3.3 The Chain Rule CHAPTER 3. RULES FOR DERIVATIVES 63

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1 CHAPTER 3. RULES FOR DERIVATIVES The Chain Rule Comments. The Chain Rule is the most interesting one we have learne so far, an it is one of the most complicate we will learn. It is one of the rules we will learn for taking the erivative of a combination of a functions. In particular, this rule will tell us how to take the erivative when one function is insie of another. Example. (Base on Hughes-Hallett, 5e, 3.3 Example ) Let G be the amount of gas, in gallons, consume by a car on a particular trip, let s be the istance travele, in miles, an let t be the amount of time that has elapse, in hours. Then G is a function of s an s is function of t. We can combine these statements an conclue that G is a function of t. Let 0.05 gallons of gas be consume for each mile travele, an suppose that the car is traveling at 30 mi /hr. How fast is gas being consume? Calculate your answer in Leibniz notation, an give units. Solution. The information we are given can be summarize this way G s =0.05 gal /mi, s = 30 mi /hr, an we want to fin this: G =? If we give these numbers an units the usual interpretation, we have this: We use 0.05 gallons by riving one mile, We rive 30 miles in one hour, How many gallons will we use in one hour? The right way to answer this question is to combine the given information with multiplication: gallons in one hour = 30 miles in one hour 0.05gallons in one mile 30 mi /hr 0.5 gal /mi =.5 gal /hr. We can summarize this calculation in Leibniz notation this way: G = G s s. Rule (Chain Rule in Leibniz Notation). If y is a function of z an z is a function of t then y = y z z. This is where we ene on Monay, February 25 Example 2. Break each of the following complicate functions into a combination of two simple functions. In each case write the result as a function of z, wherez is the insie function.

2 CHAPTER 3. RULES FOR DERIVATIVES 64 y = p t 2 +2t (b) y = 5(2t + 7) 8 (c) y = t 2 + () y = 7.2e t2 (e) y = 2 ln(3t2 + 5) Solution. (b) (c) () (e) y = p z z = t 2 +2t y =5z 8 z =2t +7 y = z z = t 2 + y = z = t 2 7.2e z y = 2 ln(z) z =3t 2 +5 Example 3. Fin the erivatives of the following functions. Use z for the insie function, an use the Leibniz notation for the chain rule. y = 5( 3t 2 +2t + 7). (b) y = 7 3 ln(t3 + t). Solution. (b) y =5z, z = 3t 2 +2t +7 y = y z z =5 z 0 ( 6t + 2) = 55( 3t 2 +2t + 7) 0 ( 6t + 2) y = 7 3 ln(z), z = t3 + t y = y z z = 7 3 z (3t2 + ) = 7 3 t 3 + t (3t2 + )

3 CHAPTER 3. RULES FOR DERIVATIVES 65 Comments. To some egree, a lot of people learn the chain rule one function at a time, as it applies to the outsie function. In other wors, they learn how the chain rule works on a case-by-case basis. There s nothing wrong with this (as long as the stuent can also use the chain rule in a new or general situation that oesn t fit any of the case-by-cases). In any case, here s how the chain rule looks for various functions that we know. Rule. The following table shows the erivative of a variety of basic functions, an then a chain rule version for each basic function. Usual version Chain rule version x x = x 2 x p x = x 2 p x x 2 = 2 2 p 2 = 2 p 2 x2 x2 x xn = nx n x2 n = n 2 n x2 x ex = e x x ln(x) = x Example 4. Fin the following erivatives (3t2 7t) p (b).9t 4.3 (c) 0t 4 () (e) e.03t ln(2t + et ) x e2 = e 2 x2 x ln ( 2)= 2 x2 Solution. In this problem I m not going to use the z notation, but you on t have to follow me. What I mean is: if you are at all confuse by the steps that take place, go ahea an use z an write out the extra steps where you can clearly label an see what happene. But, with practice people can often keep track of some stu in their hea. Below I o each problem twice, once just showing with a box where the insie stu is in each function, an basically using the box notation for the chain rule that I gave above. 3t 2 7t = 3t 2 0 7t 3t2 = (3t 2 7t) 0 (6t 7) 7t

4 CHAPTER 3. RULES FOR DERIVATIVES 66 (b) (c) () p.9t 4.3 = 2 (.9t 4.3 ) /2.9t 4.3 = 2 (.9t 4.3) /2 (.9) 0t 4 = ( 0t 4 ) 2 0t 4 = (0t 4) 2 (0) (e) e.03t = e.03t.03t ln 2t + e t = = e.03t (.03) = 2t + e 2t + t et 2t + e t (2 + et ) It may be worth mentioning that as people practice the chain rule, at least for simple examples, rather than using y an z, or even using boxes an filling them in, they ten to just focus their attention on part of the formula at a time, an say the right wors in their hea to prompt them to o the right thing at each step. Roughly, they things like take the erivative of the outsie an make sure they are looking at the outsie function while they o this. Then they say on t the insie while they write the insie again, without changing it. Then they say multiply by the erivative of the insie an o that. The results can be color coe, as shown below. 3t 2 7t = (3t 2 7t) 0 (6t 7) eriv. of outsie on t insie eriv. of insie p.9t 4.3 = 2 (.9t 4.3) /2 (.9) eriv. of outsie on t insie eriv. of insie

5 CHAPTER 3. RULES FOR DERIVATIVES 67 0t 4 = (0t 4) 2 (0) eriv. of outsie on t insie eriv. of insie e.03t = e (.03t ) (.03) eriv. of outsie on t insie eriv. of insie ln 2t + et = (2t + e t ) (2 + et ) eriv. of outsie on t insie eriv. of insie

6 CHAPTER 3. RULES FOR DERIVATIVES 68 This is where we ene on Wenesay, February 27.