Product Rule MCV4U: Calculus & Vectors Recap x . Determine the - PDF document

f u n d a m e n t a l r u l e s o f d e r i v a t i v e s f u n d a m e n t a l r u l e s o f d e r i v a t i v e s Product Rule MCV4U: Calculus & Vectors Recap √ x . Determine the derivative of f ( x ) = ( x 4 − 3 x 2 + 7 x ) 3 Let p ( x ) = x 4 − 3 x 2 + 7 x and q ( x ) = x 1 3 . 3 x − 2 Then p ′ ( x ) = 4 x 3 − 6 x + 7 and q ′ ( x ) = 1 3 . Chain Rule of Derivatives � � 1 � � 3 x − 2 � � x 4 − 3 x 2 + 7 x 4 x 3 − 6 x + 7 1 f ′ ( x ) = � + � x 3 3 J. Garvin 10 4 1 or 13 3 + 28 3 − 7 x 3 x 3 x 3 J. Garvin — Chain Rule of Derivatives Slide 1/15 Slide 2/15 f u n d a m e n t a l r u l e s o f d e r i v a t i v e s f u n d a m e n t a l r u l e s o f d e r i v a t i v e s Chain Rule Chain Rule Sometimes we need to differentiate a composite function , A comprehensive proof of the chain rule is rather f ( x ) = g ◦ h = g ( h ( x )). complicated, so consider an informal “argument” that explains the chain rule instead. In this case, there is an “inner” function, h ( x ), and an “outer” function, g ( x ). This average rate of change is defined by two quantities, ∆ i and ∆ d , which represent changes in the independent and For instance, the function f ( x ) = 2( x − 3) 2 can be thought dependent variables respectively. of as being composed of the inner function, x − 3, and the outer function, 2 x 2 . If y is a function that depends on another function u , such that y = f ( u ( x )), then values of y will change according to To differentiate these types of functions, use the Chain Rule. changes in u , just as values of u will change according to Chain Rule changes in x . If f ( x ) = g ( h ( x )), then f ′ ( x ) = g ′ ( h ( x )) · h ′ ( x ). If ∆ u , ∆ y and ∆ x represent changes in the variables, then the average rate of change in function y will be ∆ y ∆ u , and the If y = f ( u ) and u = g ( x ), then dy dx = dy du · du dx . average rate of change in function u will be ∆ u ∆ x . J. Garvin — Chain Rule of Derivatives J. Garvin — Chain Rule of Derivatives Slide 3/15 Slide 4/15 f u n d a m e n t a l r u l e s o f d e r i v a t i v e s f u n d a m e n t a l r u l e s o f d e r i v a t i v e s Chain Rule Chain Rule Since all of these changes are quantities, the relationship As ∆ x → 0, ∆ u → 0. ∆ y ∆ x = ∆ y ∆ u · ∆ u ∆ x will hold. ∆ u → 0 ∆ y lim ∆ x → 0 ∆ u lim Therefore, dy dx = ∆ x → 0 ∆ x . ∆ u → 0 ∆ u · Recall that the derivative is an expression that represents the lim lim instantaneous rate of change of a function at a specific point. ∆ u → 0 ∆ y lim ∆ x → 0 ∆ u lim ∆ y Thus, dy Note that dy ∆ u → 0 ∆ u and du dx = lim ∆ x , where ∆ x and∆ y are ( x + h ) − x = h du = dx = ∆ x → 0 ∆ x . ∆ x → 0 lim lim and f ( x + h ) − f ( x ) respectively. Therefore, dy dx = dy du · du ∆ y ∆ u · ∆ u dx as required. Substituting the above relationship, dy dx = lim ∆ x . ∆ x → 0 ∆ x → 0 ∆ y lim ∆ x → 0 ∆ u lim Using limit properties, dy dx = ∆ x → 0 ∆ x . ∆ x → 0 ∆ u · lim lim J. Garvin — Chain Rule of Derivatives J. Garvin — Chain Rule of Derivatives Slide 5/15 Slide 6/15

f u n d a m e n t a l r u l e s o f d e r i v a t i v e s f u n d a m e n t a l r u l e s o f d e r i v a t i v e s Chain Rule Chain Rule Example Example Determine the derivative of f ( x ) = 3(4 x 2 − 3 x ) 5 . 2 Determine the derivative of f ( x ) = ( x 3 − 5 x 2 ) 3 . While the function could be expanded using the Binomial Rewrite the function as f ( x ) = 2( x 3 − 5 x 2 ) − 3 . Theorem, it is faster to use the chain rule. The inner function is h ( x ) = x 3 − 5 x 2 , with derivative Let the inner function be h ( x ) = 4 x 2 − 3 x , with derivative h ′ ( x ) = 3 x 2 − 10 x . h ′ ( x ) = 8 x − 3. The outer function g ( h ) = 3 h 5 , with derivative g ′ ( h ) = 15 h 4 . The outer function is g ( h ) = 2 h − 3 , with derivative g ′ ( h ) = − 6 h − 4 . f ′ ( x ) = g ′ ( h ( x )) · h ′ ( x ) f ′ ( x ) = − 6( x 3 − 5 x 2 ) − 4 · (3 x 2 − 10 x ) = 15(4 x 2 − 3 x ) 4 (8 x − 3) or 60 − 18 x ( x − 5) 4 x 7 J. Garvin — Chain Rule of Derivatives J. Garvin — Chain Rule of Derivatives Slide 7/15 Slide 8/15 f u n d a m e n t a l r u l e s o f d e r i v a t i v e s f u n d a m e n t a l r u l e s o f d e r i v a t i v e s Chain Rule The Chain Rule with Leibniz Notation Using Leibniz notation, the chain rule is dy dx = dy du · du Example dx where y Determine the value of f ′ ( − 1) if f ( x ) = (5 x 3 − 1)(2 x 2 + x ) 4 . is a function of u and u is a function of x . Thus, u is the inner function, and y is the outer function. In this example, we need to use both the product rule and In cases where functions are (or can be) broken down into the chain rules. such relationships, it may be easier to find the individual derivatives dy du and du dx separately, then multiply and f ′ ( x ) =15 x 2 (2 x 2 + x ) 4 + (5 x 3 − 1)(4)(2 x 2 + x ) 3 (4 x + 1) substitute. f ′ ( − 1) =15( − 1) 2 (2( − 1) 2 + ( − 1)) 4 + (5( − 1) 3 − 1)(4)(2( − 1) 2 + ( − 1)) 3 (4( − 1) + 1) =87 J. Garvin — Chain Rule of Derivatives J. Garvin — Chain Rule of Derivatives Slide 9/15 Slide 10/15 f u n d a m e n t a l r u l e s o f d e r i v a t i v e s f u n d a m e n t a l r u l e s o f d e r i v a t i v e s The Chain Rule with Leibniz Notation The Chain Rule with Leibniz Notation Example Example Determine the equation of the tangent to y = √ u , if If y = 2 u 4 + 5 u and u = 5 x 2 + 1, determine dy dx . u = 3 x 2 + 4, when x = 2. The derivative of u with respect to x is du dx = 10 x . The derivative of u with respect to x is du dx = 6 x . du = 8 u 3 + 5. The derivative of y with respect to u is dy The derivative of y with respect to u is dy 1 du = 2 √ u . Therefore, the derivative of the composite function is Therefore, the derivative of the composite function is dy dx = dy du · du dx dx = dy dy du · du = (8 u 3 + 5)(10 x ) dx 1 = (8(5 x 2 + 1) 3 + 5)(10 x ) = 2 √ u · 6 x 6 x = √ 3 x 2 + 4 2 This can probably be simplified, but is in a workable state. J. Garvin — Chain Rule of Derivatives J. Garvin — Chain Rule of Derivatives Slide 11/15 Slide 12/15

f u n d a m e n t a l r u l e s o f d e r i v a t i v e s f u n d a m e n t a l r u l e s o f d e r i v a t i v e s The Chain Rule with Leibniz Notation The Chain Rule with Leibniz Notation 3(2) 2 + 4 = 4. Therefore, the point of � At x = 2, y = Even if a function is not explicitly broken up into two tangency is (2 , 4). separate functions, this same technique can be used. Calculate the slope at x = 2. Example √ 3 3 x 2 − 7 x + 4, determine dy x =2 = 6(2) If y = dx . � dy � dx 2[4] � √ u + 4 = u 1 / 3 + 4. Let u = 3 x 2 − 7 x and y = 3 = 3 2 dx = 6 x − 7 and dy du = 1 Then du 3 u − 2 / 3 . Use the slope and point of tangency to find the equation of 3 u − 2 / 3 · (6 x − 7) = (3 x 2 − 7) − 2 / 3 the tangent. Thus, dy dx = 1 · (6 x − 7). 3 y = 3 2 ( x − 2) + 4 = 3 2 x + 1 J. Garvin — Chain Rule of Derivatives J. Garvin — Chain Rule of Derivatives Slide 13/15 Slide 14/15 f u n d a m e n t a l r u l e s o f d e r i v a t i v e s Questions? J. Garvin — Chain Rule of Derivatives Slide 15/15