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Unformatted text preview: Math021, week 6 Chain Rule Theorem 6.1 (Chain Rule) Let f and g be functions. Then, ( f ◦ g ) ( x ) = f ( g ( x )) g ( x ) proof (sloppy): For all numbers x and h 6 = 0, f ◦ g ( x + h ) f ◦ g ( x ) h = f ( g ( x + h )) f ( g ( x )) h = f ( g ( x )+( g ( x + h ) g ( x ))) f ( g ( x )) g ( x + h ) g ( x ) g ( x + h ) g ( x ) h Thus, ( f ◦ g ) ( x ) = lim h → f ( g ( x )+( g ( x + h ) g ( x ))) f ( g ( x )) g ( x + h ) g ( x ) g ( x + h ) g ( x ) h = lim h → f ( g ( x )+( g ( x + h ) g ( x ))) f ( g ( x )) g ( x + h ) g ( x ) lim h → g ( x + h ) g ( x ) h = lim k → f ( g ( x )+ k ) f ( g ( x )) k lim h → g ( x + h ) g ( x ) h = f ( g ( x )) g ( x ) . Remark 6.2 The chain rule can be presented alternatively as follows: Let u = g ( x ) and y = f ( u ) (a way to present the composite function f ◦ g .). The chain rule becomes dy dx = dy du du dx . Example 6.3 Let y = (2 x + 1) 4 . Evaluate dy dx . solution: Let u = 2 x + 1, then y = u 4 so that dy du = 4 u 3 . On the other hand, u = 2 x + 1 so that du dx = 2 . 1 Combining these results by chain rule, we have dy dx = (4 u 3 )(2) = 8(2 x + 1) 3 . Alternatively, we may let g ( x ) = 2 x + 1, f ( x ) = x 4 . Then, f ◦ g ( x ) = (2 x + 1) 4 and ( f ◦ g ) ( x ) = g ( x ) f ( g ( x )) = (2)(4( g ( x ) 3 )) = 8(2 x + 1) 3 . Example 6.4 Let y = sin( x 2 + 1) . Evaluate dy dx . solution: Let u = x 2 + 1, then y = sin u so that dy du = cos u. On the other hand, u = x 2 + 1 so that du dx = 2 x. By chain rule, dy dx = dy du du dx = 2 x cos u = 2 x cos( x 2 + 1) ....
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This note was uploaded on 02/13/2012 for the course MATH 021 taught by Professor Luxnu during the Fall '10 term at HKUST.
 Fall '10
 LUXNU
 Chain Rule

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