Function Composition and s James K. Peterson Department of Biological Sciences and Department of Matematical Sciences Clemson University Marc 8, 2017
Outline 1 Function Composition and Continuity 2
Function Composition and Continuity Te composition of functions is actually a simple concept. You sove one function into anoter and calculate te result.
Function Composition and Continuity Te composition of functions is actually a simple concept. You sove one function into anoter and calculate te result. We know ow to find x 2 and u 2 for any x and u. So wat about (x 2 + 3) 2? Tis just means take u = x 2 + 3 and square it.
Function Composition and Continuity Te composition of functions is actually a simple concept. You sove one function into anoter and calculate te result. We know ow to find x 2 and u 2 for any x and u. So wat about (x 2 + 3) 2? Tis just means take u = x 2 + 3 and square it. Tat is if f (u) = u 2 and g(x) = x 2 + 3, te composition of f and g is simply f (g(x)). Let s talk about continuity first and ten we ll go on to te idea of taking te derivative of a composition. If f and g are bot continuous, ten if you tink about it, anoter way to prase te continuity of f is tat lim y x f (y) = f (x) = f (x) = f ( lim y x y).
Function Composition and Continuity Te composition of functions is actually a simple concept. You sove one function into anoter and calculate te result. We know ow to find x 2 and u 2 for any x and u. So wat about (x 2 + 3) 2? Tis just means take u = x 2 + 3 and square it. Tat is if f (u) = u 2 and g(x) = x 2 + 3, te composition of f and g is simply f (g(x)). Let s talk about continuity first and ten we ll go on to te idea of taking te derivative of a composition. If f and g are bot continuous, ten if you tink about it, anoter way to prase te continuity of f is tat lim y x f (y) = f (x) = f (x) = f ( lim y x y). So if f and g are bot continuous, we can say lim y x f (g(y)) = f ( lim g(y)) = f (g( lim y)) = f (g(x)). y x y x
Function Composition and Continuity Te composition of functions is actually a simple concept. You sove one function into anoter and calculate te result. We know ow to find x 2 and u 2 for any x and u. So wat about (x 2 + 3) 2? Tis just means take u = x 2 + 3 and square it. Tat is if f (u) = u 2 and g(x) = x 2 + 3, te composition of f and g is simply f (g(x)). Let s talk about continuity first and ten we ll go on to te idea of taking te derivative of a composition. If f and g are bot continuous, ten if you tink about it, anoter way to prase te continuity of f is tat lim y x f (y) = f (x) = f (x) = f ( lim y x y). So if f and g are bot continuous, we can say lim y x f (g(y)) = f ( lim g(y)) = f (g( lim y)) = f (g(x)). y x 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!
Function Composition and Continuity Let s do tis using te ɛ δ approac. We assume f is locally defined at te point g(x) and g is locally defined at te point x. So tere are radii r f and r g so tat f (w) is defined if w B rf (g(x)) and g(y) is defined if y B rg (x). For any ɛ, tere is a δ 1 so tat f (u) f (g(x)) < ɛ if u g(x) < δ 1 because f is continuous at x. Of course, δ 1 < r f. For te tolerance δ 1, tere is a δ 2 so tat g(y) g(x)) < δ 1 if y x < δ 2 because f is continuous at x. Of course, δ 2 < r g. Tus, y x < δ 2 implies g(y) g(x)) < δ 1 wic implies f (g(y)) f (g(x)) < ɛ. Tis sows f g is continuous at x. You sould be able to understand bot types of arguments ere!
Te next question is weter or not te composition of functions aving derivatives gives a new function wic as a derivative. And ow could we calculate tis derivative if te answer is yes? It turns out tis is true but to see if requires a bit more work wit limits.
Te next question is weter or not te composition of functions aving derivatives gives a new function wic as a derivative. And ow could we calculate tis derivative if te answer is yes? It turns out tis is true but to see if requires a bit more work wit limits. So we want to know wat d dx (f (g(x))) is. First, if g was always constant, te answer is easy. It is d dx (f (constant)) = 0 wic is a special case of te formula we are going to develop.
Te next question is weter or not te composition of functions aving derivatives gives a new function wic as a derivative. And ow could we calculate tis derivative if te answer is yes? It turns out tis is true but to see if requires a bit more work wit limits. So we want to know wat d dx (f (g(x))) is. First, if g was always constant, te answer is easy. It is d dx (f (constant)) = 0 wic is a special case of te formula we are going to develop. So let s assume g is not constant locally. So tere is a radius r > 0 so tat g(y) g(x) for all y B r (x). Anoter way of saying tis is g(x + ) g(x) 0 if < r. Ten, we want to calculate d (f (g(x))) = lim dx 0 f (g(x + )) f (g(x)). Rewrite by dividing and multiplying by g(x + ) g(x) wic is ok to do as we assume g is not constant locally and so we don t divide by 0. We get ( f (g(x))) = lim 0 f (g(x + )) f (g(x)) g(x + ) g(x) g(x + ) g(x)
g(x+) g(x) Now lim 0 = g (x) because we know g is differentiable at x.
g(x+) g(x) Now lim 0 = g (x) because we know g is differentiable at x. Now let u = g(x) and = g(x + ) g(x). Ten, g(x + ) = g(x) + = u +. Ten, we ave f (g(x+)) f (g(x)) g(x+) g(x) = f (u+) f (u). Since g is continuous because g as a derivative, as lim 0 = lim 0 (g(x + ) g(x)) = 0. Also, since f( is differentiable at ) u = g(x), we know lim 0 f (u + ) f (u) / = f (u) = f (g(x)).
g(x+) g(x) Now lim 0 = g (x) because we know g is differentiable at x. Now let u = g(x) and = g(x + ) g(x). Ten, g(x + ) = g(x) + = u +. Ten, we ave f (g(x+)) f (g(x)) g(x+) g(x) = f (u+) f (u). Since g is continuous because g as a derivative, as lim 0 = lim 0 (g(x + ) g(x)) = 0. Also, since f( is differentiable at ) u = g(x), we know lim 0 f (u + ) f (u) / = f (u) = f (g(x)). Tus, lim 0 = lim 0 f (g(x + )) f (g(x)) g(x + ) g(x) f (u + ) f (u) = lim 0 f (u + ) f (u) = f (u) = f (g(x)).
Since bot limits above exist, we know ( f (g(x + )) f (g(x)) lim 0 g(x + ) g(x) ( lim 0 f (g(x + )) f (g(x)) g(x + ) g(x) Tis result is called te. ) g(x + ) g(x) = ) ( g(x + ) g(x) lim 0 ) = f (g(x) g (x). Teorem For Differentiation If te composition of f and g is locally defined at a number x and if bot f (g(x)) and g (x) exist, ten te derivative of te composition of f and g also exists and is given by d dx (f (g(x)) = f (g(x)) g (x) Proof We reasoned tis out above. We usually tink about tis as follows d dx (f (inside)) = f (inside) inside (x).
Let s do a error term based proof. Teorem For Differentiation If te composition of f and g is locally defined at a number x and if bot f (g(x)) and g (x) exist, ten te derivative of te composition of f and g also exists and is given by d dx (f (g(x)) = f (g(x)) g (x) Proof Again, we do te case were g(x) is not a constant locally. Like before, let = g(x + ) g(x) wit u = g(x). Using error terms, since f as a derivative at u = g(x) and g as a derivative at x, we can write f (u + ) = f (u) + f (u) + E f (u +, u) were lim 0 of bot E f (u +, u) and E f (u +, u)/ are zero and g(x + ) = g(x) + g (x) + E g (x +, x) were lim 0 of bot E g (x +, x) and E g (x +, x)/ are zero.
Proof Ten, since u = g(x) and = g(x + ) g(x), we ave d (f (g(x))) = lim dx 0 f (g(x + )) f (g(x)) f (u + ) f (u) = lim 0 f (u) + E f (u +, u) = lim 0 ( f (u) + E f (u +, u) = lim 0 ) were we know 0 locally because g is not constant locally.
Proof We see lim 0 derivative at x. = lim 0 g(x+) g(x) Next, lim 0 f (u) +E f (u+,u) = g (x) because g as a = f E (u) + lim f (u+,u) 0. But E f (u +, u) lim 0 E f (u +, u) = lim = 0. 0 f Tus, lim (u) +E f (u+,u) 0 ( d (f (g(x))) = dx = f (u). Since bot limits exist, we ave lim (f (u) + E f (u +, u) 0 = f (g(x)) g (x). ) ( lim 0 )
Let s do an ɛ δ based proof. Teorem For Differentiation If te composition of f and g is locally defined at a number x and if bot f (g(x)) and g (x) exist, ten te derivative of te composition of f and g also exists and is given by d dx (f (g(x)) = f (g(x)) g (x) Proof Again, we do te case were g(x) is not a constant locally. Like before, let = g(x + ) g(x) wit u = g(x). Since g is not constant locally at x, tere is a radius r g so tat g(x ) g(x) 0 wen < r g. Again we write d (f (g(x))) = lim dx 0 f (g(x + )) f (g(x)) g(x + ) g(x) g(x + ) g(x)
Proof Ten, since u = g(x) and = g(x + ) g(x), we ave f (g(x + )) f (g(x)) g(x + ) g(x) g(x + ) g(x) = f (u + ) f (u). Since g is differentiable at x, for a given ξ 1, tere is a δ 1 so tat < δ 1 implies g (x) ξ 1 < g(x+) g(x) = < g (x) + ξ 1. And since f is differentiable at u = g(x), for a given ξ 2, tere is a δ 2 so tat < δ 2 implies f f (u+) f (u) (u) ξ 2 < < f (u) + ξ 2. f (g(x+)) f (g(x)) g(x+) g(x) g(x+) g(x) Let for convenience. Ten, if δ < min{r g, δ 1, δ 2 }, all conditions old and we ave = f (f (u) ξ 2 ) (g (x) ξ 1 ) < f < (f (u) + ξ 2 ) (g (x) + ξ 1 )
Proof Multiplying out tese terms and canceling te ξ 1 ξ 2, we ave or f (u)g (x) ξ 1 f (u) ξ 2 g (x) < f < f (u)g (x) + ξ 1 f (u) + ξ 2 g (x) ξ 1 f (u) ξ 2 g (x) < f Tus, ( f (u)g (x) ) < ξ 1 f (u) + ξ 2 g (x) f ( f (u)g (x) ) < ξ 1 f (u) + ξ 2 g (x)
Proof Coose ξ 1 = ɛ 2( f (g(x)) +1) and ξ 2 = Ten, we ave < δ implies f Tis proves te result! ɛ 2( g (x) +1). ( f (u)g (x) ) < ɛ. Comment You sould know ow to attack tis proof all tree ways!
Example Find te derivative of (t 3 + 4) 3. Solution It is easy to do 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 Find te derivative of 1/(t 2 + 4) 3. Solution Tis is also ( (ting) power ) = power (ting) power - 1 (ting) were power is 3 and ting is t 2 + 4. So we get ( 1/(t 2 + 4) 3) = 3 (t 2 + 4) 4 (2t)
Example Find te derivative 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) + 2g(x) is differentiable at any x were f and g are differentiable. 23.2 Use an ɛ N argument to sow 19a n 24b n converges if (a n ) and (b n ) converge. 23.3 Recall in te error form for differentiation, f (x + ) = f (x) + f (x) + E(x +, x) were te linear function T (x) = f (x) + f (x) is called te tangent line approximation to f at te base point x. For f (x) = 2x 3 + 3x + 2, sketc f (x) and T (x) on te same grap carefully at te points x = 1, x = 0.5 and x = 1.3. f (x+) f (x) Also draw a sample slope triangle for eac base point. Finally, draw 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 lim n f (x n ) = f (x) for any sequence (x n ) wit x n x?