3.33pt
Chain Rule MATH 311, Calculus III J. Robert Buchanan Department of Mathematics Spring 2019
Single Variable Chain Rule Suppose y = g(x) and z = f (y) then dz dx = d (f (g(x))) dx = f (g(x))g (x) = dz dy dy dx
Multi-variable Chain Rule Theorem If z = f (x(t), y(t)), where x(t) and y(t) are differentiable and f (x, y) is a differentiable function of x and y, then dz dt = f (x(t), y(t))dx x dt + f (x(t), y(t))dy y dt
Multi-variable Chain Rule Proof (1 of 2) Define g(t) = z = f (x(t), y(t)), then g (t) = g(t + t) g(t) lim t 0 t = z lim t 0 t f f x x + y y + ɛ 1 x + ɛ 2 y = lim t 0 [ f = lim t 0 x = f dx x dt + f y x t + f y dy dt t y t + ɛ 1 + ( lim ɛ 1) dx t 0 dt + ( lim ɛ 2) dy t 0 dt x t + ɛ 2 ] y t
Multi-variable Chain Rule Proof (2 of 2) Consider x = x(t + t) x(t). Since x(t) is differentiable, it is also continuous, therefore Likewise lim x = lim (x(t + t) x(t)) = 0. t 0 t 0 lim y = lim (y(t + t) y(t)) = 0, t 0 t 0 which implies Therefore lim ( x, y) = (0, 0). t 0 lim 1 t 0 = lim 1 = 0 ( x, y) (0,0) lim 2 t 0 = lim 2 = 0. ( x, y) (0,0)
Examples (1 of 2) Find dz/dt where z = x 2 + y 2 and x(t) = cos t and y(t) = t 2 + 3t + 1.
Examples (1 of 2) Find dz/dt where z = x 2 + y 2 and x(t) = cos t and y(t) = t 2 + 3t + 1. dz dt = x y ( sin t) + (2t + 3) x 2 + y 2 x 2 + y 2 cos t sin t = cos 2 t + (t 2 + 3t + 1) + (t2 + 3t + 1)(2t + 3) 2 cos 2 t + (t 2 + 3t + 1) 2
Examples (2 of 2) Suppose the output P of a factory depends on the size of the labor force l and the amount of capital k spent on production. If the output is described by P = 25k 1/2 l 3/4 and the time rate of change of the size of the labor force is 50 workers per year and the time rate of change of the capital spent is $90, 000 per year, determine the time rate of change in P.
Examples (2 of 2) Suppose the output P of a factory depends on the size of the labor force l and the amount of capital k spent on production. If the output is described by P = 25k 1/2 l 3/4 and the time rate of change of the size of the labor force is 50 workers per year and the time rate of change of the capital spent is $90, 000 per year, determine the time rate of change in P. dp dt = 25l3/4 dk 2k 1/2 dt + 75k 1/2 4l 1/4 = 1125000l3/4 k 1/2 + dl dt 1875k 1/2 2l 1/4
Generalization of Chain Rule Theorem Suppose that z = f (x, y), where f is a differentiable function of x and y and where x x(s, t) and y y(s, t) both have first-order partial derivatives. Then z s z t = z x x s + z y = z x x t + z y y s y t
x s x x t z x z z y y s s t s t y y t Tree Diagrams.
Examples (1 of 2) Suppose z = e x sin y where x = st 2 and y = s 2 t. Find z s and z t.
Examples (1 of 2) Suppose z = e x sin y where x = st 2 and y = s 2 t. Find z s and z t. z s = z x x s + z y y s = e x (sin y)t 2 + e x (cos y)2st = t 2 e st2 sin(s 2 t) + 2ste st2 cos(s 2 t) z t = z x x t + z y y t = e x (sin y)2st + e x (cos y)s 2 = 2ste st2 sin(s 2 t) + s 2 e st2 cos(s 2 t)
Examples (2 of 2) Suppose w = x 4 y + y 2 z 3 where x = se t, y = s 2 e t, and z = s sin t. Find w s and w t.
Examples (2 of 2) Suppose w = x 4 y + y 2 z 3 where x = se t, y = s 2 e t, and z = s sin t. Find w s and w t. w s = w x x s + w y y s + w z z s = 4x 3 ye t + (x 4 + 2yz 3 )2se t + 3y 2 z 2 sin t = 4(se t ) 3 (s 2 e t )e t + ((se t ) 4 + 2(s 2 e t )(s sin t) 3 )2se t + 3(s 2 e t ) 2 (s sin t) 2 sin t = 6s 5 e 3t + 7s 6 e 2t sin 3 t w t = w x x t + w y y t + w z z t = 4x 3 y( se t ) + (x 4 + 2yz 3 )s 2 e t + 3y 2 z 2 s cos t = 3s 6 e 3t + 3s 7 e 2t cos t sin 2 t + 2s 7 e 2t sin 3 t
Implicit Differentiation (1 of 3) Suppose f (x, y) = C a constant. If we assume that y is implicitly defined by x, then d dx f (x, y(x)) = d dx (C) f (x, y)dx x dx + f (x, y)dy y dx = 0 assuming f y (x, y) 0. dy dx = f x f y (x, y) (x, y)
Implicit Differentiation (2 of 3) Theorem If F(x, y, z) = C (a constant) determines an implicit, differentiable function f depending on x and y such that z = f (x, y) then z x = F x(x, y, z) F z (x, y, z) provided F z (x, y, z) 0. and z y = F y(x, y, z) F z (x, y, z)
Implicit Differentiation (3 of 3) F(x, y, z) = x x (C) F x (x, y, z)(1) + F y (x, y, z)(0) + F z (x, y, z) z = 0 as long as F z (x, y, z) 0. x z x = F x(x, y, z) F z (x, y, z)
y 2 0 2 0 2 x 2 0 z 2 Example (1 of 2) If x 3 + y 3 + z 3 + 6xyz = 1 find z/ x and z/ y.. 2
Example (2 of 2) 1 = F(x, y, z) = x 3 + y 3 + z 3 + 6xyz
Example (2 of 2) z x 1 = F(x, y, z) = x 3 + y 3 + z 3 + 6xyz = F x F z = 3x 2 + 6yz 3z 2 + 6xy
Example (2 of 2) z x z y 1 = F(x, y, z) = x 3 + y 3 + z 3 + 6xyz = F x F z = 3x 2 + 6yz 3z 2 + 6xy = F y F z = 3y 2 + 6xz 3z 2 + 6xy
Higher Order Derivatives If g(u, v) = f (x(u, v), y(u, v)) find g vu.
Higher Order Derivatives If g(u, v) = f (x(u, v), y(u, v)) find g vu. g v = f x x v + f y y v
Higher Order Derivatives If g(u, v) = f (x(u, v), y(u, v)) find g vu. g v = f x x v + f y y v g vu = u (f xx v ) + u (f yy v ) = u (f x) x v + f x u (x v) + u (f y) y v + f y u (y v) = u (f x) x v + f x x vu + u (f y) y v + f y y vu = (f xx x u + f xy y u ) x v + f x x vu + (f yx x u + f yy y u ) y v + f y y vu = f x x vu + f y y vu + f xx x u x v + f xy (x v y u + x u y v ) + f yy y u y v
Homework Read Section 12.5. Exercises: 1 25 odd, 26