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Chain Rules, Taylor's Theorem and Function Extrema

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Chain Rules, Taylor's Theorem and Function Extrema Chain rules, Taylor’s theorem and function extrema I. Chain rules • Basic version (for one parameter). Suppose that h(t) = f(x1(t), . , xd(t)), where x1, ..., xd are differentiable at t and f is differentiable at (x1(t), . , xd(t)), then d ∂f h0(t) = X (x (t), . , x (t))x0 (t). (1) ∂x 1 d j j=1 j If x1(t), ..., xd(t) are differentiable functions of t and f is differentiable on the range of (x1(t), . , xd(t)), then (1) is sometimes expressed as dh ∂f dx ∂f dx = 1 + ··· + d . dt ∂x1 dt ∂xd dt – When d = 2 and h(t) = f(x(t), y(t)), where x and y are differen- tiable at t and f is differentiable at (x(t), y(t)), (1) becomes 0 0 0 h (t) = fx(x(t), y(t))x (t) + fy(x(t), y(t))y (t). (2) – Proof of (2). Recall that f is differentiable at (x0, y0) means that f(x, y) = f(x0, y0) + fx(x0, y0)(x − x0) + fy(x0, y0)(y − y0) +1(x − x0) + 2(y − y0), (3) where lim(x,y)→(x0,y0) 1 = 0 and lim(x,y)→(x0,y0) 2 = 0. Apply (3) with (x, y) = (x(t + ∆), y(t + ∆)) and (x0, y0) = (x(t), y(t)) and (2) holds by taking the limits as ∆ → 0. • Other versions. – Suppose that z(u, v) = f(x, y), where x = x(u, v) and y = y(u, v). Suppose that xu, xv, yu and yv exist and f is differentiable. Then ∂z ∂f ∂x ∂f ∂y ∂z ∂f ∂x ∂f ∂y = + and = + . ∂u ∂x ∂u ∂y ∂u ∂v ∂x ∂v ∂y ∂v – Suppose that u = f(x, y, z), where x = x(s, t), y = y(s, t) and z = z(s, t). Suppose that the partial derivatives of x, y and z with respect to s and t exist and f is differentiable. Then ∂u ∂f ∂x ∂f ∂y ∂f ∂z = + + ∂s ∂x ∂s ∂y ∂s ∂z ∂s 1 and ∂u ∂f ∂x ∂f ∂y ∂f ∂z = + + . ∂t ∂x ∂t ∂y ∂t ∂z ∂t Example 1. Suppose that h(t) = x2 + xy, where x = t and y = t2. Find h0(1). Example 2. Suppose that u = xy + 2z, where x = st, y = s + t and z = t2. Find ∂u/∂s. Example 3. Supposet that z = u + f(uv), where f is differentiable. Show that u(∂z/∂u) − v(∂z/∂v) = u. II. Taylor’s theorem • The multivariate version of Taylor’s theorem follows from the uni- variate Taylor’s theorem and the chain rule. To obtain the Taylor expansion of f(x) at a, let g(t) = f(a + t(x − a)), then g(0) = f(a) and g(1) = f(x). Then the univariate version of Taylor’s theorem says that ! n g(k)(0)tk g(n+1)(c)tn+1 g(t) = X + , k! (n + 1)! k=1 where c is between t and 0. Put t = 1 and we have the multivariate version of Taylor’s theorem. • Example 4. Suppose that f, fx and fy are differentiable on B((x0, y0), δ) for some δ > 0, where B((x0, y0), δ) = {(x, y): k(x, y) − (x0, y0)k < δ} and (x, y) ∈ B((x0, y0), δ). Show that R (x, y) f(x, y) = f(x , y ) + f (x , y )(x − x ) + f (x , y )(y − y ) + 2 , 0 0 x 0 0 0 y 0 0 0 2 where 2 R2(x, y) = fxx(x∗, y∗)(x − x0) + 2fxy(x∗, y∗)(x − x0)(y − y0) 2 +fyy(x∗, y∗)(y − y0) (4) and the point (x∗, y∗) lies on the segment with endpoints (x0, y0) and (x, y). • Taylor’s theorem can be applied to control the error of a linear ap- proximation if all the second-order partial derivatives are bounded. 2 III. Extrema of bivariate functions • Absolute extrema. – f(x, y) has an absolute maximum at (x0, y0) means f(x, y) ≤ f(x0, y0) for every (x, y) in the domain of f. – f(x, y) has an absolute minimum at (x0, y0) means f(x, y) ≥ f(x0, y0) for every (x, y) in the domain of f. • Relative extrema. Suppose that f is defined on a set containing B((x0, y0), δ) = {(x, y): k(x, y) − (x0, y0)k < δ} for some δ > 0. – f(x, y) has an relative maximum at (x0, y0) means f(x, y) ≤ f(x0, y0) for every (x, y) in B((x0, y0), δ). – f(x, y) has an relative minimum at (x0, y0) means f(x, y) ≥ f(x0, y0) for every (x, y) in B((x0, y0), δ). domain of f. • Partial derivative criteria for relative extrema (Theorem 11.11). Sup- pose that f is defined on a set containing B((x0, y0), δ) = {(x, y): k(x, y) − (x0, y0)k < δ} for some δ > 0. If f has a relative maximum or minimum at (x0, y0) and fx and fy both exist at (x0, y0), then fx(x0, y0) = fy(x0, y0) = 0. • If fx(x0, y0) = fy(x0, y0) = 0 or fx(x0, y0) and fy(x0, y0) do not both exist, then we say there is a critical point at (x0, y0). • Remark. If f is differentiable at (x0, y0), then f(x, y) ≈ L(x, y), where L(x, y) = f(x0, y0) + fx(x0, y0)(x − x0) + fy(x0, y0)(y − y0). We cannot have L(x, y) ≥ 0 on B((x0, y0), δ) or L(x, y) ≤ 0 on B((x0, y0), δ) unless fx(x0, y0) = fy(x0, y0) = 0. • Second partials test (Theorem 11.12). Suppose that fx(x0, y0) = fy(x0, y0) = 0 and all the second-order partial derivatives of f are continuous on B((x0, y0), δ) = {(x, y): k(x, y) − (x0, y0)k < δ} for 2 some δ > 0. Let D = fxxfyy − fxy. Then we have the following results. (a) If D(x0, y0) > 0, then (i) a relative minimum occurs at (x0, y0) if fxx(x0, y0) > 0 or fyy(x0, y0) > 0; 3 (ii) a relative maximum occurs at (x0, y0) if fxx(x0, y0) < 0 or fyy(x0, y0) < 0. (b) If D(x0, y0) < 0, then a saddle point occurs at (x0, y0). • We say a saddle point occurs at (x0, y0) if there is a critical point at (x0, y0) but there is no relative extremum at (x0, y0). 2 • The function D = fxxfyy − fxy is called the discriminant of f. • Remark. Under the conditions in the second partial test, when (x, y) ≈ (x0, y0), the R2(x, y) in (4) is close to 2 U(x, y) = fxx(x0, y0)(x − x0) + 2fxy(x0, y0)(x − x0)(y − y0) 2 +fyy(x0, y0)(y − y0) and U(x, y) f(x, y) ≈ f(x , y ) + . 0 0 2 2 Let D0 = fxx(x0, y0)fyy(x0, y0) − fxy(x0, y0). Then we have the fol- lowing results. (a) If D0 > 0, then (i) U(x, y) > 0 for (x, y) 6= (x0, y0) if fxx(x0, y0) > 0 or fyy(x0, y0) > 0; (ii) U(x, y) < 0 for (x, y) 6= (x0, y0) if fxx(x0, y0) < 0 or fyy(x0, y0) < 0. (b) If D0 < 0, then U(x, y) > 0 for some (x, y) ≈ (x0, y0) and U(x, y) < 0 for some other (x, y) ≈ (x0, y0). Example 5. Suppose that f(x, y) = x2 − y2. Find all the point(s) where the relative extrema or saddle point(s) of f occur. Answer: no relative extrema; saddle point at (0, 0) Example 6. Supppose that f(x, y) = x2 + y2 + 2x. Find all the point(s) where the relative extrema or saddle point(s) of f occur. Answer: relative minimum at (−1, 0). • Extreme value theorem for a bivariate function (Theorem 11.13). Sup- pose that f is a bivariate function that is continuous on S, where S is closed and bounded. Then f attains both its maximum and minimum on S. 4 – S is bounded means that there exists a number M > 0 such that S ⊂ {(x, y): k(x, y)k ≤ M}. – S is closed means that if limn→∞ k(xn, yn) − (x0, y0)k = 0 and ∞ {(xn, yn)}n=1 is a sequence in S, then (x0, y0) is also in S. – Extreme value theorem implies that the maximum and minimum of f on a closed and bounded set S can occur only at critical points or the boundary of S. – A point (x0, y0) is in the boundary of S means that for every δ > 0, B((x0, y0), δ) contains both some points in S and some points not in S. Example 7. Find the maximum and minimum of f(x, y) = x2 + y on S = {(x, y) : 0 ≤ x ≤ 1 and 2 ≤ y ≤ 3}. 5.
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