) = (0, 0) f (x, y) = x2 + y 2 0 if (x, y) = (0, 0). Show that Dj fi (x, y) exists for all i, j and all x, y, but f is not continuous, and therefore not differentiable, at (0, 0). Show that all partial derivatives of f exist everywhere, and f is continuous everywhere, but f (0, 0) still does not exist. Exercise 4.3. Let f : U Rn R, and fix v Rn . Prove that if f is differentiable at x, then (4.1) Here f (x) = (D1 f (x), . . . , Dn f (x)) is the gradient vector of f at x. Dv f (x) = v, f (x). Date: January 18, 2012 S. Kaliszewski, Department of Mathematics and Statistics, Arizona State University. ...
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This document was uploaded on 03/11/2012.

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