CHAPTER 20 The Product And Quotient Rules

CHAPTER 20The Product and Quotient RulesWe have developed rules for the derivatives of the sum or difference oftwo functions that work as followsid hf (x) g(x) f 0 (x) g0 (x)dxid hf (x) g(x) f 0 (x) g0 (x).dxBut what about the derivative of product of two functions, or the quotient oftwo functions? This chapter answers these questions by deriving two newrules forid hf (x) · g(x) anddx d f (x).dx g(x)These two new rules will be called the product rule and the quotient rule,respectively.Let’s begin by deriving the product rule. Given two functionsh f (x) andidg(x), we aim to work out the derivative of their product, that is, dxf (x)g(x) .By Definition 16.1, the derivative of a function F(x) isid hF(z) F(x)F(x) lim.z! xdxz xWe are interested in the case where F(x) f (x)g(x), which isid hf (z)g(z) f (x)g(x)f (x)g(x) lim.z!xdxz xWe will now work this limit out carefully. In this limit, the denominatorz x approaches zero, so we have to get rid of it somehow. In the followingcomputation it gets absorbed into the definition of f 0 (x) and g0 (x).

242The Product and Quotient RulesSo let us begin our computation. As noted above, our first step isid hf (z)g(z) f (x)g(x)f (x)g(x) lim.z!xdxz xLet’s insert a little space int this expression to give ourselves room to work.id hf (z)g(z) f (x)g(z) f (x)g(z) f (x)g(x)f (x)g(x) lim.z! xdxz xTo the space just created, add zero in the form of 0 f (x)g(z) f (x)g(z).This is an allowable step because adding in 0 doesn’t alter the limit’s value.id hf (z)g(z) f (x)g(z) f (x)g(z) f (x)g(x)f (x)g(x) lim.z! xdxz xIn the numerator, factor out g(z) from the first two terms, and f (x) from thesecond two, as shown below. Then split the fraction and apply limit laws: if (z) f (x) g(z) f (x) g(z) g(x)d hf (x)g(x) limz! xdxz xµ f (z) f (x)g(z) g(x) limg(z) f (x)z! xz xz xf (z) f (x)g(z) g(x) lim· lim g(z) lim f (x) · limz! xz!xz!xz!xz xz x f 0 (x)g(x) f (x)g0 (x) .In the last step we used the facts that limlim f (x) f (x), andz! xg ( z ) g ( x )lim z xz! xz! xf ( z ) f ( x )z x f 0 (x), lim g(z) g(x),z! x0 g (x). (We assume here that g (x) exists, so,0by Theorem 18.1, g(x) is continuous, and hence lim g(z) g(x).) §z! xdWe have just determined that dxf (x)g(x) f 0 (x)g(x) f (x)g0 (x). Nowthat we have done this computation—and we believe it—we will never haveto do it again. It becomes our latest rule.Rule 7 (The Product Rule)id hf (x)g(x) f 0 (x)g(x) f (x)g0 (x)dxThe derivative of a product equals the derivative of the first function timesthe second function, plus the first function times the derivative of the second:iiid hd hd hf (x)g(x) f (x) · g(x) f (x) ·g(x) .dxdxdxIn applying the product rule to f (x)g(x), you also have to do the derivativesof f and g, using whatever rules apply to them.

243Example 20.1Find the derivative of 4x3 e x .This is a product (4x3 ) · (e x ) of two functions, so we use the product rule.d h 3 xid 3§ xd x§4x e 4x · e 4x3 ·edxdxdx 12x2 · e x 4x3 · e x 4e x 3x2 x3 . Example 20.2 Find the derivative of y x2 3 x2 5x 7 .This is a product of two functions, so we use the product rule.i i i d h 2d h 2d h 2x 3 x2 5x 7 x 3 · x2 5x 7 x2 3 ·x 5x 7dxdxdx 2 2 2x x 5x 7 x 3 (2x 5)This is the derivative, but some simplification is possible: 2x3 10x2 14x 2x3 5x2 6x 15 4x3 15x2 8x 15 .Here our approach was to take the derivative (using the product rule) andthen simplify. Alternatively, you could first simplify (by multiplying) andthen take the derivative:i 2 id h 2d h 4322x 3 x 5x 7 x 5x 7x 3x 15x 21dxdx 4x3 15x2 14x 6x 15 4x3 15x2 8x 15 .Example 20.3 Very often you’ll have a choice of rules, and one choicemay lead to a simpler computation than the other. Consider finding thederivative of y 3x5 . This is a product, so you could use the product rule:d h 5id § 5d 5§3x 3 · x 3·x 0 · x5 3 · 5x4 dxdxdxBut it’s much easier to just use the constant multiple rule:d h 5id h 5i3x 3x 3 · 5x4 15x4 .dxdx15x4 .

244The Product and Quotient Ruleshid f ( x)Next we derive the rule for dxg( x) . (You may opt to skip the derivationon a first reading and go straight to the conclusion at the bottom of thispage.) Our computation begins with the definition of the derivative andproceed by adding zero in a clever way, as we did for the product rule. d f (x)dx g(x) f (z) f (x) g(z) g(x)lim(definition of derivative)z! xz xµ µ f (z) f (x)f (x) f (x) g(z) g(z)g(x) g(z)lim(insert space)z! xz xµ µ f (z) f (x)f (x) f (x) g(z) g(z)g(x) g(z)lim(add zero)z! xz xµ 111f (z) f (x) f (x) g(z)g(x) g(z)lim(factor)z! xz xµ 1g(z) g(x)f (z) f (x) f (x)g(z)g(x)g(z)lim(combine fractions)z! xz x 1 f (x) f (z) f (x) g(z) g(x)g(z) g(x)g(z) limz! xz xµ f (z) f (x) 1f (x) g(z) g(x) lim z! xz xg(z) g(x)g(z)z x limz! x(regroup)(split fraction)f (z) f (x)1f (x)g(z) g(x)· lim lim· limz! x g(z) z! x g(x)g(z) z! xz xz x f 0 (x)1f (x) g0 (x)g(x) g(x)g(x)(limitlaws)(evaluate limits) f 0 (x)g(x) f (x)g0 (x) g(x)g(x)g(x)g(x)(get common denominator) f 0 (x)g(x) f (x)g0 (x).g(x)2(combine fractions)We have our new rule.Rule 8 (The Quotient Rule) d f (x)f 0 (x)g(x) f (x)g0 (x) dx g(x)g(x)2

245Compare the differences and similarities between the product rule andthe quotient rule. There are similarities, but quotient rule is more complex.It has a minus instead of a plus, and you must divide by g(x)2 .Product Rule:Quotient Rule:iiid hd hd hf (x)g(x) f (x) · g(x) f (x) ·g(x)dxdxdx d f (x)dx g(x) iid hd hf (x) · g(x) f (x) ·g(x)dxdx 2g(x)Example 20.4 Find the derivative ofx2 1.x2 xThis function is a quotient, so we apply the quotient rule to find its derivative. 2d x 1dx x2 x i i d h 2d h 2x 1 x2 x x2 1x xdxdx 22x x 2 (2x 0) x x x2 1 (2x 1) 2x2 xThis is the derivative, but a few extra steps of algebra simplify our answer. 3 2x 2x2 2x3 x2 2x 1 2x2 x x2 2x 1 2 .x2 xExample 20.5 Find the derivative ofx5.exThis function is a quotient, so we apply the quotient rule to find its derivative. 5 d xdx e xd h 5i xd h xix e x5edxdx (e x )2 5x4 e x x5 e x.ex exThis is the derivative, but some simplification is possible: x4 e x (5 x) ex exx4 (5 x).ex

246The Product and Quotient RulesExample 20.6 The quotient rule can be computationally expensive, sodon’t use it if you don’t have to. As an example, consider differentiating3 x2 4 x 5. This is a quotient, so you could use the quotient rule. But the2denominator is constant, so a better choice would be to factor it out usingthe constant multiple rule: id 3x2 4x 51 d h 21 ·3x 4x 5 (6x 4) 3x 2 .dx22 dx2If you used the quotient rule, your work would go like this: 2d 3x 4x 5dx2 i d h id h 23x 4x 5 · 2 3x2 4x 52dxdx 22 2 (6x 4) · 2 3x 4x 5 · 0 4(6x 4) · 2 3x 2 .4Though it gave the same answer, the quotient rule was an overkill. Workenough exercises that you see your choices and their consequences.Example 20.7 Find the derivative of x 10 .We will do this two ways. First, we can simply apply the power rule.d h 10 ix 10x 10 1 10x 11 .dxOur second step uses the quotient rule.d h i 10d h 10 i 1 x 1·xhid 10d1dxdxx 2dxdx x10x10 0 · x10 10x9 10x9 20 10x 11 .x20Besides reminding us that a problem can be done several ways, thisdexample raises an important point. In deriving the power rule dx[x n ] nx n 1in Chapter 17, we proved it only for positive integer values of n. But weasserted then that it actually holds for all real values of n, positive ornegative, and that that we would see this in due time. Example 20.7 suggeststhat we can use the quotient rule to show that the power rule holds fornegative integer values of n. Exercise 13 below asks for the details.

247 Example 20.8 Find all x for which the tangent to y xe x at x, f (x) hasslope 0.Solution: The slope of the tangent line is given by the derivative, so as inthe previous example our first task is to find the derivative of y xe x .This requires the product rule:ydy 1 · e x x · e x e x (1 x).dxThis means the tangent has slope 0where e x (1 x) 0. Because e x 0for all x, the only way we can havee x (1 x) 0 is if x 1. 1y xe xxAnswer: The tangent to y xe x hasslope 0 at x 1. (See the graph onthe right.)Exercises for Chapter 20In problems 1–12 find the indicated derivative.1.3.5.7.9.11. id h 42 x 3 x 3 x2 xdxd h xixedx d x2 xdx x 5 dex xdx x3 x2 1 dxdx x3 x2 1 d1dx x2 12.4.6.8.10.12.d h xp ie xdxd h 2x ie(Hint: e2 x e x e x .)dx d x2 3 x 4pdxx 5 3 d x x2 1dxx d x 1dx x 1 x depdxxd13. In Chapter 17 we proved that the power rule dx[ x n ] nx n 1 works for positiveinteger values of n. Combine this fact with the quotient rule to show that thepower rule also holds for negative integer values of n. (See Example 20.7.)p14. Find the equation of the tangent line to the graph of f ( x) e x x at point (1, f (1)).115. Find the equation of the tangent line to the graph of f ( x) p at point (4, f (4)).x16. Find all points ( x, y) on the graph of y x 1where the tangent has slope 0.x 3

248The Product and Quotient Rules17. Find all points ( x, y) on the graph of y xwhere the tangent line is horizontal.ex18. Two functions f ( x) and g( x) are graphed below.4Estimate h0 (1.5) if:(a) h( x) f ( x) g( x).y433221f ( x)(b) h( x) .g ( x) 4 3 2 1 1g ( x)(c) h( x) .f ( x) 31 2 3 4xy f ( x ) 2y1 4 3 2 1 11 2 3 4x 2y g( x ) 3 4 419. A function f ( x) is graphed below.Find g0 (3) if:(a) g( x) x2 f ( x).x2(b) g( x) .f ( x) 2(c) g( x) f ( x) .43yy f ( x)21 7 6 5 4 3 2 1 11 2 3 4 5 6 7x 2 3 420. Information about f ( x), g( x), f 0 ( x) and g0 ( x) is given in the table below.Find h0 (3) if:x01234 3f ( x) 4 2011(a) h( x) x x f ( x).0f(x)21130.5g ( x)(b) h( x) .g ( x)109740f ( x)0g ( x)0 0.5 1 3 4(c) h( x) f ( x) g( x).50 1 4 4

249Exercise Solutions for Chapter 201.3.5.7.9.11. i d h 42 x 3 x 3 x2 x 8 x3 3 3 x2 x 2 x4 3 x (6 x 1)dxd h xixe 1 · e x xe x e x xe xdx (2 x 1)( x 5) x2 x (1 0) 2 x2 11 x 5 x2 x x2 10 x 5d x2 x dx x 5( x 5)2( x 5)2( x 5)2 ( e x 1) x3 x2 1 ( e x x) 3 x2 2 xdex x 2dx x3 x2 1x3 x2 1 1 · x3 x2 1 x 3 x2 2 xdx 2 x3 x2 1 2 232dx x x 1x3 x2 1x3 x2 1 0 · x2 1 1 · (2 x 0)d1 2 x 2 222dx x2 1x 1x 1d13. In Chapter 17 we proved that the power rule dx[ x n ] nx n 1 works for positiveinteger values of n. Combine this fact with the quotient rule to show that thepower rule also holds for negative integer values of n.Let n be a negative integer, so n m for a the positive integer m. Then thedpower rule as proved in Chapter 17 says dx[ x m ] mx m 1 . This problem is askingdnn 1us to show dx [ x ] nx . To confirm this with the quotient rule, observe that d h nid h m id10 · x m 1 · mx m 1 mx m 1x x mx m 1 nx n 1 .mdxdxdx xx2 m( x m )2115. Find the equation of the tangent line to the graph of f ( x) p at point (4, f (4)).xThe derivative f 0 ( x) will tell us the slope of this line. Finding f 0 ( x) will beeasier with the product rule than the quotient rule. Since f ( x) x 1/2 , we get1111f 0 ( x) x 1/2 1 x 3/2 3/2 p 3 . When x 4, the tangent line has222x2 x110slope is thus m f (4) p 3 , and passes through the point (4, f (4)) 162 4(4, 1/2). Now use the point-slope formula for the equation of a line.Answer:y y01y 21y 2 y m( x x0 )1 ( x 4)1611 x 16413 x 164