3.4 Worksheet: Proof of the Chain Rule NAME

Transcription

1 Mat Workseet: Proof of te Cain Rule NAME Te Cain Rule So far we are able to differentiate all types of functions. For example: polynomials, rational, root, and trigonometric functions. We are even able to compute te derivative of suc functions wen tey are being added, subtracted, multiplied or divided by one anoter. However, tere are still some functions tat we cannot reasonably differentiated witout using te definition of te derivative. For example, k(x) = (x + 1) 100 k(x) = cos(x 3 + 1) k(x) = e 1 ex Te issue wit tese functions is tat tey are a composition of more tan one function. For example, k(x) = (x + l) 100 is te composition of f(x) = x 100 and g(x) = x + 1 so tat k(x) = (f g)(x). Find f(x) and g(x) suc tat k(x) = (f g)(x) for: k(x) = cos(x 3 + 1) k(x) = e 1 ex Te Cain Rule elps us to differentiate tese functions tat are te composition of oter differentiable functions. Te Cain Rule If g is differentiable at x and f is differentiable at g(x), ten te composite function k = f g defined by k(x) = f(g(x)) is differentiable at x and k is given by te product. k (x) = f (g(x)) g (x) Historical Note: Remember James Gregory ( )? He was te dude wo would ave discovered calculus if e would ave just been a little more organized. Anyway, e was also te first person to formulate te Cain Rule! 1

2 Warm-up Before we move to examining te proof of tis fact, let s use it to try to determine some difficult derivatives. For eac of te following, determine k (x): k(x) = (x + 1) 100 k(x) = x 2 + 3x k(x) = cos (x 3 + 1) k(x) = e 1 ex k(x) = sin (sin (sin (x))) k(x) = sin 3 (x). As we see from tis, Te Cain Rule is quite useful, and lets us differentiate functions tat would be very difficult to differentiate using te definition of te derivative! As responsible investigators, toug, we sould make sure tat we not only understand ow to use te tools we are given, but also tat we know wy tey work te way tey do. In te remainder of tis workseet we tus turn to examining te proof of Te Cain Rule. 2

3 Assumptions and Definitions Suppose k(x) = f g(x). Our goal is to determine, for any a in te range of g, wat te value of k (a) is. Fill in te blanks in te following sentence: Since k(a) = f(g(a)), te natural assumptions to make are tat g is differentiable at f is differentiable at. and We now build te fancy tool tat we will need to use in te proof: Define g(a + ) g(a) g ε g () := (a) wen 0 0 wen = 0 f(g(a) + ) f(g(a)) f ε f () := (g(a)) wen 0 0 wen = 0 Tere is a lot to digest in te above equations, so take a second to assess wat tey say. Togeter, we will now sow tat ε g is continuous at zero. proof: (i) Recall tat, by te definition of continuity at a point, ε g is continuous at 0 if and only if ε g () = ε g (0). (ii) Notice tat ε g (0) = 0 by te definition of ε g, so ε g is continuous at 0 if and only if. (iii) Recall tat for functions a(x) and b(x), if a(x) = b(x) for all x c, ten x c a(x) = x c b(x). (iv) Notice tat, by te definition of ε g, for 0, ε g () = [ ] g(a + ) g(a) ε g() = g (a). 0 g(a + ) g(a) g (a). Tus, we see tat (v) Finally, (using te Laws of Limits) notice tat [ ] g(a + ) g(a) g (a) =. ( ) g(a + ) g(a) But, = g (a) since g is differentiable at a, so g(a + ) g(a) g (a) = =. (vi) Combining (iv) and (v) we see x 0 ε g () = 0. So, by (ii), we see ε g is continuous at 0. 3

4 A very similar argument can be given to establis tat ε f is continuous at zero. Take a few minutes to see if you can piece tis argument togeter using te previous argument as an example. Te next tings to notice about ε g and ε f are te equalities ε g () = g(a + ) g(a) g (a) (1) ε f () = f(g(a) + ) f(g(a)) f (g(a)) (2) To establis tese equalities, one must ceck tat tey old bot wen 0 and wen = 0. Take a moment to establis tem yourself. We can rearrange tese equalities to get te following: g(a + ) = (g (a) + ε g ()) + g(a) (3) f(g(a) + ) f(g(a)) = (f (g(a)) + ε f ()) (4) Take a moment to verify tat equation (3) is just a rearrangement of equation (1), and equation (4) is just a rearrangement of equation (2). Te expression (g (a) + ε g ()) will sow up a lot in te following calculation. For brevity, ten, we put d := (g (a) + ε g ()), and restate equation (3) as g(a + ) = g(a) + d (5) 4

5 Facts Establising a few facts now will allow us to sorten te proof of Te Cain Rule we are about to examine. Te first of tese is: d = g (a). (6) Take a minute to establis tis equality. You will need to use four facts: (i) Te definition of d, (ii) te it law involving a sum of functions, (iii) te continuity of ε g, and (iv) te definition of ε g. Look up any of tese tat you do not know and ten establis te equality. proof: Te next fact to establis is: ε f (d) = 0. (7) proof: To establis tis, you will use te fact you just establised in conjunction wit te particular it law tat deals wit composite functions (p. 119 of Stewart), and te it law dealing wit a product of functions. If necessary, look tese it laws up, ten establis te equality. 5

6 Te Proof Given te assumptions and definitions above, we now compute te derivative of k at a. You will find below a classic two column proof. In tis two column proof, te left column contains a series of equalities, wile te rigt column contains justifications for eac equality. Some equalities and some justifications ave been left out. Please supply te missing components. k (a) = k(a + ) k(a) k(a + ) k(a) f(g(a + )) f(g(a)) = From te definition of te derivative of k at a. From te definition of k. f(g(a + )) f(g(a)) f(g(a) + d) f(g(a)) = From equation (4) Since, for 0, d(f (g(a)) + ε f (d)) d(f (g(a)) + ε f (d)) = d(f (g(a)) + ε f (d)) = d (f (g(a)) + ε f (d)) 0 Since te it of a product is te product of te its. d (f (g(a)) + ε f (d)) = g (a)((f (g(a)) + ε f (d))) 0 g (a)( (f (g(a)) + ε f (d))) = g (a) f (g(a)) By following troug te above cain of equalities. 6