Quotient Rule

12 Key excerpts on "Quotient Rule"

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  Single Variable Calculus

  Early Transcendentals, Metric Edition

  • James Stewart, Daniel K. Clegg, Saleem Watson, , James Stewart, James Stewart, Daniel K. Clegg, Saleem Watson(Authors)
  • 2020(Publication Date)
  • Cengage Learning EMEA

    (Publisher)

  Copyright 2021 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it. 188 CHAPTER 3 Differentiation Rules The Quotient Rule and the other differentiation formulas enable us to compute the derivative of any rational function, as the next example illustrates. EXAMPLE 4 Let y - x 2 1 x 2 2 x 3 1 6 . Then y9 - s x 3 1 6d d dx s x 2 1 x 2 2d 2 s x 2 1 x 2 2d d dx s x 3 1 6d s x 3 1 6d 2 - s x 3 1 6ds2x 1 1d 2 s x 2 1 x 2 2ds3x 2 d s x 3 1 6d 2 - s2x 4 1 x 3 1 12x 1 6d 2 s3x 4 1 3x 3 2 6x 2 d s x 3 1 6d 2 - 2x 4 2 2x 3 1 6x 2 1 12 x 1 6 s x 3 1 6d 2 ■ EXAMPLE 5 Find an equation of the tangent line to the curve y - e x ys1 1 x 2 d at the point (1, 1 2 e). SOLUTION According to the Quotient Rule, we have dy dx - s1 1 x 2 d d dx se x d 2 e x d dx s1 1 x 2 d s1 1 x 2 d 2 - s1 1 x 2 de x 2 e x s2xd s1 1 x 2 d 2 - e x s1 2 2x 1 x 2 d s1 1 x 2 d 2 - e x s1 2 xd 2 s1 1 x 2 d 2 So the slope of the tangent line at (1, 1 2 e) is dy dx Z x-1 - 0 This means that the tangent line at (1, 1 2 e) is horizontal and its equation is y - 1 2 e. (See Figure 4.) ■ NOTE Don’t use the Quotient Rule every time you see a quotient. Sometimes it’s easier to rewrite a quotient first to put it in a form that is simpler for the purpose of differenti- ation. For instance, although it is possible to differentiate the function Fs xd - 3x 2 1 2 sx x using the Quotient Rule, it is much easier to perform the division first and write the func- tion as Fs xd - 3x 1 2x 21y2 before differentiating. Figure 3 shows the graphs of the function of Example 4 and its deriva- tive.

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  Calculus

  Concepts and Contexts, Enhanced Edition

  • James Stewart(Author)
  • 2018(Publication Date)
  • Cengage Learning EMEA

    (Publisher)

  Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it. SECTION 3.2 THE PRODUCT AND Quotient RuleS 187 so As , also, because is differentiable and therefore continuous. Thus, using the Limit Laws, we get The Quotient Rule If and are differentiable, then In words, the Quotient Rule says that the derivative of a quotient is the denominator times the derivative of the numerator minus the numerator times the derivative of the denominator, all divided by the square of the denominator. The Quotient Rule and the other differentiation formulas enable us to compute the derivative of any rational function, as the next example illustrates. Using the Quotient Rule Let . Then Find an equation of the tangent line to the curve at the point . SOLUTION According to the Quotient Rule, we have 1 x 2 e x e x 2 x 1 x 2 2 e x 1 x 2 1 x 2 2 dy dx 1 x 2 d dx e x e x d dx 1 x 2 1 x 2 2 ( 1, 1 2 e ) y e x 1 x 2 EXAMPLE 6 v x 4 2 x 3 6 x 2 12 x 6 x 3 6 2 2 x 4 x 3 12 x 6 3 x 4 3 x 3 6 x 2 x 3 6 2 x 3 6 2 x 1 x 2 x 2 3 x 2 x 3 6 2 y x 3 6 d dx x 2 x 2 x 2 x 2 d dx x 3 6 x 3 6 2 y x 2 x 2 x 3 6 EXAMPLE 5 v d dx f x t x t x d dx f x f x d dx t x t x 2 t f d dx u v v lim x l 0 u x u lim x l 0 v x v lim x l 0 v v v du dx u d v dx v 2 v t x v l 0 x l 0 d dx u v lim x l 0 u v x lim x l 0 v u x u v x v v v In prime notation: f t t f f t t 2 We can use a graphing device to check that the answer to Example 5 is plausible. Figure 3 shows the graphs of the function of Example 5 and its derivative. Notice that when grows rapidly (near ), is large. And when grows slowly, is near . 0 y y y 2 y 1.5 _1.5 _4 4 yª y FIGURE 3 Copyright 2019 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part.

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  Calculus

  Single Variable

  • Howard Anton, Irl C. Bivens, Stephen Davis(Authors)
  • 2022(Publication Date)
  • Wiley

    (Publisher)

  88 CHAPTER 2 The Derivative Derivative of a Quotient Just as the derivative of a product is not generally the product of the derivatives, so the derivative of a quotient is not generally the quotient of the derivatives. The correct relationship is given by the following theorem. Theorem 2.4.2: Quotient Rule If f and g are both differentiable at x and if g(x)  = 0, then f / g is differentiable at x and d dx  f (x) g(x)  = g(x) d dx [ f (x)] − f (x) d dx [g(x)] [g(x)] 2 (2) Formula (2) can be expressed as  f g   = g · f  − f · g  g 2 Proof d dx  f (x) g(x)  = lim h→0 f (x + h) g(x + h) − f (x) g(x) h = lim h →0 f (x + h) · g(x) − f (x) · g(x + h) h · g(x) · g(x + h) Adding and subtracting f (x) · g(x) in the numerator yields d dx  f (x) g(x)  = lim h →0 f (x + h) · g(x) − f (x) · g(x) − f (x) · g(x + h) + f (x) · g(x) h · g(x) · g(x + h) = lim h →0  g(x) · f (x + h) − f (x) h  −  f (x) · g(x + h) − g(x) h  g(x) · g(x + h) = lim h →0 g(x) · lim h →0 f (x + h) − f (x) h − lim h →0 f (x) · lim h →0 g(x + h) − g(x) h lim h →0 g(x) · lim h →0 g(x + h) =  lim h →0 g(x)  · d dx [ f (x)] −  lim h →0 f (x)  · d dx [g(x)] lim h →0 g(x) · lim h →0 g(x + h) = g(x) d dx [ f (x)] − f (x) d dx [g(x)] [g(x)] 2 [See the note at the end of the proof of Theorem 2.4.1 for an explanation of the last step.] In words, the derivative of a quotient of two functions is the denominator times the deriva- tive of the numerator minus the numerator times the derivative of the denominator, all divided by the denominator squared. Sometimes it is better to sim- plify a function first than to apply the Quotient Rule imm- ediately. For example, it is easier to differentiate f(x) = x 3/2 + x √ x by rewriting it as f(x) = x + √ x as opposed to using the quo- tient rule. Example 3 Find y  (x) for y = x 3 + 2x 2 − 1 x + 5 .

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  Mathematical Applications for the Management, Life, and Social Sciences

  • Ronald Harshbarger, James J. Reynolds(Authors)
  • 2018(Publication Date)
  • Cengage Learning EMEA

    (Publisher)

  ■ For a quotient of two functions, we might be tempted to take the derivative of the numer- ator divided by the derivative of the denominator, but this is incorrect. With the exam- ple f (x) 5 x 3 > x (which equals x 2 if x notequal.alt1 0), this approach would give 3x 2 > 1 5 3x 2 as the derivative, rather than 2x. Thus finding the derivative of a function that is the quotient of two functions requires the Quotient Rule. Product Rule EXAMPLE 1 Quotient Rule Copyright 2019 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it. SECTION 9.5 The Product Rule and the Quotient Rule 585 If f (x) 5 u(x) v(x) , where u and v are differentiable functions of x, with v(x) notequal.alt1 0, then f 9(x) 5 v(x) # u9(x) 2 u(x) # v9(x) [v(x)] 2 Quotient Rule The preceding formula says that the derivative of a quotient is the denominator times the derivative of the numerator minus the numerator times the derivative of the denominator, all divided by the square of the denominator. To see that this rule is reasonable, again consider the function f (x) 5 x 3 > x, x notequal.alt1 0. Using the Quotient Rule, with u(x) 5 x 3 and v(x) 5 x, we get f 9(x) 5 x(3x 2 ) 2 x 3 (1) x 2 5 3x 3 2 x 3 x 2 5 2x 3 x 2 5 2x We see that f 9(x) 5 2x is the correct derivative. The proof of the Quotient Rule is left for the student in the exercises in this section. Quotient Rule (a) If f (x) 5 x 2 2 4x x 1 5 , find f 9(x). (b) If f (x) 5 x 3 2 3x 2 1 2 x 2 2 4 , find the instantaneous rate of change of f (x) at x 5 3.

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  Biocalculus

  Calculus, Probability, and Statistics for the Life Sciences

  • James Stewart, Troy Day, James Stewart(Authors)
  • 2015(Publication Date)
  • Cengage Learning EMEA

    (Publisher)

  Thus, using the Limit Laws, we get d dx S u v D -v lim D x l 0 D u D x 2 u lim D x l 0 D v D x v lim D x l 0 s v 1 D v d -v du dx 2 u d v dx v 2 The Quotient Rule If f and t are differentiable, then d dx F f s x d t s x d G -t s x d d dx f f s x dg 2 f s x d d dx f t s x dg f t s x dg 2 In words, the Quotient Rule says that the derivative of a quotient is the denominator times the derivative of the numerator minus the numerator times the derivative of the denominator, all divided by the square of the denominator. The Quotient Rule and the other differentiation formulas enable us to compute the derivative of any rational function, as the next example illustrates. EXAMPLE 4 | Let y -x 2 1 x 2 2 x 3 1 6 . Then y 9 -s x 3 1 6 d d dx s x 2 1 x 2 2 d 2 s x 2 1 x 2 2 d d dx s x 3 1 6 d s x 3 1 6 d 2 -s x 3 1 6 ds 2 x 1 1 d 2 s x 2 1 x 2 2 ds 3 x 2 d s x 3 1 6 d 2 -s 2 x 4 1 x 3 1 12 x 1 6 d 2 s 3 x 4 1 3 x 3 2 6 x 2 d s x 3 1 6 d 2 -2 x 4 2 2 x 3 1 6 x 2 1 12 x 1 6 s x 3 1 6 d 2 ■ In prime notation: S f t D 9 -t f 9 2 f t 9 t 2 1.5 _1.5 _4 4 yª y FIGURE 3 We can use a graphing device to check that the answer to Example 4 is plau-sible. Figure 3 shows the graphs of the function of Example 4 and its deriva-tive. Notice that when y grows rapidly (near 2 2 ), y 9 is large. And when y grows slowly, y 9 is near 0 . Copyright 2016 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.

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  Calculus

  Single Variable

  • Carl V. Lutzer, H. T. Goodwill(Authors)
  • 2011(Publication Date)
  • Wiley

    (Publisher)

  Helpful Tip#2: The patterns in the 2nd column of the table are the same as in the 1st column, except that we’ve added the sound “co” to those functions that lacked it, and removed the sound from those that had it. Section 3.5 The Product Rule & Quotient Rule 197 z Connection to Percent Rate of Change To understand percent rate of change at an intuitive level, consider this: spending an extra $1000 per month would be a hardship for many college students, but its impact on a multinational company is much less significant because it makes so much more money. Similarly, whether f 0 (t) = 1000 is considered “large” depends on the value of f (t). We make this quantitatively precise by calculating the ratio f 0 (t)/f (t), which is the percent rate of change. The Product Rule and Quotient Rule are nicely and simply related by an idea called percent rate of change. Simply put, the percent rate of change in f is the ratio f 0 (t)/f (t). Provided that both f (t) and g(t) are nonzero: d dt (fg) = (fg)  f 0 f + g 0 g  | {z } and d dt  f g  =  f g   f 0 f - g 0 g  | {z } . (current value)×(sum of percent rates of change) (current value)×(difference of percent rates of change) Note: The derivative of a product is a sum, and the derivative of a quotient is a difference. This relationship (products to sums, and quotients to differences) has a lot to do with the rules of logarithms. You’ll be able to see why later, when we talk about a technique called logarithmic differentiation. z Common Difficulties with the Product Rule and Quotient Rule People sometimes forget that the derivative of a quotient is not the quotient of the derivatives, but it’s easy to check. For example, suppose that g(t) = t and f (t) = t. Then as long as t 6= 0, we have 1 = f (t) g(t) . Since this quotient is constant, its derivative is zero, but the quotient of the derivatives is f 0 (t) g 0 (t) = 1 1 = 1 6= 0. Similarly, the derivative of a product is not the product of the derivatives.

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  Applied Calculus

  • Geoffrey Berresford, Andrew Rockett(Authors)
  • 2015(Publication Date)
  • Cengage Learning EMEA

    (Publisher)

  All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it. 130 Chapter 2 Derivatives and Their Uses The function in this Example was a quotient, so it was perhaps natural to use the Quotient Rule. It is easier, however, to simplify the original function first, x 2 1 1 x 5 x 2 x 1 1 x 5 x 1 x 2 1 and then differentiate by the Power Rule. So, the first derivative of x 1 x 2 1 is 1 2 x 2 2 , and differentiating again gives 2 x 2 3 , agreeing with the answer found by the Quotient Rule. PRACTICE PROBLEM 3 Find f – ( x ) if f ( x ) 5 x 1 1 x . Solution on page 135 > Calculating higher derivatives merely requires repeated use of the same differen-tiation rules that we have been using. EXAMPLE 3 FINDING A SECOND DERIVATIVE USING THE Quotient Rule Find d 2 dx 2 a x 2 1 1 x b . Solution d dx a x 2 1 1 x b 5 x (2 x ) 2 ( x 2 1 1) x 2 First derivative, using the Quotient Rule 5 2 x 2 2 x 2 2 1 x 2 5 x 2 2 1 x 2 Simplifying Differentiating this answer gives the second derivative of the original: d dx a x 2 2 1 x 2 b 5 x 2 (2 x ) 2 2 x ( x 2 2 1) x 4 Second derivative (derivative of the derivative) 5 2 x 3 2 2 x 3 1 2 x x 4 5 2 x x 4 5 2 x 3 Simplifying Answer: d 2 dx 2 a x 2 1 1 x b 5 2 x 3 The Quotient Rule was introduced on page 116. Always try to simplify before differentiating. Take Note EXAMPLE 4 EVALUATING A SECOND DERIVATIVE If f ( x ) 5 1 1 x , find f – a 1 4 b . First differentiate, then evaluate Solution f ( x ) 5 x 2 1 y 2 f ( x ) in power form f 9 ( x ) 5 2 1 2 x 2 3 y 2 Differentiating once Copyright 2016 Cengage Learning.

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  Anton's Calculus

  Early Transcendentals

  • Howard Anton, Irl C. Bivens, Stephen Davis(Authors)
  • 2018(Publication Date)
  • Wiley

    (Publisher)

  Applying the product rule yields ds dt = d dt [(1 + t) √ t] = (1 + t) d dt [ √ t] + √ t d dt [1 + t] = 1 + t 2 √ t + √ t = 1 + 3t 2 √ t DERIVATIVE OF A QUOTIENT Just as the derivative of a product is not generally the product of the derivatives, so the derivative of a quotient is not generally the quotient of the derivatives. The correct relation- ship is given by the following theorem. Formula (2) can also be expressed as ( f g ) ′ = g ⋅ f ′ − f ⋅ g ′ g 2 2.4.2 theorem (The Quotient Rule) If f and g are both differentiable at x and if g(x) ≠ 0, then f ∕ g is differentiable at x and d dx [ f (x) g(x) ] = g(x) d dx [ f (x)] − f (x) d dx [g(x)] [g(x)] 2 (2) proof d dx [ f (x) g(x) ] = lim h→0 f (x + h) g(x + h) − f (x) g(x) h = lim h →0 f (x + h) ⋅ g(x) − f (x) ⋅ g(x + h) h ⋅ g(x) ⋅ g(x + h) Adding and subtracting f (x) ⋅ g(x) in the numerator yields d dx [ f (x) g(x) ] = lim h →0 f (x + h) ⋅ g(x) − f (x) ⋅ g(x) − f (x) ⋅ g(x + h) + f (x) ⋅ g(x) h ⋅ g(x) ⋅ g(x + h) = lim h →0 [ g(x) ⋅ f (x + h) − f (x) h ] − [ f (x) ⋅ g(x + h) − g(x) h ] g(x) ⋅ g(x + h) = lim h →0 g(x) ⋅ lim h →0 f (x + h) − f (x) h − lim h →0 f (x) ⋅ lim h →0 g(x + h) − g(x) h lim h →0 g(x) ⋅ lim h →0 g(x + h) = [ lim h →0 g(x) ] ⋅ d dx [ f (x)] − [ lim h →0 f (x) ] ⋅ d dx [g(x)] lim h →0 g(x) ⋅ lim h →0 g(x + h) = g(x) d dx [ f (x)] − f (x) d dx [g(x)] [g(x)] 2 [See the note at the end of the proof of Theorem 2.4.1 for an explanation of the last step.] 2.4 The Product and Quotient Rules 111 In words, the derivative of a quotient of two functions is the denominator times the derivative of the numerator minus the numerator times the derivative of the denomi- nator, all divided by the denominator squared. Sometimes it is better to simplify a function first than to apply the quo- tient rule immediately. For example, it is easier to differentiate f (x) = x 3∕2 + x √ x by rewriting it as f (x) = x + √ x as opposed to using the Quotient Rule. Example 3 Find y ′ (x) for y = x 3 + 2x 2 − 1 x + 5 .

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  Fast Start Differential Calculus

  • Daniel Ashlock(Author)
  • 2022(Publication Date)
  • Springer

    (Publisher)

  108 3. LIMITS, DERIVATIVES, RULES, AND THE MEANING OF THE DERIVATIVE Knowledge Box 3.14 The reciprocal rule 1 f .x/ 0 D f 0 .x/ f 2 .x/ Example 3.59 Find the derivative of h.x/ D 1 x 2 C 1 . Solution: Apply the reciprocal rule to the function for which the denominator is f .x/ D x 2 C 1. h 0 .x/ D x 2 C 1 0 .x 2 C 1/ 2 D 2x .x 2 C 1/ 2 ˙ Example 3.60 Find the derivative of r.x/ D 1 e x C x 2 . Solution: Apply the reciprocal rule to the function for which the denominator is f .x/ D e x C x 2 . q 0 .x/ D e x C x 2 0 .e x C x 2 / 2 D .e x C 2x/ .e x C x 2 / 2 D e x C 2x .e x C x 2 / 2 ˙ 3.4. THE PRODUCT, QUOTIENT, RECIPROCAL, AND CHAIN RULES 109 3.4.1 FUNCTIONAL COMPOSITION AND THE CHAIN RULE The composition of two functions results from applying one to the other. If the functions are f .x/ and g.x/, then their composition is written f .g.x//. Let’s look at a few examples. Example 3.61 If f .x/ D x C 7 and g.x/ D x 2 , then f .g.x// D x 2 C 7 while g.f .x// D .x C 7/ 2 ˙ Example 3.62 If f .x/ D sin.x/ and g.x/ D e x , then f .g.x// D sin .e x / while g.f .x// D e sin.x/ ˙ The order in which two functions are composed matters a lot—the results are not symmetric. The chain rule is used to compute the derivative of a composition of functions. Knowledge Box 3.15 The chain rule .f .g.x/// 0 D f 0 .g.x// g 0 .x/ In the composition f .g.x// we call f .x/ the outer function and g.x/ the inner function. Example 3.63 Compute the derivative of h.x/ D e 2x . 110 3. LIMITS, DERIVATIVES, RULES, AND THE MEANING OF THE DERIVATIVE Solution: Apply the chain rule to the functional composition for which the outer function is f .x/ D e x , and the inner function is g.x/ D 2x. For these, f 0 .x/ D e x and g 0 .x/ D 2. So: h 0 .x/ D e 2x 2 D 2e 2x ˙ Example 3.64 Compute the derivative of q.x/ D sin x 2 . Solution: Apply the chain rule to the functional composition for which the outer function is f .x/ D sin.x/, and the inner function is g.x/ D x 2 .

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  Calculus for The Life Sciences

  • Sebastian J. Schreiber, Karl J. Smith, Wayne M. Getz(Authors)
  • 2014(Publication Date)
  • Wiley

    (Publisher)

  Armed with these elementary differentiation rules, we might guess that the derivative of a product is the product of the derivatives. The following simple example, however, shows this not the case. Let f (x) = x and g(x) = x 2 , and consider their product p(x) = f (x)g(x) = x 3 Because f  (x) = 1 and g  (x) = 2x, the product of the derivatives is f  (x)g  (x) = (1)(2x) = 2x whereas the actual derivative of p(x) = x 3 is p  (x) = 3x 2 . Hence, our na¨ ıve guess is wrong! It is also easy to show that the derivative of a quotient is not the quotient of the derivatives. The goal of this section is to uncover the correct rules for differenti- ation for products and quotients of functions. Product rule To derive a rule for products, we appeal to our geometric intuition by considering ar- eas where x > 0 and f (x) and g(x) are assumed to be increasing, positive differen- tiable functions of x. Note that the algebraic steps stand alone—without considering area or making the assumptions we made in the previous sentence. Let p(x)    area of rectangle = f (x)    length g(x)    width This product of p can be represented as the area of a rectangle: y g(x) f (x) p(x) x 210 Chapter 3 Derivative Rules and Tools Next, we find p(x + x)    area of larger rectangle = f (x + x)    length g(x + x)    width y (This is drawn with x  0.) g(x) g(x  x) f (x  x) p(x  x) f (x) p(x) x The next step gives us the area of the “inverted L-shaped” region: p(x + x) − p(x) = f (x + x)g(x + x) − f (x)g(x) The key to the proof of the product rule is to rewrite this difference.

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  Calculus: An Applied Approach, Brief, International Metric Edition

  • Ron Larson(Author)
  • 2016(Publication Date)
  • Cengage Learning EMEA

    (Publisher)

  2. State the Quotient Rule (page 120) . For examples of the Quotient Rule, see Examples 4, 5, and 6. 3. Describe a real-life example of how the Quotient Rule can be used to analyze the rate of change of systolic blood pressure (page 123, Example 8) . Copyright 2018 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. WCN 02-300 124 Chapter 2 Differentiation Exercises 2.4 See CalcChat.com for tutorial help and worked-out solutions to odd-numbered exercises. Using the Product Rule In Exercises 1–10, use the Product Rule to find the derivative of the function. See Examples 1, 2, and 3. 1. f ( x ) = ( 2 x -3 )( 1 -5 x ) 2. g ( x ) = ( 4 x -7 )( 3 x + 1 ) 3. f ( x ) = ( 6 x -x 2 )( 4 + 3 x ) 4. f ( x ) = ( 5 x -x 3 )( 2 x + 9 ) 5. f ( x ) = x ( x 2 + 3 ) 6. f ( x ) = x 2 ( 3 x 3 -1 ) 7. h ( x ) = parenleft.alt4 2 x -3 parenright.alt4 ( x 2 + 7 ) 8. f ( x ) = ( 3 -x ) parenleft.alt4 4 x 2 -5 parenright.alt4 9. g ( x ) = ( x 2 -4 x + 3 )( x -2 ) 10. g ( x ) = ( x 2 -2 x + 1 )( x 3 -1 ) Using the Quotient Rule In Exercises 11–20, use the Quotient Rule to find the derivative of the function. See Examples 4 and 6. 11. h ( x ) = x x -5 12. h ( x ) = x 2 x + 3 13. f ( t ) = 2 t 2 -3 3 t + 1 14. f ( x ) = 7 x + 3 4 x -9 15. f ( t ) = t + 6 t 2 -8 16. g ( x ) = 4 x -5 x 2 -1 17. f ( x ) = x 2 + 6 x + 5 2 x -1 18. f ( x ) = 4 x 2 -x + 2 3 -4 x 19. f ( x ) = 6 + ( 2 H20862 x ) 3 x -1 20. f ( x ) = 5 -( 1 H20862 x 2 ) x + 2 Using the Constant Multiple Rule In Exercises 21–30, find the derivative of the function. See Example 7. Original Function Rewrite Differentiate Simplify 21. f ( x ) = x 3 + 6 x 3 uni25A0 uni25A0 uni25A0 22. f ( x ) = x 3 + 2 x 2 10 uni25A0 uni25A0 uni25A0 23. y = 7 x 2 5 uni25A0 uni25A0 uni25A0 24. y = 2 x 4 9 uni25A0 uni25A0 uni25A0 25. y = 7 3 x 3 uni25A0 uni25A0 uni25A0 26. y = 4 5 x 2 uni25A0 uni25A0 uni25A0 27. y = 4 x 2 -3 x 8 radical.alt2 x uni25A0 uni25A0 uni25A0 28.

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  Applied Calculus

  for Business, Life, and Social Sciences

  • Denny Burzynski(Author)
  • 2014(Publication Date)
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  That is, d ___ dx  u(x) ____ v(x)  ≠ u′(x) _____ v′(x) Example 3 (bottom) (top)′ (top) (bottom)′ 2.2 Differentiating Products and Quotients 155 For example, if f (x) = x 3 + 2x − 1 __________ x 2 − 1 then f ′ (x) ≠ 3x 2 + 2 _______ 2x There is a significant difference between the quotient of derivatives, f ′(x) ____ g′(x) , and derivative of a quotient,  f (x) ___ g(x)  ′ . The Quotient Rule gives us that  f (x) ____ g(x)  ′ = g(x) ⋅ f ′(x) − f (x) ⋅ g′(x) ____________________ [g(x)] 2 Comparing expressions, it is hard to imagine that the left and right sides below could ever be equal: g(x) ⋅ f ′(x) − f (x) ⋅ g′(x) ____________________ [g(x)] 2 ≠ f ′(x) _____ g′(x) Thus,  f (x) ____ g(x)  ′ ≠ f ′(x) _____ g′(x) Example 4 illustrates this difference with an application of a revenue function. Two quantities can be compared using division. Division in- dicates how many times more one quantity is than another. For example, 15 __ 3 = 5 indicates that 15 is 5 times as big as 3. Suppose that analyst A uses the revenue model R A (t) to predict the annual revenue for a company t years from now. For her revenue function, she computes R A (3) = 513,684 and R′ A (3) = 4,623. Analyst B uses the revenue model R B (t) to predict the annual revenue of the same company also t years from now. For his revenue function, he computes R B (3) = 512,145 and R′ B (3) = 4,097. a. What information is contained in the quotient R A (3) _____ R B (3) ≈ 1.003 b. What information is contained in the quotient R′ A (3) _____ R′ B (3) = 4,623 _____ 4,097 ≈ 1.13 c. What information is contained in the quotient  R A (t) _____ R B (t)  ′ ≈ 0.001 when t = 3 Example 4 156 Chapter 2 Differentiation: The Language of Change Solution a. This quotient indicates that 3 years from now, the revenue predicted by analyst A will be about 1.003 times as big as that predicted by analyst B.

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