4.4. CHAIN RULE I 351

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1 4.4. CHAIN RULE I Chain Rule I The chain rule is perhaps the most important of the ifferentiation rules. It is immensely rich in application, an very elegantly state when notation is chosen wisely. In this section we will look closely at the rule itself, an the unerlying intuition of the rule. The mechanics of applying the rule are of utmost importance, but as with other calculus principles, unerstaning the intuition ais in etermining when an how to apply the chain rule. Because of the importance an theoretical richness of the chain rule, the reaer is encourage to revisit this section from time to time, to reinforce this most important topic. In its simplest form, the chain rule ictates how we must calculate erivatives of compositions of functions, i.e., functions of the form h(x) f(g(x)), especially when we know how to calculate f an g. With it we will be able to calculate erivatives for a much wier class of functions, fin slopes on implicit curves, an to fin so-calle relate rates relationships among variables. We will first state the chain rule using our prime notation first introuce by English mathematician Brook Taylor (685 73) an then re-write it in terms of the more powerful an explanatory Leibniz notation. (Taylor s name appears prominently in Chapter, which bears his name.) 4.4. Chain Rule in Prime Notation Theorem 4.4. (Chain Rule) Suppose that h(x) f(g(x)), where g (a) exists an f (g(a)) exists. Then h (a) [f (g(a))]g (a). (4.28) Another way of writing this in Taylor s prime notation is (f(g(x))) f (g(x))g (x). (4.29) Note that there is an outer function, namely f, an an inner function g. So the chain rule is sometimes state that we compute the erivative of the outer function with respect to the inner function, that is f (g(x)), an then multiply by the erivative of the inner function, i.e., by g (x). That is a common, colloquial way of expressing the chain rule. It shoul be mentione that multiplying by the erivative of the inner function is the step that is most commonly forgotten in such erivative problems. Example 4.4. Compute h (x) if h(x) (x 2 + 3x) 2. Solution: Besies the unwiely limit efinition, thus far there are two possible methos for computing f (x) here: Expan the function first, an then compute the erivative, as we woul nee to o if we ha little more than the power rule to rely upon: h(x) x 4 + 6x 3 + 9x 2 h (x) 4x 3 + 8x 2 + 8x. Use the chain rule. Here the outer function is f(x) x 2 (squaring the input), an the inner function is g(x) x 2 + 3x. Note that f (x) 2x while g (x) 2x + 3. Using the chain rule we woul have h (x) [f (g(x))]g (x) [ 2(g(x)) ] g (x) [ 2(x 2 + 3x) ] (2x + 3). Note that this gives h (x) 2(2x 3 + 9x 2 + 9x) 4x 3 + 8x 2 + 8x as before.

2 352 CHAPTER 4. THE DERIVATIVE In Chapter 2 we mae some use of a visual evice that utilize empty parentheses to illustrate the actions of functions. For the above erivative, we can look at the following: outer function is ( ) 2 its erivative is 2( ) inner function is ( ) 2 + 3( ) its erivative is 2( ) + 3 The chain rule for this problem coul then rea: [ ((x) 2 + 3(x)) 2] 2((x) 2 + 3(x)) [2(x) + 3] 2(x 3 + 3x) (2x + 3). As we will see, there are simpler ways of looking at the chain rule than labeling an outer function f(x) an an inner function g(x), calculating f an g, an evaluating at g(x) an x, respectively. Still, there are avantages over expaning the function first. For instance, expansion might not be so easy. Moreover the final answer using the chain rule is somewhat factore which helps in, say, constructing its sign chart. The next example will emonstrate some of these avantages more ramatically. Example Fin h (x) if h(x) (x x + 9) 55. Solution: Certainly we o not want to multiply this out to use earlier rules. Instea we just notice that this is a composition of two functions, with the outer function being f(x) x 55 an the inner function being g(x) x x+9. Now f (x) 55x 54, while g (x) 3x Thus h (x) [f (g(x)]g (x) 55(g(x)) 54 g (x) 55(x x + 9) 54 (3x ). Even if we ha somehow expane the original, 65-egree polynomial first, an then calculate the erivative, it is unlikely we woul have notice that our resulting 64-egree polynomial answer factors so nicely. The chain rule has much to say about erivatives of functions which contain trigonometric functions in their structures. Below are two examples where the outer function an inner function are squaring an sine functions, respectively an then vice-versa. Example Fin h (x) if h(x) sin 2 x. Solution: Note that h(x) (sinx) 2, so the outer function is f(x) x 2, while the inner function is sinx. Next note that f (x) 2x an g (x) cosx. Thus h (x) [f (g(x))]g (x) [2(g(x))]g (x) 2 sinx cosx, (sin x)2 2(sinx) cosx. or alternatively, Example Suppose instea h(x) sinx 2. Here f(x) sin x, g(x) x 2, f (x) cosx, g (x) 2x. Hence h (x) [f (g(x))]g (x) cosg(x) g (x) cosx 2 2x 2xcosx 2, sin(x2 ) cos(x 2 ) 2x 2xcosx 2. or alternatively, (It is customary to write the polynomial factor before the trigonometric function factor in the final answer, so it is clearer what terms are insie the trigonometric function, an which are multiplying the trigonometric function.)

3 4.4. CHAIN RULE I 353 Note how it is crucial to ientify the outer function an the inner function. It is also important that the inner function g(x) is inputte into the erivative f of the outer function. While we ientifie the outer an inner functions by name, in fact naming an f(x) an g(x) is unwiely an unnecessary. Before long (though usually not immeiately) the pattern of these computations oes become natural enough. (Observe in particular the secon methos in each of the two examples given above.) To show that the chain rule makes sense from the limit-efinition stanpoint as a erivative rule, we next offer a partial proof of the chain rule. Note how the way the limit is re-written reflects the ultimate statement of the chain rule. It also reflects much of the intuition of the rule Partial Proof of Chain Rule We will not completely prove the chain rule in this context because of some technical ifficulties which arise when proving the rule in its most general form. However, we will look at a proof in perhaps the most common case, which is the case that g(x + x) g(x) 0 properly (so that we also have ( δ > 0)[0 < x < δ g(x + x) g(x 0]. In such a case we can safely ivie an multiply by g(x + x) g(x) in the limit efinition of the erivative to get f(g(x + x)) f(g(x)) [f(g(x))] lim x 0 x lim x 0 f(g(x + x)) f(g(x)) g(x + x) g(x) } {{ } (I) g(x + x) g(x) } x {{} (II) Now we claim that this limit is f (g(x))g (x). The secon term (II) clearly has limit g (x), by efinition. Uner the assumption that g(x + x) g(x) 0 properly as x 0, we can substitute g(x) g(x + x) g(x) 0, an rewrite the limit (I) 38 f(g(x) + g(x)) f(g(x)) lim f (g(x)). g(x) 0 g(x) Note that the computation above is also correct even if g(x) 0 + or g(x) 0 properly, because the existence of the two-sie limit represente by f (g(x)) is assume to exist in our statement of the chain rule theorem. Thus f(g(x)) (I) (II) f (g(x)) g (x), q.e Leibniz Notation an the Chain Rule Before we see how the chain rule is state with Leibniz notation, first we will make some observations about that notation. For example, the following three formulas say the same thing how the square of a quantity changes with respect to the quantity albeit with ifferent variables: 2 2x, t 2 t 2t, u 2 u 2u. See Figure 4.7 for a graphical interpretation of this fact. It is important that in each equation the variables matche. (Note, for instance, that u 2 / 2u, as we shall soon see, the problem 38 When we rewrite f(g(x + x)) f(g(x) + g(x)), we were justifie because g(x + x) (g(x + x) g(x)) + g(x) g(x) + g(x) g(x) + g(x)..

4 354 CHAPTER 4. THE DERIVATIVE x 2 t 2 u x t u Figure 4.7: Three ientical graphs: x 2 versus x, t 2 versus t, an u 2 versus u. All have the same slope though these are ubbe 2 t2 u2 2x, 2t an 2u respectively at each t u fixe horizontal axis value. One such value is represente by a black ot on each of the three graphs. being that the variables involve o not match, so the original power rule, namely xn nxn or equivalently un u nun, oes not apply.) Now suppose we have a ifferentiable function u u(x) an want to take the erivative of (u(x)) 2. The outer function is f(x) x 2 (i.e., squaring what is insie), while the inner function is u u(x). Consier how we fin the erivative of (u(x)) 2 (a chain rule problem) with Taylor s prime notation (4.29), an with Leibniz s notation: Taylor: Leibniz: ( (u(x)) 2 ) 2(u(x)) u (x) (4.30) u 2 u2 u u u 2u. (4.3) The two notations say the same thing, but the Leibniz notation has several avantages, two of which we point out here: Resemblance to algebraic manipulations: it appears that we simply ecompose u2 by iviing an multiplying by u, yieling erivatives that mae sense an coul be calculate by known rules (because the variables matche): the first by the power rule, an the secon by whatever metho coul give us u/. Variable of ifferentiation appears explicitly: we know when we are taking the erivative with respect to x, an when it is with respect to u. Compare (4.3) to our partial proof from the last subsection. Before giving more computational examples, we will look at another argument for the valiity of the chain rule. We begin with a very simple example. Example Suppose we have a vehicle which always achieves a fuel efficiency rating of 35 mile/gallon, an each gallon costs.40. Then we can ask what is the cost per mile. We can think of this situation as total cost C being a function of total gallons consume g, i.e., C C(g) an total gallons as a function of total miles, i.e., g g(m). Ultimately cost is then a (composite) function of miles, i.e., C C(g(m)). Now cost per mile will be cost per gallon

5 4.4. CHAIN RULE I 355 times gallons per mile. In other wors, the rate of change in C with respect to total miles m is $.40 gallon }{{} C g gallon $0.04/mile. (4.32) 35 mile }{{}}{{} g C m m This example is simple because these rates o not change. Still, it is reasonable that even if these rates C/g an g/m only hol for an instant in time, uring that instant C m will still be the prouct C g g m as above. (This is not a proof, but an argument for reasonableness.) For that reason, we can similarly o the following (but where the instant is a particular value of x): (x 3 + ) 2 (x3 + ) 2 (x 3 + ) (x3 + ) 2(x 3 + ) 3x 2. (4.33) That (x 3 + ) 2 /(x 3 + ) 2(x 3 + ) is not much ifferent from the argument in Figure 4.7, page 354, except the horizontal axis woul be (x 3 + ) while the vertical woul be (x 3 + ) 2. The erivative with respect to x which is a hien variable upon which the others epen is foun by compensation, that is, the rate of change of (x 3 + ) 2 with respect to x is foun by first fining its rate of change with respect to (x 3 + ), an then multiplying by a compensating factor which is the rate of change of (x 3 +) with respect to x. The iea is very similar to (4.32). To review (4.33), there we want to know how (x 3 + ) 2 changes with respect to x, so we first ask how oes (x 3 + ) 2 change with respect to (x 3 + ), i.e., how oes the square of a quantity change with respect to that quantity (power rule) an multiply by how (x 3 + ) changes as x changes. 39 With this kin of argument we can exten the power rule to have a chain rule version, u n un u }{{} u, i.e., u n nun u. (4.34) Example With (4.34) we can quickly calculate the following erivatives, which woul be more ifficult without the chain rule: (29x x 2 ) 4 4(29x x 2 ) 3 (29x x2 ) 4(29x x 2 ) 3 (29 2x). (x + 2) 9 9(x + 2) 8 (x + 2) 9(x + 2) 8 9(x + 2) 8. Note that occasionally the erivative u of the inner function is just. 39 This is somewhat similar to compensating for unmatche units in physics or chemistry problems. If we travel 60 miles in 75 minutes, an we want our average spee in miles/hour, we can first fin miles/minute, an then multiply by a compensating factor which relates minutes to hours: spee 60 mile 75 minute 0.8 mile minute 60 minute 48 mile hour {z } hour. compensating factor Since the question was how the istance relates to hours, we first i the easy computation relating istance to minutes, an then compensate by multiplying how minutes relate to hours.

6 356 CHAPTER 4. THE DERIVATIVE (5x 9)8 8(5x 9) 7 (5x 9) 8(5x 9)7 5 40(5x 9) 7. When we compute (5x 9)8, we coul have written a Leibniz-style ecomposition/expansion, (5x 9) 8 (5x 9)8 (5x 9) (5x 9) 8(5x 9) 7 5, (4.35) an this is quite correct. However it is not stanar practice to write the mile step. Inee most authors prefer to avoi having complicate expressions in the enominator of a ifferential operator, preferring enominators as in, t, u, etc. In this text we will still occasionally write as in (4.35) for clarity (which is akin to using a truth table to show a style of argument is vali), but more often we will just state the chain rule version (4.34) of the power rule with the correct terms in place of the general u. The reaer is encourage in the strongest possible terms to always write the first step of power rule with chain rule problems, as shown in the next examples below. This will avoi common errors, especially as erivative computations become more an more complex, an such practice will reinforce the proper use of the chain rule version of the power rule, (4.34). We now list several further examples of this rule, showing in the first step what the rule says literally for each problem, before the inner function s erivative is compute. (2x + 9)3 3(2x + 9) 2 (2x + 9) 3(2x + 9)2 2 6(2x + 9) 2. sin2 x (sin x)2 2(sin x) sin x 2 sinxcosx. Note that the final answer is unerstoo to mean 2(sinx)(cos x). (x2 + cosx) 4 4(x 2 + cosx) 3 (x2 + cosx) 4(x 2 + cosx) 3 (2x sin x). ( 3x 2 + 6x + 7 ) 7 ( 7 3x 2 + 6x + 7 ) 6 ( 3x 2 + 6x + 7 ) 7 ( 3x 2 + 6x + 7 ) 6 (6x + 6) 7(3x 2 + 6x + 7) 6 6(x + ) 42(3x 2 + 6x + 7)(x + ). Note that the chain rule version of the power rule, un nun u, oes not contraict the earlier power rule that xn nxn. For instance, when u is equal to x, we can write 9 9x8 9x8 9x 8, which agrees with our original power rule, which here woul give us x9 9x8. Thus the chain rule version of the power rule in fact generalizes the original power rule Chain Rule Derivatives of Sine an Cosine Now we look at erivatives of sinu an cosu with respect to x, assuming u u(x), i.e., that sin x u is actually a function of x. Recall (4.8) an (4.9) from page 37: cosx an cos x sinx. The chain rule versions of the erivatives of sine an cosine follow easily from the Leibniz notation: sin u cosu sin u u cosu u u cosu u, u u sinu,

7 4.4. CHAIN RULE I 357 As interesting as the above equations are to behol, in practice one usually uses summary versions: sin u cosu cosu u, (4.36) sinu u. (4.37) Thus erive, we are free to use (4.36) an (4.37) where applicable. In the next example we show two methos of computing a particular erivative: using a Leibniz-style ecomposition, an applying (4.36) irectly. Example If f(x) sin x 2, that is f(x) sin(x 2 ), then we can compute f (x) the following two ways: sin x 2 sin x 2 sin x2 2 2 cosx2 2x 2xcosx 2, (4.38) cosx 2 2 cosx2 2x 2xcosx 2. (4.39) In the secon metho (4.39) for computing the erivative above, we i just insert u x 2 into (4.36), but the justification for that can also be seen in the first metho (4.38) with the Leibnizstyle ecomposition. Note also that we compute this same erivative in Example 4.4.4, page 352 using the prime notation. Example Compute f(z) z if f(z) cos(z 3 + sin z). Solution: Here the names of the variables have change, but the principle of the chain rule is the same. Again we will compute this two ways, the secon using (4.37), except with z in place of x: f (z) z cos(z3 + sin z) cos(z3 + sin z) (z 3 + sinz) (z3 + sin z) z f (z) z cos(z3 + sin z) sin(z 3 + sin z) (z3 + sin z) z sin(z 3 + sin z) (3z 2 + cosz), sin(z 3 + sinz) (3z 2 + cosz). Accepte practice is to compute the above erivative using the latter metho, an so that is the metho stuents shoul eventually strive to reprouce. As in an earlier iscussion involving the power rule, one can think of this example as using the thought pattern that says the erivative of cosine is minus sine..., multiplie by the erivative of what is insie the cosine. That is perhaps an over-simplification, an shoul be informe by awareness of what we get from the Leibniz-style expansion. To be clear on what (4.36) an (4.37) say, an why these shoul hol, consier the following abstract equations, which are in fact restatements of (4.36): 40 sin u w sin θ ξ sin x t sin u u sin θ θ sin x u u cosu w w, θ ξ t cosθ θ ξ, cosx t.

8 358 CHAPTER 4. THE DERIVATIVE Note that in all the cases, the ecomposition s first factor let us use the known erivative formula for sine because the variables matche an then we compensate for introucing the new variable s erivative (as a fraction of sorts) with the secon factor. We now point out that it is quite common for the chain rule to apply more than once in a particular problem. Our next example below shows a case of a function within a function within a function, an the example following will be a sum of two functions, each requiring a chain rule. [ sin 3 (4x) ]. Example Compute Solution: Note that the function can be rewritten [sin4x] 3. Now we compute the erivative, first applying the power rule version of the chain rule (4.34), page 355, an then (4.36), page 357. Then we show the same computation using the Leibniz-style ecomposition. [sin 4x] 3 3[sin4x] 2 [sin 4x] 3 sin 4x 3 sin 2 4x cos4x (4x) 3 sin 2 4xcos4x 4 2 sin 2 4xcos4x. [sin 4x]3 [sin 4x] sin 4x (4x) (4x) 3[sin4x] 2 cos4x 4 2 sin 2 4xcos4x. So the Leibniz-style ecomposition will work for longer chains of functions within functions. But so will the abbreviate chain rules which say u3 3u2 u sin u, an cosu u, which was the first approach in the computations above: after applying the power rule, the inner erivative calle another chain rule. 4 Again, it is best to strive for the abbreviate approach in practice, though both approaches are worth stuying (an of course the ecomposition explains the abbreviate approach). Example Fin f (x) if f(x) sin 2 x + cos 2 x. Solution: The larger structure of this function is that of a sum of two functions, so first we use the sum rule, which tells us to a (an thus first compute) the erivatives of sin 2 x an cos 2 x. These are then both chain rules. f (x) [ sin 2 x + cos 2 x ] [ sin 2 x ] + [ cos 2 x ] [ (sin x) 2 ] + [ (cos x) 2 ] sin x + 2(cosx) cosx 2(sin x) 2 sinxcos x + 2 cosx( sin x) 2 sinxcos x 2 cosxsin x Just to be sure, it shoul be pointe out that when we write for instance cos x, we mean that the t t is outsie of the cosine function, i.e., we mean (cos x). Note that many texts assume this meaning without t making it explicit with the ot, an simply write cos x. As a matter of style, it is assume the erivative t is not part of the argument of the cosine function in such a case. t 4 This phenomenon of rules calling other rules occurs repeately throughout the rest of the textbook. We will see it occasionally in this section, an it will become the norm in later sections.

9 4.4. CHAIN RULE I 359 Actually this is what we shoul hope woul be the answer, for the original function we are taking the erivative of is actually constant: [ sin 2 x + cos 2 x ] [] 0. It happens frequently in calculus that it is avantageous to algebraically rewrite a function before taking its erivative. In fact we i that each time we took a erivative of sin 2 x (sin x) 2, the latter notation being more obvious in illustrating the composition (function insie of a function) structure of the original function. For calculus to be consistent (which it is, so no nee to fear!), we shoul be able to rewrite the function an get the same erivative, as long as we rewrite the function correctly. The ifferentiation (erivative computing) rules are eventually sufficient to compute the erivative no matter how the function is rewritten, but some forms of a given function are easier to eal with than others Power Rule for Rational Powers With the chain rule we have enough theoretical evelopment to show that the power rule actually hols for any constant power which is a rational number p/q (where p, q Z, an of course q 0). Recall that the set of all rational numbers was enote Q, for quotients, i.e., fractions, of integers. We alreay prove the power rule for powers n {0,, 2, 3, }, an that result is use in the proof for rational powers which we leave the proof until the en of this section, so we can expeitiously come to examples. But first we state the theorem. Theorem (Power Rule for Rational Powers) For any r Q {0} (i.e., nonzero rational numbers), r rxr, (4.40) u r rur u. (4.4) Example 4.4. For example, the following (which was an exercise with ifference quotient limits in Section 4.) yiels quickly to the power rule: x /2 2 x/2 2 x /2 2 x. In fact, this particular erivative occurs often enough that it, along with its chain rule version, eserves special attention (an shoul be committe to memory): Example Fin f (x) for f(x) x 2 +. f (x) x2 + 2 x 2 + x 2 x, (4.42) u 2 u u. (4.43) ( x 2 + ) 2 x 2 + (2x) x x2 +.

10 360 CHAPTER 4. THE DERIVATIVE Note that in the above example the outer function was the square root, while the inner function is x 2 +. One coul write (though again, it is not stanar practice): x2 + x 2 + (x 2 + ) (x2 + ) 2 x 2 + 2x x x2 +. Example Suppose f(x) x + x. Then the Leibniz ecomposition woul look like f(x) x + x (x + x) (x + ( x) 2 x + x + ) 2. x Again an especially with practice one woul usually not write the Leibniz ecomposition in the first step, but shoul instea write x + x 2 x + x (x + ( x) 2 x + x + ) 2. x A common mistake in the above example is to think of x as the inner function, since geometrically it somehow appears to be innermost. In fact the inner function is actually the whole of x + x. Example Suppose f(x) 2 (x 3 9x + 7) 7. Then f (x) [ 2(x 3 9x + 7) 7] 2 ( 7)(x 3 9x + 7) 8 (x3 9x + 7) 4(x 3 9x + 7) 8 (3x 2 9) 4(3x2 9) (x 3 9x + 7) 8. In the previous example we were able to use the chain rule version (4.4) of the power rule once we wrote the function as a power of a polynomial, albeit negative an with a multiplicative constant along for the rie. The next example calls for a rewriting (for simplicity), an then calls the chain rule twice. Example f(x) 3 x + x f (x) 3 [ 3 x + x ( x + x ] ) /3 ( x + 4/3 x ) ( x + [ 4/3 x 3 + 9) + 3 ( x + [ 4/3 x 3 + 9) + ( x + ) x x x 2 ] 2 x (x3 + 9) ]

11 4.4. CHAIN RULE I 36 In the calculation above, we first rewrote the expression as a 3 power, then use the chain rule version of the power rule, an use the chain rule again in calculating the erivative of that inner function Proof Of Power Rule for Ratoinal Powers We now ivert temporarily to prove the power rule for rational numbers r Q r rxr, from which the chain rule version also follows. The technique use will recur several times throughout the text. Fortunately it is somewhat self-explanatory, assuming knowlege of the general chain rule. The proof is in two steps, the first being a proof in the case of negative integer powers, from which we can eventually recover all rational power cases. Proof: Now we will use the chain rule to show that the power rule, xr rx r, hols also for any rational power r p/q, with p, q nonzero integers. (The case p 0 is trivial an the case q 0 is meaningless.) The proofs below are inclue for completeness, an also because they foreshaow a metho we will use extensively later in the text, that metho being implicit ifferentiation. First we will show that the power rule hols for y x n for any negative integer exponents n. In such cases we can write y x m for a positive integer exponent m (namely n). But then y x m. Furthermore we alreay showe in Section 4. (Example 4..4, page 300) that the erivative efinition gives us y y /y 2 (though the variable use in the proof there was x). Using this an the chain rule, we get y x m [ y ] [xm ] y 2 y mxm y y2 mx m (x n ) 2 ( n)x n nx 2n n nx n, q.e.. It is important to that we interpret the first implication correctly. Recall that y x n, so y is a function of x. But then so is y an, in fact, y an x m are the same functions of x. Hence, if y an x m were graphe versus x, the graphs woul be the same, so the slopes at each x-value woul be the same. Therefore y an x m have the same erivative with respect to x. Now we will use the chain rule in a similar way to compute the erivatives of rational powers of x. Suppose y x p/q, where p, q Z {0} are nonzero integers, an that

12 362 CHAPTER 4. THE DERIVATIVE r p/q is in simplifie form. Then we can raise both sies of y x p/q to the power q to get 42 qy y q x p y q p q y pxp y p q y q x p p ( x p/q) q x p p q q x p q p x p p q x p q, which can be rewritten y rxr. Thus y x r implies the form that we sought to prove for the erivative. In fact, once we have logarithms we can efine y x r for all r R, an fin again that the erivative is given by the same formula as in our power rules here Further Examples Example Suppose charge q is fixe an q 2 is moving away from charge q, an r is the istance between them. When r 0.m, we have F 0.6N. Fin F r when r 0.m an when r 0.3m. Solution: Recall Coulomb s Law: F k qq2 r. Since F k qq2 2 r (kq 2 q 2 )r 2, an 0.6 kq q 2 (0.) (kq q 2 ) 00 kq q F kq q 2 r r 2. (Recall that k, q an q 2 are all constants.) From this we can euce With that, we have F r [ ] r r 2 [ 0.006r 2 ] 2(0.006)r r 3 r F r 0.02 r 3. F r F r 0.02 r , 0.02 r Note that for r > 0 we have F r ecrease in F itself slows. < 0, so F ecreases as r increases. Furthermore, the rate of 42 Note that p or q (but not both, since p/q is simplifie) coul be negative, but what we are about to o is justifie by the previous result that the power rule also works for negative integer exponents.

13 4.4. CHAIN RULE I 363 Example Supply an eman for an item are usually relate to its price: eman will be high if the price is low; proucers will supply a large number of an item if the price is high. The price elasticity of eman is given by the formula 43 ε ( ) P q q P, (4.44) where P is the price an q is the quantity for a given eman equation which relates P an q. Deman is sai to be elastic if ε >, inelastic if ε <, an unitary if ε. Describe the elasticity of the following eman curve equations: (a) q 00 3 P (b) q 200 P Solution: (a) For q 00 q 3P, we have P 3, an so ( ) ( ) P q ε q P P 00 3 P 3 P 00 3 P 3 P 300 P. We can either solve irectly for the three cases P/(300 P) <, >, or we can introuce F(P) P 300 P an see where this is negative, positive an zero in short, construct a sign chart for F(P): F(P) P 300 P P (300 P) 2P P 300 P. we can see this is zero when 2P P 300 P 50. This correspons to ε, so eman is unitary (ε ) when P 50. Before constructing the sign chart, we notice that q 00 3P requires that P < 300, lest q be negative. Furthermore, we can assume P > 0 as well. With these constraints in min, we prouce our sign chart: 2(P 50) 300 P Function: F(P) Test P Sign Factors: / / Sign F(P): From the chart we get F(P) < 0 on 0 < P < 50, F(P) 0 at P 50 an F(P) > 0 on 50 < P < 300. Equivalently, we get (inelastic) ε < 0 0 < P < 50, (unitary) ε 0 P 50, (elastic) ε > 0 50 < P < This formula is the erivative version of an alternative efinition, in which ε is efine as ( q)/q ( P)/P 00% ( q)/q 00% ( P)/P percent change in q percent change in P. Using the first expression in the line above, an replacing with, we see this is approximately (P/q) (q/p). However, this quantity being almost always negative, economists ten to verbalize its absolute value instea. Since P an q are nonnegative (or at least it stretches the imagination to fin scenarios where they are negative), using absolute values with the erivative s output generally leaves a nonnegative value for ε.

14 364 CHAPTER 4. THE DERIVATIVE (b) For the case q 200 P, we have ( ) ( ) P q ε q P P 200 P P P 200 P P P 200 P P 2(200 P). P (200 P) P We can use the same technique as we i in (a) to fin where ε <, >,, though this time we will opt for solving these irectly. Note that we must have 0 < P < 200 ue to the efinition of q. Uner that assumption, we have: P ε P P P 3p 2400 p 800, P ε < P < P < P 3p < 2400 p < 800, P ε > P > P > P 3p > 2400 p > 800, Remembering our constraint 0 < P < 200, we can write in fact that (inelastic) ε < 0 0 < P < 800, (unitary) ε 0 P 800, (elastic) ε > < P < 200. It shoul be note that we coul multiply both sies of the inequalities by P because we were operating uner the assumption that 0 < P < 200, an this implies P > 0. If we are unsure of the sign of an expression we wish to multiply by to manipulate an inequality, it is safer to use a sign chart metho, which avois multiplying both sies of an inequality by a variable quantity (of unknown sign). Also, in a case like (b) we see nearly the same computation complete three times, an the sign chart somewhat ecreases this reunancy. However, the observant stuent might combine both ieas, fining where ε an simply testing the other regions in simpler cases such as the above. Example Suppose an object slies on a straight, frictionless track. Suppose further that one en of a spring is attache to a fixe point at the en of the track, an that the other en of the spring is attache to the sliing object. If the object is otherwise free to slie, its motion will be perioic, of a type known as simple harmonic motion. Suppose that the equilibrium point for object attache to the spring is at position x 0, A is the maximum positive isplacement of the object, x(t) is the position of the object at any given time t, an x(0) A, we can give the position by

15 4.4. CHAIN RULE I 365 ( ) 2π x Acos T t, where T is the perio of the motion, that is, the time require for one complete cycle before the motion repeats. This equation woul hol, for instance, if the object were pulle to position A an then release at time t 0. A 0 A x Fin the velocity an acceleration for any time t > 0. Solution: As usual we will efine velocity v(t) t x(t), an acceleration a(t) tv(t). From the chain rule we get the following computations (note 2π/T is a constant): v(t) [ ( )] 2π Acos t T t ( ) 2π Asin T t t ( ) 2π Asin T t 2π T ) v(t) 2πA T sin ( 2π T t [ ] 2π T t a(t) t v(t) [ 2πA ( 2π t T sin T t, 2πA T cos )] 2πA ( ) 2π T cos T t t ( ) 2π T t 2π T ( ) 2 ( ) 2π 2π a(t) A sin T T t. ( ) 2π T t It is common in the stuy of oscillating functions to efine ω 2π/T so that the above computations can be summarize x Acos(ωt) t v Aω sin(ωt) t a Aω 2 cos(ωt). In such a case, the perio is given by T 2π/ω. Note also that a(t) ω 2 x(t). This is a efining feature of simple harmonic oscillators, namely that the force is proportional to but in the opposite irection of the position. (Recall force mass acceleration.) This is the case with most springs, which conform to Hooke s Law, F kx, where k > 0 is a constant, x is the position relative to the equilibrium (rest) position of the spring an F is the force it exerts Name for Robert Hooke( ), an English inventor, professor, natural philosopher an architect. His 678 statement of his law in wors rea, Ut tensio, sic vis (Latin), meaning, As the extension, so the force. Hooke was a contemporary of Newton s, an was very much involve in many of the significant iscussions of his time. For instance, Newton gave him some creit for reviving Newton s interest in astronomical mechanics, leaing to Newton s evelopment of the inverse square law of gravitational attraction, which mirrors Coulomb s electrostatic law (4.2), page 329, though there are no cases of gravitational repulsion in Newton s law (only attraction).

16 366 CHAPTER 4. THE DERIVATIVE Exercises For 6, use the power rule to compute the erivative, after re-writing the problem. In particular, you can use (axn ) a nx n [ x ] [ ] x [ ] 3 x 4 [ ] 6 t x [ 2t 2 ] [ ] 9y Recall (ab) n a n b n anytime a, b y 0. For 7 0, compute the given erivative. [ ( 9x) ] [ 27(3x 2 0x + 55) 2]. [ 2x5 ] 0. (3x + )2 (Do this two ways: using the chain rule, an by first expaning the square. Compare your answers.) For 4, compute f (x).. f(x) (x + 5) f(x) (2x + 5) f(x) (x 4 x + ) f(x) x + x + x. Compute the following erivatives. sin z cosθ t 7 t sin(cosx) cosx For 9 23, compute the given erivative. sin x sin x sin(cosx) cos3 x cos(x + cosx) 24. Fin h (9) if h(x) f(g(x)), g(9) 5, g (9) 2, an f (5) On the unit circle, y 2 x 2. If we take either the upper semicircle or the lower semicircle, then y is also a function of x. Fin the tangent line to the graph at the point (3/5, 4/5) by fining y two ways: (a) Using y x 2 for the upper semicircle, an the chain rule. (b) By applying to both sies of y 2 x 2, as we i in the proof of Theorem 4.4.2, an then solving for y, an plugging into that expression (x, y) (3/5, 4/5).

17 4.4. CHAIN RULE I Using sec x (cosx), (a) erive secx sec xtan x. (b) Use (a) an the chain rule to compute sec x Show that if k, q, q 2 are constant in Coulomb s law (4.2), page 329, then F r 2 F r. 28. As a charge q 2 moves away from a stationary chage q, the instantaneous rate of change of the Coulomb force F with respect to r is measure to be 5N/m. (a) Fin the magnitue of the force between the charges when they are 0.3m apart. (See Example 4.4.6, page 362.) (b) If we are given that k N m 2 /C, an q C, fin the equation for the force between the two particles when separate by a istance r. (c) Fin F r from (b) when r 0.25M. 29. The gravitational force of attraction between any two masses is irectly proportional to the prouct of the masses an inverseley proportional to the square of the istance between them. In particular, F G m m 2 r 2, (4.45) where G N(m) 2 /(kg) 2 is the universal gravitational constant, m, m 2 are in kg, an r is in m. Fin the rate of change of F with respect to istance r (i.e., fin F/r) when r 0.5m. 30. An object moves along the x-axis accoring to the equation s(t) t+2 sint. Fin those values of t when the object is moving to the left, i.e., fin where s (t) < 0, within the interval t [0, 2π]. 3. Suppose that the cost in thousans of ollars to manufacture x thousans of items is given by C(x) 0.003x x x + 70, where x [0, 0]. (a) Fin the (proxy) marginal cost function, C (x). (b) Determine the marginal cost at x 4 (in units of thousans of ollars per thousans of items, which simplifies to ollers per item). 45 (c) Fin the average rate of change of cost in going from x 3.9 to x 4.. () Compare the answers to parts (b) an (c). 32. Consier the following alternating current circuit: V R The current I where the resistor R an an inuctor L are present is given by I L V R2 + (2πfL) 2, where V is the voltage an f is the current s frequency of alternation (in cycles/secon, or Hertz, also written Hz). Assuming that V, R an f are constants, fin an equation for the instantaneous rate of change of current I with respect to inuctance L. 45 Here marginal cost woul represent the cost of the next thousan items.