Chain Rule MCV4U: Calculus & Vectors Recap 2 x 5 7 x . 3 - 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 Chain Rule MCV4U: Calculus & Vectors Recap √ 2 x 5 − 7 x . 3 Determine the derivative of f ( x ) = 12 The inner function is h ( x ) = 2 x 5 − 7 x , so h ′ ( x ) = 10 x 4 − 7. 3 , so g ′ ( x ) = 4 h − 2 1 3 . The outer function is g ( h ) = 12 h Quotient Rule of Derivatives f ′ ( x ) = 4(2 x 5 − 7 x ) − 2 3 (10 x 4 − 7 x ) J. Garvin 40 x 4 − 28 or 2 ( x (2 x 4 − 7)) 3 J. Garvin — Quotient Rule of Derivatives Slide 1/13 Slide 2/13 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 Quotient Rule Quotient Rule Rewrite f ( x ) = p ( x ) q ( x ) as f ( x ) = p ( x )[ q ( x )] − 1 and use the Sometimes a function can be written as the quotient of two other functions. chain rule to differentiate. To determine the derivative of a function with the form f ( x ) = p ( x ) f ′ ( x ) = p ′ ( x )[ q ( x )] − 1 + p ( x )( − 1)[ q ( x )] − 2 q ′ ( x ) q ( x ), use the Quotient Rule. × [ q ( x )] 2 p ′ ( x )[ q ( x )] − 1 − p ( x )[ q ( x )] − 2 q ′ ( x ) � � = Quotient Rule [ q ( x )] 2 = p ′ ( x ) q ( x ) − p ( x ) q ′ ( x ) If f ( x ) = p ( x ) q ( x ), then f ′ ( x ) = p ′ ( x ) q ( x ) − p ( x ) q ′ ( x ) , [ q ( x )] 2 [ q ( x )] 2 provided that q ( x ) � = 0. dx = v du dx − u dv If y = u v , then dy dx , v � = 0. v 2 Note the similarity to the product rule in the numerator. J. Garvin — Quotient Rule of Derivatives J. Garvin — Quotient Rule of Derivatives Slide 3/13 Slide 4/13 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 Quotient Rule Quotient Rule Example Example Determine the derivative of f ( x ) = 3 x 2 − 1 Determine the derivative of y = 4 x 2 − 6 2 x + 3 . √ x . Let u = 4 x 2 − 6 and v = √ x = x Let p ( x ) = 3 x 2 − 1 and q ( x ) = 2 x + 3. 1 2 . Then p ′ ( x ) = 6 x and q ′ ( x ) = 2. 1 Then du dx = 8 x and dv dx = 2 √ x . f ′ ( x ) = (6 x )(2 x + 3) − (3 x 2 − 1)(2) (8 x )( √ x ) − (4 x 2 − 6) � � 1 2 √ x (2 x + 3) 2 dy dx = [ √ x ] 2 = 12 x 2 + 18 x − 6 x 2 + 2 or 6 x 2 + 3 (2 x + 3) 2 √ = 6 x 2 + 18 x + 2 x 3 (2 x + 3) 2 J. Garvin — Quotient Rule of Derivatives J. Garvin — Quotient Rule of Derivatives Slide 5/13 Slide 6/13

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 Quotient Rule An Alternative to the Quotient Rule Example Since we derived the general formula for the quotient rule using the chain rule, an alternative to the quotient rule is to The concentration, C (g/L), of a chemical released into a 6 t use both the product rule and the chain rule instead. stream after t days is given by C ( t ) = 2 t 2 + 9. At what rate This may be useful, because it is not necessary to memorize is the concentration changing after 4 days? a separate formula for quotients. Let p ( t ) = 6 t and q ( t ) = 2 t 2 + 9. Derivatives derived using each technique may appear different, but should be the same after simplification. Then p ′ ( t ) = 6 and q ′ ( t ) = 4 t . f ′ ( x ) = 6(2 t 2 + 9) − 6 t (4 t ) (2 t 2 + 9) 2 = 6(2(4) 2 + 9) − 6(4)(4(4)) (2(4) 2 + 9) 2 = − 138 1681 J. Garvin — Quotient Rule of Derivatives J. Garvin — Quotient Rule of Derivatives Slide 7/13 Slide 8/13 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 An Alternative to the Quotient Rule An Alternative to the Quotient Rule Example Example Use the product and chain rules to show that the derivative At what point(s), if any, is the tangent to y = 3 x − 4 of f ( x ) = 3 x 2 − 1 2 x + 3 is f ′ ( x ) = 6 x 2 + 18 x + 2 x 2 − 5 x . horizontal? (2 x + 3) 2 Rewrite the function as y = (3 x − 4)( x 2 − 5 x ) − 1 . Rewrite the function as f ( x ) = (3 x 2 − 1)(2 x + 3) − 1 . The slope of the tangent is given by dy dx . f ′ ( x ) = (6 x )(2 x + 3) − 1 + (3 x 2 − 1)( − 1)(2 x + 3) − 2 (2) dx = 3( x 2 − 5 x ) − 1 + (3 x − 4)( − 1)( x 2 − 5 x ) − 2 (2 x − 5) dy = (6 x )(2 x + 3) − 1 − 2(3 x 2 − 1)(2 x + 3) − 2 = 3( x 2 − 5 x ) − (3 x − 4)(2 x − 5) = (6 x )(2 x + 3) − 6 x 2 + 2 ( x 2 − 5 x ) 2 (2 x + 3) 2 = − 3 x 2 + 8 x − 20 = 6 x 2 + 18 x + 2 ( x 2 − 5 x ) 2 (2 x + 3) 2 J. Garvin — Quotient Rule of Derivatives J. Garvin — Quotient Rule of Derivatives Slide 9/13 Slide 10/13 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 An Alternative to the Quotient Rule An Alternative to the Quotient Rule Since the slope of a horizontal line is 0, set dy dx = 0 and solve for x . 0 = − 3 x 2 + 8 x − 20 ( x 2 − 5 x ) 2 = − 3 x 2 + 8 x − 20 Since 8 2 − 4( − 3)( − 20) < 0, there are no real solutions to this equation. Therefore, there are no points on y = 3 x − 4 x 2 − 5 x where the tangent is horizontal. J. Garvin — Quotient Rule of Derivatives J. Garvin — Quotient Rule of Derivatives Slide 11/13 Slide 12/13

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 — Quotient Rule of Derivatives Slide 13/13