Mathematics
Product Rule
The product rule in mathematics is a formula used to find the derivative of a product of two functions. It states that the derivative of the product of two functions is equal to the first function times the derivative of the second function, plus the second function times the derivative of the first function. This rule is a fundamental tool in calculus for finding the derivatives of products of functions.
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6 Key excerpts on "Product Rule"
eBook - PDFBiocalculus
Calculus, Probability, and Statistics for the Life Sciences
- James Stewart, Troy Day, James Stewart(Authors)
- 2015(Publication Date)
- Cengage Learning EMEA(Publisher)
u Î√ Î√ √ u√ u Îu Î√ √ Îu Îu FIGURE 1 The geometry of the Product Rule Recall that in Leibniz notation the defi-nition of a derivative can be written as dy dx -lim D x l 0 D y D x In prime notation: s f t d 9 -f t 9 1 t f 9 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. 196 CHAPTER 3 | Derivatives EXAMPLE 1 | Differentiate y -x 2 sin x . SOLUTION Using the Product Rule, we have dy dx -x 2 d dx s sin x d 1 sin x d dx s x 2 d -x 2 cos x 1 2 x sin x ■ EXAMPLE 2 | Differentiate the function f s t d -s t s a 1 bt d . SOLUTION 1 Using the Product Rule, we have f 9 s t d -s t d dt s a 1 bt d 1 s a 1 bt d d dt ( s t ) -s t ? b 1 s a 1 bt d ? 1 2 t 2 1 y 2 -b s t 1 a 1 bt 2 s t -a 1 3 bt 2 s t SOLUTION 2 If we first use the laws of exponents to rewrite f s t d , then we can proceed directly without using the Product Rule. f s t d -a s t 1 bt s t -at 1 y 2 1 bt 3 y 2 f 9 s t d -1 2 at 2 1 y 2 1 3 2 bt 1 y 2 which is equivalent to the answer given in Solution 1. ■ Example 2 shows that it is sometimes easier to simplify a product of functions before differentiating than to use the Product Rule. In Example 1, however, the Product Rule is the only possible method. EXAMPLE 3 | If h s x d -x t s x d and it is known that t s 3 d -5 and t 9 s 3 d -2 , find h 9 s 3 d .
- Kevin Corner, Leslie Jackson, William Embleton(Authors)
- 2013(Publication Date)
- Thomas Reed(Publisher)
13 ADVANCED CALCULUS: METHODS OF DIFFERENTIATION Differentiation – The Product Rule When two separate functions of a variable are multiplied they can only be differentiated if the Product Rule is applied. The Product Rule Let y = u · v where u and v are functions of the same variable, x , for instance. Then to differentiate y the following rule applies: d y d x = u · d v d x + v · d u d x or sometimes d y d x = u · d v + v · d u Advanced Calculus: Methods of Differentiation • 315 Example (set all questions this way to aid memory) Let y = x 3 · sin( x ) Solution u = x 3 v = sin( x ) u x = 3 x 2 d d cos( x ) d v = d x so d y d x = x 3 · cos( x ) + 3 x 2 · sin( x ) If possible the answer should be simplified as far as possible. So d y d x = x 2 ( x · cos( x ) + 3 sin( x )) Questions Differentiate the following: 1. y = 3 x 2 · cos( x ) 2. y = sin( x ) · cos( x ) 3. y = √ x · cos( x ) 4. y = ( 3 x 2 + 2 ) · ( x 2 − 2 ) 5. y = 5e 2 x · sin( x ) 6. y = e 7 x · cos( x ) 7. y = x 3 · ln( 2 x ) 8. y = 3 √ x · e 4 x 9. y = e x · ln( x ) 10. y = e 2 x ( 4 x 2 − 4 x + 1 ) 11. y = 3 √ x 7 · ln( 5 x ) 12. y = 6e 5 x · sin( x ) Answers 1. d y d x = 6 x cos( x ) − 3 x 2 sin( x ) = 3 x ( 2 cos( x ) − x sin( x )) 2. d y d x = cos 2 ( x ) − sin 2 ( x ) 3. d y d x = 1 2 x − 1 2 cos( x ) − √ x sin( x ) = cos( x ) − 2 x sin( x ) 2 √ x 4. d y d x = ( 6 x ) ( x 2 − 2 ) + ( 3 x 2 + 2 ) ( 2 x ) = 6 x 3 − 12 x + 6 x 3 + 4 x = 12 x 3 − 8 x 5. d y d x = 10e 2 x sin( x ) + 5e 2 x cos( x ) = 5e 2 x ( 2 sin( x ) + cos( x )) 6. d y d x = 7e 7 x cos( x ) − e 7 x sin( x ) = e 7 x ( 7 cos( x ) − sin( x )) 316 • Mathematics 7. d y d x = 3 x 2 ln( 2 x ) + x 3 · 1 x = x 2 ( 3 ln( 2 x ) + 1 ) 8. d y d x = 3 √ x 4e 4 x + 3 2 x − 1 2 · e 4 x = 3e 4 x ( 8 x + 1 ) 2 √ x 9. d y d x = e x ln( x ) + e x · 1 x = e x ln( x ) + 1 x 10.
eBook - PDFSingle 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)
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 3.3 Derivatives of Trigonometric Functions 191 63. Extended Product Rule The Product Rule can be extended to the product of three functions. (a) Use the Product Rule twice to prove that if f , t, and h are differentiable, then s f thd9 - f 9 th 1 f t9h 1 f th9. (b) Taking f - t - h in part (a), show that d dx f f s xdg 3 - 3f f s xdg 2 f 9 s xd (c) Use part (b) to differentiate y - e 3x . 64. (a) If Fs xd - f s xd ts xd, where f and t have derivatives of all orders, show that F99 - f 99t 1 2 f 9 t9 1 f t99. (b) Find similar formulas for F999 and F s4d . (c) Guess a formula for F snd . 65. Find expressions for the first five derivatives of f s xd - x 2 e x . Do you see a pattern in these expressions? Guess a formula for f snd s xd and prove it using mathematical induction. 66. Reciprocal Rule If t is differentiable, the Reciprocal Rule says that d dx F 1 ts xd G - 2 t9 s xd f ts xdg 2 (a) Use the Quotient Rule to prove the Reciprocal Rule. (b) Use the Reciprocal Rule to differentiate the function in Exercise 14. (c) Use the Reciprocal Rule to verify that the Power Rule is valid for negative integers, that is, d dx sx 2n d - 2nx 2n21 for all positive integers n. was increasing at about $2250 per year (a little above the national average of about $1810 yearly). Use the Product Rule and these figures to estimate the rate at which total personal income was rising in Boulder in 2015. Explain the meaning of each term in the Product Rule. 60. A manufacturer produces bolts of a fabric with a fixed width.
- James Stewart, Daniel K. Clegg, Saleem Watson, , James Stewart, James Stewart, Daniel K. Clegg, Saleem Watson(Authors)
- 2020(Publication Date)
- Cengage Learning EMEA(Publisher)
SECTION 3.2 The Product and Quotient Rules 187 Example 2 shows that it is sometimes easier to simplify a product of functions before differentiating than to use the Product Rule. In Example 1, however, the Product Rule is the only possible method. EXAMPLE 3 If f s xd - sx ts xd, where ts4d - 2 and t9 s4d - 3, find f 9 s4d. SOLUTION Applying the Product Rule, we get f 9 s xd - d dx fsx ts xdg - sx d dx f ts xdg 1 ts xd d dx fsx g - sx t9 s xd 1 ts xd ∙ 1 2 x 21y2 - sx t9 s xd 1 ts xd 2 sx So f 9 s4d - s4 t9 s4d 1 ts4d 2 s4 - 2 ∙ 3 1 2 2 ∙ 2 - 6.5 ■ ■ The Quotient Rule We find a rule for differentiating the quotient of two differentiable functions u - f s xd and v - ts xd in much the same way that we found the Product Rule. If x, u, and v change by amounts Dx, Du, and Dv, then the corresponding change in the quotient uyv is D S u v D - u 1 Du v 1 Dv 2 u v - su 1 Dudv 2 usv 1 Dvd v sv 1 Dvd - v Du 2 u Dv vsv 1 Dvd so d dx S u v D - lim D x l0 Dsuyvd D x - lim D x l0 v Du D x 2 u Dv D x vsv 1 Dvd As D x l 0, Dv l 0 also, because v - ts xd is differentiable and therefore continuous. Thus, using the Limit Laws, we get d dx S u v D - v lim D x l0 Du D x 2 u lim D x l0 Dv D x v lim D x l0 sv 1 Dvd - v du dx 2 u dv dx v 2 The Quotient Rule If f and t are differentiable, then d dx F f s xd ts xd G - ts xd d dx f f s xdg 2 f s xd d dx f ts xdg f ts xdg 2 In prime notation the Quotient Rule is written as S f t D 9 - t f 9 2 f t9 t 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. 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).
eBook - PDF- James Stewart, Daniel K. Clegg, Saleem Watson, , James Stewart, James Stewart, Daniel K. Clegg, Saleem Watson(Authors)
- 2020(Publication Date)
- Cengage Learning EMEA(Publisher)
Also, since f is differentiable at x, it is continuous at x by Theorem 2.2.4, and so lim h l 0 f s x 1 hd - f s xd. (See Exercise 1.8.65.) n In words, the Product Rule says that the derivative of a product of two functions is the first function times the derivative of the second function plus the second function times the derivative of the first function. EXAMPLE 6 Find F9 s xd if Fs xd - s6x 3 ds7x 4 d. SOLUTION By the Product Rule, we have F9 s xd - s6x 3 d d dx s7x 4 d 1 s7x 4 d d dx s6x 3 d - s6x 3 ds28x 3 d 1 s7x 4 ds18x 2 d - 168x 6 1 126x 6 - 294x 6 n Notice that we could verify the answer to Example 6 directly by first multiplying the factors: Fs xd - s6x 3 ds7x 4 d - 42x 7 ? F9 s xd - 42s7x 6 d - 294x 6 But later we will meet functions, such as y - x 2 sin x, for which the Product Rule is the only possible method. EXAMPLE 7 If hs xd - x ts xd and it is known that ts3d - 5 and t9 s3d - 2, find h9 s3d. SOLUTION Applying the Product Rule, we get h9 s xd - d dx f x ts xdg - x d dx f ts xdg 1 ts xd d dx f xg - x t9 s xd 1 ts xd s1d Therefore h9 s3d - 3t9 s3d 1 ts3d - 3 2 1 5 - 11 n 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. SECTION 2.3 Differentiation Formulas 139 ■ The Quotient Rule The following formula tells us how to differentiate the quotient of two functions. The Quotient Rule If f and t are differentiable, then d dx F f s xd ts xd G - ts xd d dx f f s xdg 2 f s xd d dx f ts xdg f ts xdg 2 In prime notation the Quotient Rule is written as S f t D 9 - t f 9 2 f t9 t 2 PROOF Let Fs xd - f s xdy ts xd.
eBook - PDFApplied Calculus
for Business, Life, and Social Sciences
- Denny Burzynski(Author)
- 2014(Publication Date)
- XYZ Textbooks(Publisher)
Solution If we identify u (x) = 5x 2 + 4 and v (x) = x 3 + 11, we have the product f (x) = u (x)v (x), and f ′ (x) = u′(x) ⋅ v(x) + u(x) ⋅ v′(x) = (10x) ⋅ (x 3 + 11) + (5x 2 + 4) ⋅ (3x 2 ) The Product Rule If u(x) and v(x) are differentiable functions, then d ___ dx [u(x) ⋅ v(x)] = d ___ dx [u(x)] ⋅ v(x) + u(x) ⋅ d ___ dx [v(x)] (first)′ (second) (first) (second)′ VIDEO EXAMPLES SECTION 2.2 Example 1 (first)′ (second) (first) (second)′ (first)′ (second) (first) (second)′ Differentiating Products and Quotients 152 Chapter 2 Differentiation: The Language of Change = 10x 4 + 110x + 15x 4 + 12x 2 = 25x 4 + 12x 2 + 110x Thus, f ′ (x) = 25x 4 + 12x 2 + 110x. Note In addition to illustrating the Product Rule, Example 1 illustrates two impor- tant facts. 1. The derivative of a product could also be found by performing the multiplica- tion, then differentiating term-by-term using the sum rule. f (x) = (5x 2 + 4)(x 3 + 11) = 5x 5 + 4x 3 + 55x 2 + 44 so that f ′ (x) = 25x 4 + 12x 2 + 110x as before. It is not always practical or possible to perform the multiplication. For exam- ple, it is impossible to multiply out (5x + 4) 7 (x 3 + 11) −2/3 . 2. A note of caution: the derivative of a product is not the product of the individual derivatives. That is, d ___ dx [u(x)v(x)] ≠ u′ (x) ⋅ v′ (x) For example, if f (x) = (5x 2 + 4)(x 3 + 11) then f ′ (x) ≠ (10x)(3x 2 ) For the function g(t) = (t 2 − 4)(t 2 + 1), find a. the values of t for which g (t) = 0. b. the values of t for which g ′ (t) = 0. Solution a. We set g (t) equal to 0, and solve for t. (t 2 − 4)(t 2 + 1) = 0 (t + 2)(t − 2)(t 2 + 1) = 0 t + 2 = 0 or t − 2 = 0 or t 2 + 1 = 0 t = −2 or t = 2 or no real solution The function g(t) is 0 when t = −2, and when t = 2. b. First we differentiate g with respect to t using the Product Rule: g ′ (t) = d __ dt (t 2 − 4)(t 2 + 1) + (t 2 − 4) d __ dt (t 2 + 1) = (2t)(t 2 + 1) + (t 2 − 4)(2t) = 2t 3 + 2t + 2t 3 − 8t = 4t 3 − 6t Example 2
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