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Unformatted text preview: (3/19/08) Section 2.6 Derivatives of products and quotients Overview: In this section, we derive formulas for derivatives of functions that are constructed by taking products and quotients of other functions, and we use these formulas to study rates of change in a variety of applications. Topics: • The Product Rule • Relatedrate problems • The Quotient Rule • The derivative of y = x n for integers n ≥ 2 and Mathematical Induction The rate of change of the area of a rectangle Imagine that the sides of the rectangle in Figure 1 are changing, so that the width w = w ( t ), height h = h ( t ), and area A ( t ) = w ( t ) h ( t ) of the rectangle are functions of the time t . We want a formula for the rate of change of the area in terms of w , h , and their rates of change. w h w h Δ w Δ h ( I ) ( II ) ( III ) FIGURE 1 FIGURE 2 We consider a nonzero change Δ t in the time from t to t + Δ t and let Δ w and Δ h be the corresponding changes in the width and height. If Δ w and Δ h are positive, then at time t + Δ t , the width is w + Δ w and the height is h + Δ h , as in Figure 2, and the change Δ A in the area from t to t + Δ t is the area of the three rectangles labeled ( I ) , ( II ), and ( III ) in Figure 2. Example 1 Express the areas of rectangles ( I ) , ( II ), and ( III ) in Figure 2 in terms of w,h, Δ w , and Δ h . Solution Rectangle ( I ) is w units wide and Δ h units high and has area w Δ h . Rectangle ( II ) is Δ w units wide and h units high and has area h Δ w . Rectangle ( III ) is Δ w units wide and Δ h units high and has area Δ w Δ h . square The results of Example 1 show that if Δ w and Δ h are positive, then Δ A = w Δ h + h Δ w + Δ w Δ h. (1) To verify (1) in general, we derive it algebraically. The area of a rectangle of width w and height h is wh , and the area of a rectangle of width w + Δ w and height h + Δ h is ( w + Δ w )( h + Δ h ), so the change in the area is Δ A = ( w + Δ w )( h + Δ h ) wh = ( wh + w Δ h + h Δ w + Δ w Δ h ) wh = w Δ h + h Δ w + Δ w Δ h. 137 p. 138 (3/19/08) Section 2.6, Derivatives of products and quotients This establishes equation (1) . We divide both sides of it by the change Δ t in the time to obtain Δ A Δ t = w Δ h Δ t + h Δ w Δ t + Δ w bracketleftbigg Δ h Δ t bracketrightbigg . (2) We suppose that w = w ( t ) and h = h ( t ) have derivatives at t . Then w prime ( t ) = lim Δ t → Δ w Δ t and h prime ( t ) = lim Δ t → Δ h Δ t . Also Δ w → 0 as Δ t → 0 since, by Theorem 1 of Section 2.5, w = w ( t ) is continuous at t . Hence (2) gives A prime ( t ) = lim Δ t → Δ A Δ t = lim Δ t → ( w Δ h Δ t + h Δ w Δ t + Δ w Δ h Δ t ) = w ( t ) h prime ( t ) + h ( t ) w prime ( t ) + (0) h prime ( t ) = w ( t ) h prime ( t ) + h ( t ) w prime ( t ) ....
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This note was uploaded on 09/13/2010 for the course MATH Math 20A taught by Professor Eggers during the Summer '08 term at UCSD.
 Summer '08
 Eggers
 Math, Derivative, Formulas

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