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
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 2 + 3 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!
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)).
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.
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 )
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)
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!
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 2 + 4. So we get 1/t 2 + 4) 3) = 3 t 2 + 4) 4 2t)
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 3 + 18t) Homework 23 23.1 Use an ɛ δ argument to sow 7f x) + 2gx) is ifferentiable at any x were f an g are ifferentiable. 23.2 Use an ɛ N argument to sow 19a n 24b n converges if a n) an b n) converge. 23.3 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?