The Power Rule

6 Key excerpts on "The Power Rule"

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  eBook - PDF

  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)

  ■ We illustrate The Power Rule using various notations in Example 1. The Binomial Theorem is given on Reference Page 1. 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. 176 CHAPTER 3 Differentiation Rules EXAMPLE 1 (a) If f s xd - x 6 , then f 9 s xd - 6x 5 . (b) If y - x 1000 , then y9 - 1000x 999 . (c) If y - t 4 , then dy dt - 4t 3 . (d) d dr sr 3 d - 3r 2 ■ What about power functions with negative integer exponents? In Exercise 69 we ask you to verify from the definition of a derivative that d dx S 1 x D - 2 1 x 2 We can rewrite this equation as d dx s x 21 d - s21d x 22 and so The Power Rule is true when n - 21. In fact, we will show in the next section [Exercise 3.2.66(c)] that it holds for all negative integers. What if the exponent is a fraction? In Example 2.8.3 we found that d dx sx - 1 2 sx which can be written as d dx s x 1y2 d - 1 2 x 21y2 This shows that The Power Rule is true even when the exponent is 1 2 . In fact, we will show in Section 3.6 that it is true for all real exponents n. The Power Rule (General Version) If n is any real number, then d dx s x n d - nx n21 EXAMPLE 2 Differentiate: (a) f s xd - 1 x 2 (b) y - s 3 x 2 SOLUTION In each case we rewrite the function as a power of x. (a) Since f s xd - x 22 , we use The Power Rule with n - 22: f 9 s xd - d dx s x 22 d - 22x 2221 - 22x 23 - 2 2 x 3 (b) dy dx - d dx (s 3 x 2 ) - d dx s x 2y3 d - 2 3 x s2y3d21 - 2 3 x 21y3 ■ Figure 3 shows the function y in Example 2(b) and its derivative y9. Notice that y is not differentiable at 0 (y9 is not defined there).

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  Finite Mathematics and Applied Calculus

  • Stefan Waner, Steven Costenoble(Authors)
  • 2017(Publication Date)
  • Cengage Learning EMEA

    (Publisher)

  This pattern generalizes to any power of x . Theorem 11.1 The Power Rule If n is any constant and f 1 x 2 5 x n , then f r 1 x 2 5 nx n 2 1 . Quick Examples 1. If f 1 x 2 5 x 2 , then f r 1 x 2 5 2 x 1 5 2 x . 2. If f 1 x 2 5 x 3 , then f r 1 x 2 5 3 x 2 . 3. If f 1 x 2 5 x , rewrite ✱ as f 1 x 2 5 x 1 , so f r 1 x 2 5 1 x 0 5 1 . 4. If f 1 x 2 5 1 , rewrite as f 1 x 2 5 x 0 , so f r 1 x 2 5 0 x 2 1 5 0 . Precalculus Review For this chapter you should be familiar with the algebra reviewed in Sections 0.3 and 0.4 . ✱ To use The Power Rule, we rewrite expressions like this in power form: constant times x n . See Sec-tion 0.2 in the Precalculus Review to brush up on negative and frac-tional exponents. Pay particular attention to radical, positive exponent, and power forms. Copyright 2018 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. WCN 02-300 11.1 Derivatives of Powers, Sums, and Constant Multiples 823 The proof of The Power Rule involves first studying the case when n is a positive integer and then studying the cases of other types of exponents (negative integer, rational number, irrational number). You can find a proof at the Website. Using The Power Rule for Negative and Fractional Exponents Calculate the derivatives of the following: a. f 1 x 2 5 1 x b. f 1 x 2 5 1 x 2 c. f 1 x 2 5 ! x Solution a. Rewrite the function in power form as f 1 x 2 5 x 2 1 . Then f r 1 x 2 5 1 2 1 2 x 2 2 5 2 1 x 2 . b. Rewrite the function in power form as f 1 x 2 5 x 2 2 . Then f r 1 x 2 5 1 2 2 2 x 2 3 5 2 2 x 3 . c. Rewrite the function in power form as f 1 x 2 5 x 0.5 . Then f r 1 x 2 5 0.5 x 2 0.5 5 0.5 x 0.5 . Alternatively, rewrite f 1 x 2 as x 1 > 2 , so that f r 1 x 2 5 1 2 x 2 1 > 2 5 1 2 x 1 > 2 5 1 2 ! x . Caution We cannot apply The Power Rule to terms in the denominators or under square roots. For example: 1. The derivative of 1 x 2 is NOT 1 2 x ; it is 2 2 x 3 .

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  eBook - PDF

  Applied Calculus

  • Stefan Waner, Steven Costenoble(Authors)
  • 2017(Publication Date)
  • Cengage Learning EMEA

    (Publisher)

  This pattern generalizes to any power of x. Theorem 4.1 The Power Rule If n is any constant and f 1 x 2 5 x n , then f r 1 x 2 5 nx n21 . Quick Examples 1. If f 1 x 2 5 x 2 , then f r 1 x 2 5 2x 1 5 2x. 2. If f 1 x 2 5 x 3 , then f r 1 x 2 5 3x 2 . 3. If f 1 x 2 5 x, rewrite ✱ as f 1 x 2 5 x 1 , so f r 1 x 2 5 1x 0 5 1. 4. If f 1 x 2 5 1, rewrite as f 1 x 2 5 x 0 , so f r 1 x 2 5 0x 21 5 0. Precalculus Review For this chapter you should be familiar with the algebra reviewed in Sections 0.3 and 0.4. ✱ To use The Power Rule, we rewrite expressions like this in power form: constant times x n . See Sec- tion 0.2 in the Precalculus Review to brush up on negative and frac- tional exponents. Pay particular attention to radical, positive exponent, and power forms. Copyright 2018 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. WCN 02-300 4.1 Derivatives of Powers, Sums, and Constant Multiples 305 The proof of The Power Rule involves first studying the case when n is a positive integer and then studying the cases of other types of exponents (negative integer, rational number, irrational number). You can find a proof at the Website. Using The Power Rule for Negative and Fractional Exponents Calculate the derivatives of the following: a. f 1 x 2 5 1 x b. f 1 x 2 5 1 x 2 c. f 1 x 2 5 !x Solution a. Rewrite the function in power form as f 1 x 2 5 x 21 . Then f r 1 x 2 5 1 21 2 x 22 5 2 1 x 2 . b. Rewrite the function in power form as f 1 x 2 5 x 22 . Then f r 1 x 2 5 1 22 2 x 23 5 2 2 x 3 . c. Rewrite the function in power form as f 1 x 2 5 x 0.5 . Then f r 1 x 2 5 0.5x 20.5 5 0.5 x 0.5 . Alternatively, rewrite f 1 x 2 as x 1>2 , so that f r 1 x 2 5 1 2 x 21>2 5 1 2x 1>2 5 1 2 !x . Caution We cannot apply The Power Rule to terms in the denominators or under square roots. For example: 1. The derivative of 1 x 2 is NOT 1 2x ; it is 2 2 x 3 . See Example 1(b). 2. The derivative of "x 3 is NOT "3x 2 ; it is 1.5x 0.5 .

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  Calculus: Single and Multivariable

  • Deborah Hughes-Hallett, William G. McCallum, Andrew M. Gleason, Eric Connally, Daniel E. Flath, Selin Kalaycioglu, Brigitte Lahme, Patti Frazer Lock, David O. Lomen, David Lovelock, Guadalupe I. Lozano, Jerry Morris, David Mumford, Brad G. Osgood, Cody L. Patterson, Douglas Quinney, Karen R. Rhea, Ayse Arzu Sahin, Ad(Authors)
  • 2017(Publication Date)
  • Wiley

    (Publisher)

  132 Chapter 3 SHORT-CUTS TO DIFFERENTIATION The Power Rule For any constant real number ,   (  ) =  −1 . Problem 118 (available online) asks you to show that this rule holds for negative integral powers; such powers can also be differentiated using the quotient rule (Section 3.3). In Section 3.6 we indicate how to justify The Power Rule for powers of the form 1∕. Example 1 Use The Power Rule to differentiate (a) 1  3 , (b)  1∕2 , (c) 1 3 √  . Solution (a) For  = −3:   ( 1  3 ) =   ( −3 ) = −3 −3−1 = −3 −4 = − 3  4 . (b) For  = 1∕2:   (  1∕2 ) = 1 2  (1∕2)−1 = 1 2  −1∕2 = 1 2 √  . (c) For  = −1∕3:   ( 1 3 √  ) =   (  −1∕3 ) = − 1 3  (−1∕3)−1 = − 1 3  −4∕3 = − 1 3 4∕3 . Example 2 Use the definition of the derivative to justify The Power Rule for  = −2: Show   ( −2 ) = −2 −3 . Solution Provided  ≠ 0, we have   (  −2 ) =   ( 1  2 ) = lim ℎ→0 ⎛ ⎜ ⎜ ⎝ 1 (+ℎ) 2 − 1  2 ℎ ⎞ ⎟ ⎟ ⎠ = lim ℎ→0 1 ℎ [  2 − ( + ℎ) 2 ( + ℎ) 2  2 ] (Combining fractions over a common denominator) = lim ℎ→0 1 ℎ [  2 − ( 2 + 2ℎ + ℎ 2 ) ( + ℎ) 2  2 ] (Multiplying out) = lim ℎ→0 −2ℎ − ℎ 2 ℎ( + ℎ) 2  2 (Simplifying numerator) = lim ℎ→0 −2 − ℎ ( + ℎ) 2  2 (Dividing numerator and denominator by ℎ) = −2  4 (Letting ℎ → 0) = −2 −3 . The graphs of  −2 and its derivative, −2 −3 , are shown in Figure 3.4. Does the graph of the derivative have the features you expect? 3.1 POWERS AND POLYNOMIALS 133  −2  −2 −3  Figure 3.4: Graphs of  −2 and its derivative, −2 −3 Justification of   (   ) =  − , for  a Positive Integer To calculate the derivatives of  2 and  3 , we had to expand ( + ℎ) 2 and ( + ℎ) 3 . To calculate the derivative of   , we must expand ( + ℎ)  .

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  Calculus

  Single Variable

  • Deborah Hughes-Hallett, William G. McCallum, Andrew M. Gleason, Eric Connally, Daniel E. Flath, Selin Kalaycioglu, Brigitte Lahme, Patti Frazer Lock, David O. Lomen, David Lovelock, Guadalupe I. Lozano, Jerry Morris, Brad G. Osgood, Cody L. Patterson, Douglas Quinney, Karen R. Rhea, Ayse Arzu Sahin, Adam H. Spiegler(Authors)
  • 2017(Publication Date)
  • Wiley

    (Publisher)

  132 Chapter 3 SHORT-CUTS TO DIFFERENTIATION The Power Rule For any constant real number ,   (  ) =  −1 . Problem 118 (available online) asks you to show that this rule holds for negative integral powers; such powers can also be differentiated using the quotient rule (Section 3.3). In Section 3.6 we indicate how to justify The Power Rule for powers of the form 1∕. Example 1 Use The Power Rule to differentiate (a) 1  3 , (b)  1∕2 , (c) 1 3 √  . Solution (a) For  = −3:   ( 1  3 ) =   ( −3 ) = −3 −3−1 = −3 −4 = − 3  4 . (b) For  = 1∕2:   (  1∕2 ) = 1 2  (1∕2)−1 = 1 2  −1∕2 = 1 2 √  . (c) For  = −1∕3:   ( 1 3 √  ) =   (  −1∕3 ) = − 1 3  (−1∕3)−1 = − 1 3  −4∕3 = − 1 3 4∕3 . Example 2 Use the definition of the derivative to justify The Power Rule for  = −2: Show   ( −2 ) = −2 −3 . Solution Provided  ≠ 0, we have   (  −2 ) =   ( 1  2 ) = lim ℎ→0 ⎛ ⎜ ⎜ ⎝ 1 (+ℎ) 2 − 1  2 ℎ ⎞ ⎟ ⎟ ⎠ = lim ℎ→0 1 ℎ [  2 − ( + ℎ) 2 ( + ℎ) 2  2 ] (Combining fractions over a common denominator) = lim ℎ→0 1 ℎ [  2 − ( 2 + 2ℎ + ℎ 2 ) ( + ℎ) 2  2 ] (Multiplying out) = lim ℎ→0 −2ℎ − ℎ 2 ℎ( + ℎ) 2  2 (Simplifying numerator) = lim ℎ→0 −2 − ℎ ( + ℎ) 2  2 (Dividing numerator and denominator by ℎ) = −2  4 (Letting ℎ → 0) = −2 −3 . The graphs of  −2 and its derivative, −2 −3 , are shown in Figure 3.4. Does the graph of the derivative have the features you expect? 3.1 POWERS AND POLYNOMIALS 133  −2  −2 −3  Figure 3.4: Graphs of  −2 and its derivative, −2 −3 Justification of   (   ) =  − , for  a Positive Integer To calculate the derivatives of  2 and  3 , we had to expand ( + ℎ) 2 and ( + ℎ) 3 . To calculate the derivative of   , we must expand ( + ℎ)  .

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  eBook - ePub

  What is Calculus?

  From Simple Algebra to Deep Analysis

  • R Michael Range(Author)
  • 2015(Publication Date)
  • WSPC

    (Publisher)

  differentiable functions. In fact, except for a minor—though most critical—additional argument, the same proofs will work in that most general case.

  6.1      Elementary Rules

  We begin with the simplest rules, whose verification is straightforward.

  Rule 0 (Power Rule). If n ≥ 0 is an integer, then (

  xn

  )′ = nx

  n −1

  . This is the rule we established already at the end of Section 5 .

  Rule I (Linearity).

  (1)  D (cf ) = c D (f ) for any constant c .

  (2)  D (f ± g ) = D (f ) ± D (g ).

  Rules 0 and I allow us to easily find the derivative of any polynomial P .

  Examples.

  In general, if , then

  is a polynomial of degree one less than the degree of P .

  The verification of Rule I is straightforward. We prove Rule I.2, and leave Rule I.1 to the reader. Consider the factorizations f (x ) − f (a ) =

  qf

  (x )(x − a ) and g (x ) − g (a ) =

  qg

  (x )(x − a ). Then

  It follows that (f + g )′(a ) = [

  qf

  +

  qg

  ](a ) =

  qf

  (a ) +

  qg

  (a ) = f ′(a ) + g ′(a ). The proof with − instead of + works exactly the same way.

  Rule II (Chain Rule). Recall that for two functions f and g , the composition f g of f and g is defined by evaluating first g and then inserting the output into f , i.e., (f g )(x ) = f (g (x )). Since we allow the functions to be rational, one must limit the input x to values a for which g is defined and so that f is defined at b = g (a ). (If both f and g are polynomials , there is no restriction on x .) The chain rule then states that

  By using functional notation, the chain rule can be written D (f g ) = (D (f ) g ) · D (g ). The crux of the matter is that the derivative of a composition is the product of the derivatives . The proof is very simple and natural. As before, we write f (y ) − f (b ) =

  qf

  (y )(y − b ) and g (x ) − g (a ) =

  qg

  (x )(x − a ), where

  qf

  and

  qg

  are the appropriate rational factors, and substitute y = g (x ) and b = g (a ) to obtain

  Since

  qf

  (g (x ))

  qg

  (x ) is a rational function defined at a , it follows that

  as claimed.

  Examples. i) Suppose F (x ) = (3x 3 − 5x 2 + 2)10 . We could expand F into standard polynomial form by the binomial theorem and apply rules 0 and I to find the derivative. However, this involves a messy algebraic computation, and the simple structure of F and of its derivative would be lost. Instead, we note that F is the composition F = f g of the simpler functions f (y ) = y 10 and g (x ) = 3x 3 − 5x 2

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