Chain Rule

4 Key excerpts on "Chain 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)

  ■ James Gregory The first person to formulate the Chain Rule was the Scottish mathe- matician James Gregory (1638 –1675), who also designed the first practical reflecting telescope. Gregory discov- ered the basic ideas of calculus at about the same time as Newton. He became the first Professor of Mathe- matics at the University of St. Andrews and later held the same position at the University of Edinburgh. But one year after accepting that position, he died at the age of 36. 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 3.4 The Chain Rule 201 EXAMPLE 1 Find F9 s xd if Fs xd - sx 2 1 1 . SOLUTION 1 (using Formula 1): At the beginning of this section we expressed F as Fs xd - s f 8 tds xd - f s ts xdd where f sud - su and ts xd - x 2 1 1. Since f 9 sud - 1 2 u 21y2 - 1 2 su and t9 s xd - 2x we have F9 s xd - f 9 s ts xdd  t9 s xd - 1 2 sx 2 1 1  2x - x sx 2 1 1 SOLUTION 2 (using Formula 2): If we let u - x 2 1 1 and y - su , then F9 s xd - dy du du dx - 1 2 su s2xd - 1 2 sx 2 1 1 s2xd - x sx 2 1 1 ■ When using Formula 2 we should bear in mind that dyy dx refers to the derivative of y when y is considered as a function of x (the derivative of y with respect to x), whereas dyy du refers to the derivative of y when considered as a function of u (the derivative of y with respect to u). For instance, in Example 1, y can be considered as a function of x ( y - sx 2 1 1 ) and also as a function of u ( y - su ). Note that dy dx - F9 s xd - x sx 2 1 1 whereas dy du - f 9 sud - 1 2 su NOTE In using the Chain Rule we work from the outside to the inside.

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  Calculus Simplified

  • Oscar E. Fernandez(Author)
  • 2019(Publication Date)
  • Princeton University Press

    (Publisher)

  Takeaway: (3.20) is a useful and intuitive way to remember the Chain Rule but not a proof of the Chain Rule. Finally, a note about u ’s. Since function compositions are so ubiquitous, deriva-tive rules are often stated in what many call “ u form.” For example, you will often find the Power Rule stated as d dx ( u n ) = nu n − 1 u (3.21) in calculus textbooks. This “ u form” makes the Power Rule more widely applicable than the u = x version in Theorem 3.4. For instance, it allows us to solve Example 3.26 in one line: d dx ( 3 x − 1 ) 2 = 2 ( 3 x − 1 ) 1 ( 3 x − 1 ) = 2 ( 3 x − 1 )( 3 ) = 6 ( 3 x − 1 ) . 3.11 Differentiation Shortcuts: The Quotient Rule The rule for differentiating a quotient of two functions (i.e., f ( x )/ g ( x ) ) can be de-rived using the Product and Chain Rules (see Exercise 50). Here is the formula that results. Theorem 3.7 The Quotient Rule. Suppose f and g are both differen-tiable, and g ( x ) = 0. Then, d dx f ( x ) g ( x ) = f ( x ) g ( x ) − f ( x ) g ( x ) [ g ( x ) ] 2 . EXAMPLE 3.29 Differentiate h ( x ) = x 2 − 1 x 3 + 1 . Solution h ( x ) = ( x 2 − 1 ) ( x 3 + 1 ) − ( x 2 − 1 )( x 3 + 1 ) ( x 3 + 1 ) 2 Quotient Rule with f ( x ) = x 2 − 1 and g ( x ) = x 3 + 1 = ( 2 x )( x 3 + 1 ) − ( x 2 − 1 )( 3 x 2 ) ( x 3 + 1 ) 2 Sum/Difference and Power Rules = − x ( x 3 − 3 x − 2 ) ( x 3 + 1 ) 2 . Simplifying EXAMPLE 3.30 Differentiate h ( x ) = x + 1 x . 3.12 Transcendental Functions • 69 Solution h ( x ) = ( x + 1 ) ( x ) − ( x + 1 )( x ) x 2 Quotient Rule with f ( x ) = x + 1 and g ( x ) = x = ( 1 )( x ) − ( x + 1 )( 1 ) x 2 = − 1 x 2 . Sum/Difference and Power Rules; simplifying Related Exercises 25, 28, 31, and 33. Tips, Tricks, and Takeaways This last example could have been solved without the Quotient Rule by simplifying the function first: x + 1 x = x x + 1 x = 1 + 1 x = 1 + x − 1 , x = 0. Applying the Sum and Power Rules then yields the same derivative ( − x − 2 ).

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  Calculus

  Concepts and Contexts, Enhanced Edition

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

    (Publisher)

  dx dt dy dt 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. 204 CHAPTER 3 DIFFERENTIATION RULES Figure 2 shows the curve and its tangent line. The tangent line is horizontal when , which occurs when (and ), that is, when or . (Note that the entire curve is given by .) Thus the curve has horizontal tangents at the points , which we could have guessed from Figure 2. The tangent is vertical when (and ), that is, when , , , or . The corresponding four points on the curve are . If we look again at Figure 2, we see that our answer appears to be reasonable. How to Prove the Chain Rule Recall that if and x changes from a to , we defined the increment of y as According to the definition of a derivative, we have So if we denote by the difference between the difference quotient and the derivative, we obtain But If we define to be 0 when , then becomes a continuous function of . Thus, for a differentiable function f , we can write and is a continuous function of . This property of differentiable functions is what enables us to prove the Chain Rule. PROOF OF THE Chain Rule Suppose is differentiable at a and is differ-entiable at . If is an increment in x and and are the corresponding increments in u and y , then we can use Equation 8 to write where as . Similarly where as . If we now substitute the expression for from Equation 9 into Equation 10, we get so As , Equation 9 shows that .

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  The How and Why of One Variable Calculus

  • Amol Sasane(Author)
  • 2015(Publication Date)
  • Wiley

    (Publisher)

  DIFFERENTIATION 141 Theorem 4.5 (Chain Rule) . Suppose that (1) f : ( a , b ) → R , (2) g : ( A , B ) → R , (3) f (( a , b )) ⊂ ( A , B ) , (4) f is differentiable at c ∈ ( a , b ) , and (5) g is differentiable at f ( c ) ∈ ( A , B ) . Then g ◦ f : ( a , b ) → R is differentiable at c, and ( g ◦ f ) ( c ) = g ( f ( c )) · f ( c ) . a b c f A B f ( c ) g g ◦ f g ( f ( c )) Example 4.9. We will see later on that sin x = cos x , x ∈ R . Using the Chain Rule, we obtain d dx sin( x 2 ) = cos( x 2 ) · 2 x , x ∈ R , and d dx sin 1 x = cos 1 x · − 1 x 2 , x = 0. The Chain Rule can be applied repeatedly (forming a ‘chain’ of derivatives—hence the name of the rule!): for example, d dx sin sin( x 2 ) = cos sin( x 2 ) · cos( x 2 ) · 2 x , x ∈ R . The above examples show the great power associated with this rule. ♦ Proof of the Chain Rule. For x = c , we have ( g ◦ f )( x ) − ( g ◦ f )( c ) x − c = g ( f ( x )) − g ( f ( c )) x − c , and we would like to show that as x → c , the above converges to g ( f ( c )) · f ( c ) . It is tempting to multiply and divide by f ( x ) − f ( c ) and write g ( f ( x )) − g ( f ( c )) x − c = g ( f ( x )) − g ( f ( c )) f ( x ) − f ( c ) · f ( x ) − f ( c ) x − c and say ‘as x → c , f ( x ) − f ( c ) x − c → f ( c ) , f ( x ) → f ( c ) , and g ( f ( x )) − g ( f ( c )) f ( x ) − f ( c ) → g ( f ( c )) , completing the proof’ . 142 THE HOW AND WHY OF ONE VARIABLE CALCULUS However, even for x = c , and near c , it may happen that f ( x ) − f ( c ) = 0 (for example when f ≡ f ( c )) , and so division by f ( x ) − f ( c ) is not possible then, rendering the above invalid. So instead, we proceed as follows. Since g is differentiable at f ( c ) , lim y → f ( c ) g ( y ) − g ( f ( c )) y − f ( c ) − g ( f ( c )) = 0. So if we define ϕ : ( A , B ) → R by ϕ ( y ) = ⎧ ⎪ ⎨ ⎪ ⎩ g ( y ) − g ( f ( c )) y − f ( c ) − g ( f ( c )) if y = f ( c ) , 0 if y = f ( c ) , then ϕ is continuous at f ( c ) (in fact on ( A , B ) ).

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