Function Composition and Chain Rules

Transcription

1 Function Composition an Cain Rules James K. Peterson Department of Biological Sciences an Department of Matematical Sciences Clemson University November 2, 2018 Outline Function Composition an Continuity Cain Rule

2 Te composition of functions is actually a simple concept. You sove one function into anoter an calculate te result. We know ow to fin x 2 an u 2 for any x an u. So wat about x 2 + 3) 2? Tis just means take u = x an square it. Tat is if f u) = u 2 an gx) = x 2 + 3, te composition of f an g is simply f gx)). Let s talk about continuity first an ten we ll go on to te iea of taking te erivative of a composition. If f an g are bot continuous, ten if you tink about it, anoter way to prase te continuity of f is tat y x f y) = f x) = f x) = f y). y x So if f an g are bot continuous, we can say y x f gy)) = f y x gy)) = f g y)) = f gx)). y x So te composition of continuous functions is continuous. Heave a big sig of relief as smootness as not been lost by pusing one smoot function into anoter smoot function! Let s o tis using te ɛ δ approac. We assume f is locally efine at te point gx) an g is locally efine at te point x. So tere are raii rf an rg so tat f w) is efine if w gx)) an gy) is efine if Brf y Brg x). For any ɛ, tere is a δ1 so tat f u) f gx)) < ɛ if u gx) < δ1 because f is continuous at x. Of course, δ1 < rf. For te tolerance δ1, tere is a δ2 so tat gy) gx)) < δ1 if y x < δ2 because f is continuous at x. Of course, δ2 < rg. Tus, y x < δ2 implies gy) gx)) < δ1 wic implies f gy)) f gx)) < ɛ. Tis sows f g is continuous at x. You soul be able to unerstan bot types of arguments ere!

3 Te next question is weter or not te composition of functions aving erivatives gives a new function wic as a erivative. An ow coul we calculate tis erivative if te answer is yes? It turns out tis is true but to see if requires a bit more work wit its. So we want to know wat x f gx))) is. First, if g was always constant, te answer is easy. It is x f constant)) = 0 wic is a special case of te formula we are going to evelop. So let s assume g is not constant locally. So tere is a raius r > 0 so tat gy) gx) for all y Br x). Anoter way of saying tis is gx + ) gx) 0 if < r. Ten, we want to calculate f gx))) = x 0 f gx + )) f gx)). Rewrite by iviing an multiplying by gx + ) gx) wic is ok to o as we assume g is not constant locally an so we on t ivie by 0. We get f gx + )) f gx)) gx + ) gx) f gx))) = 0 gx + ) gx) Now 0 gx+) gx) = g x) because we know g is ifferentiable at x. Now let u = gx) an = gx + ) gx). Ten, gx + ) = gx) + = u +. Ten, we ave f gx+)) f gx)) f u+) f u) gx+) gx) =. Since g is continuous because g as a erivative, as 0 = 0gx + ) gx)) = 0. Also, since f is ifferentiable at ) u = gx), we know 0 f u + ) f u) / = f u) = f gx)). Tus, 0 = 0 f gx + )) f gx)) f u + ) f u) = gx + ) gx) 0 f u + ) f u) = f u) = f gx)).

4 Since bot its above exist, we know f gx + )) f gx)) 0 gx + ) gx) 0 f gx + )) f gx)) gx + ) gx) Tis result is calle te Cain Rule. gx + ) gx) ) 0 ) = gx + ) gx) ) = f gx)) g x). Teorem Cain Rule For Differentiation If te composition of f an g is locally efine at a number x an if bot f gx)) an g x) exist, ten te erivative of te composition of f an g also exists an is given by x f gx)) = f gx)) g x) We reasone tis out above. We usually tink about tis as follows x f insie)) = f insie) insie x). Let s o a error term base proof. Teorem Cain Rule For Differentiation If te composition of f an g is locally efine at a number x an if bot f gx)) an g x) exist, ten te erivative of te composition of f an g also exists an is given by x f gx)) = f gx)) g x) Again, we o te case were gx) is not a constant locally. Like before, let = gx + ) gx) wit u = gx). Using error terms, since f as a erivative at u = gx) an g as a erivative at x, we can write f u + ) = f u) + f u) + Ef u +, u) were 0 of bot Ef u +, u) an Ef u +, u)/ are zero an gx + ) = gx) + g x) + Eg x +, x) were 0 of bot Eg x +, x) an Eg x +, x)/ are zero.

5 Ten, since u = gx) an = gx + ) gx), we ave f gx + )) f gx)) f gx))) = x 0 f u + ) f u) = 0 f u) + Ef u +, u) = 0 = 0 f u) + Ef u +, u) ) were we know 0 locally because g is not constant locally. We see 0 erivative at x. gx+) gx) = 0 = g x) because g as a Next, 0 f u) +Ef u+,u) = f Ef u+,u) u) + 0. But Ef u +, u) 0 = 0 Ef u +, u) = 0. Tus, 0 f u) +Ef u+,u) = f u). Since bot its exist, we ave f gx))) = x 0 = f gx)) g x). ) f u) + Ef u +, u) 0 )

6 Let s o an ɛ δ base proof. Teorem Cain Rule For Differentiation If te composition of f an g is locally efine at a number x an if bot f gx)) an g x) exist, ten te erivative of te composition of f an g also exists an is given by x f gx)) = f gx)) g x) Again, we o te case were gx) is not a constant locally. Like before, let = gx + ) gx) wit u = gx). Since g is not constant locally at x, tere is a raius rg so tat gx) gx) 0 wen < rg. Again we write f gx + )) f gx)) gx + ) gx) f gx))) = x 0 gx + ) gx) Ten, since u = gx) an = gx + ) gx), we ave f gx + )) f gx)) gx + ) gx) gx + ) gx) = f u + ) f u). Since g is ifferentiable at x, for a given ξ1, tere is a δ1 so tat < δ1 implies g x) ξ1 < gx+) gx) = < g x) + ξ1. An since f is ifferentiable at u = gx), for a given ξ2, tere is a δ2 so tat < δ2 implies f f u+) f u) u) ξ2 < < f u) + ξ2. f gx+)) f gx)) gx+) gx) Let gx+) gx) = f for convenience. Ten, if δ < min{rg, δ1, δ2}, all conitions ol an we ave f u) ξ2) g x) ξ1) < f < f u) + ξ2) g x) + ξ1)

7 Multiplying out tese terms an cancelling te ξ1ξ2, we ave or f u)g x) ξ1f u) ξ2g x) < f < f u)g x) + ξ1f u) + ξ2g x) ξ1f u) ξ2g x) < f Tus, f u)g x) ) < ξ1f u) + ξ2g x) f f u)g x) ) < ξ1 f u) + ξ2 g x) Coose ξ1 = ɛ 2 f gx)) +1) Ten, we ave < δ implies f Tis proves te result! ɛ an ξ2 = 2 g x) +1). f u)g x) ) < ɛ. Comment You soul know ow to attack tis proof all tree ways!

8 Example Fin te erivative of t 3 + 4) 3. Solution It is easy to o tis if we tink about it tis way. ting) power ) = power ting) power - 1 ting) Tus, t 3 + 4) 3) = 3 t 3 + 4) 2 3t 2 ) Example Fin te erivative of 1/t 2 + 4) 3. Solution Tis is also ting) power ) = power ting) power - 1 ting) were power is 3 an ting is t So we get 1/t 2 + 4) 3) = 3 t 2 + 4) 4 2t)

9 Example Fin te erivative of 6t 4 + 9t 2 + 8) 6. Solution 6t 4 + 9t 2 + 8) 6 ) = 6 6t 4 + 9t 2 + 8) 5 24t t) Homework Use an ɛ δ argument to sow 7f x) + 2gx) is ifferentiable at any x were f an g are ifferentiable Use an ɛ N argument to sow 19a n 24b n converges if a n) an b n) converge Recall in te error form for ifferentiation, f x + ) = f x) + f x) + Ex +, x) were te linear function T x) = f x) + f x) is calle te tangent line approximation to f at te base point x. For f x) = 2x 3 + 3x + 2, sketc f x) an T x) on te same grap carefully at te points x = 1, x = 0.5 an x = 1.3. f x+) f x) Also raw a sample slope triangle for eac base point. Finally, raw in te error function as a vertical line at te base points. Use multiple colors!! 23.4 If f is continuous at x, wy is it true tat n f x n) = f x) for any sequence x n) wit x n x?