233lectures-2013-cute - 2 1 So F 0(x = f 0(g(x g 0(x = e x...

This preview shows page 1 - 4 out of 63 pages.

Definition (Definition of Derivative) Let f : R R be a real valued function f ( x ). Then the derivative f prime ( x ) is given by the following formula if the limit exists: f prime ( x ) = lim h 0 f ( x + h ) - f ( x ) h Example f ( x ) = x 2 lim h 0 f ( x + h ) - f ( x ) h = lim h 0 bracketleftbigg ( x + h ) 2 - x 2 h = x 2 + 2 hx + h 2 - x 2 h bracketrightbigg = lim h 0 bracketleftbigg 2 hx + h 2 h = 2 x + h bracketrightbigg = 2 x Constant and Sum Rules Let c R and f ( x ), g ( x ) be real valued functions. Then: ( cf ( x )) prime = cf prime ( x ) . ( f ( x ) + g ( x )) prime = f prime ( x ) + g prime ( x ) . Example If F ( x ) = 10 x 2 + 7 x , then F prime ( x ) = 10(2 x ) + 7 = 20 x + 7 . Exponential Rule For f ( x ) = Ce kx , f prime ( x ) = kCe kx Example If f ( x ) = e 3 x - x 2 , find f prime ( x ). f prime ( x ) = ( e 3 x - x 2 ) prime = ( e 3 x ) prime - ( x 2 ) prime = 3 e 3 x - 2 x Product and quotient rules ( f ( x ) g ( x )) prime = f prime ( x ) g ( x ) + f ( x ) g prime ( x ) bracketleftbigg f ( x ) g ( x ) bracketrightbigg prime = f prime ( x ) g ( x ) - f ( x ) g prime ( x ) ( g ( x )) 2 Example Compute f prime ( x ) for f ( x ) = e x 1+ x 2 . Solution. d y dx = (1 + x 2 ) d dx ( e x ) - e x d dx (1 + x 2 ) (1 + x 2 ) 2 = (1 + x 2 ) e x - e x (2 x ) (1 + x 2 ) 2 = e x (1 - x ) 2 (1 + x 2 ) 2 . Chain Rule ( f g ) prime ( x ) = f prime ( g ( x )) · g prime ( x ) Example Find F prime ( x ) if F ( x ) = e x 2 +1 . Solution: Let f ( x ) = e x and g ( x ) = x 2 + 1. Then F ( x ) = ( f g )( x ). So, F prime ( x ) = f prime ( g ( x )) · g prime ( x ) = e x 2 +1 (2 x ) Derivatives of classical functions sin prime ( x ) = cos( x ) cos prime ( x ) = - sin( x ) ( e x ) prime = e x ln prime ( x ) = 1 x , where ln( x ) is the natural log. log prime a ( x ) = 1 x ln( a ) , where log a ( x ) is the log in base a . d dx (sin - 1 ( x )) = 1 1 - x 2 d dx (tan - 1 ( x )) = 1 1+ x 2
Definition (Antiderivatives) A function F is called an antiderivative of f on an interval I if F prime ( x ) = f ( x ) for all x in I . Example Let f ( x ) = x 2 . Then an antiderivative F ( x ) for x 2 is F ( x ) = x 3 3 . Theorem If F is an antiderivative of f on an interval I , then the most general antiderivative of f on I is F ( x ) + C where C is an arbitrary constant. Table of Anti-differentiation Formulas Let F ( x ), G ( x ) be the antiderivative respectively for the functions f ( x ), g ( x ). Function Particular antiderivative c · f ( x ) c · F ( x ) f ( x ) + g ( x ) F ( x ) + G ( x ) x n ( n negationslash = - 1) x n +1 n +1 1 x ln | x | e x e x cos x sin x Table of Anti-differentiation Formulas Function Particular antiderivative sin x - cos x sec 2 x tan x sec x tan x sec x 1 1 - x 2 sin - 1 x 1 1+ x 2 tan - 1 x Area under y = x 2 from 0 to 1. Example Use rectangles to estimate the area under the parabola y = x 2 from 0 to 1. Area estimate using right end points R 4 = 1 4 · parenleftbigg 1 4 parenrightbigg 2 + 1 4 · parenleftbigg 1 2 parenrightbigg 2 + 1 4 · parenleftbigg 3 4 parenrightbigg 2 + 1 4 · 1 2 = 15 32 Note area A < 15 32 = . 46875 Area estimate using left end points L 4 = 1 4 · 0 2 + 1 4 · parenleftbigg 1 4 parenrightbigg 2 + 1 4 · parenleftbigg 1 2 parenrightbigg 2 + 1 4 · parenleftbigg 3 4 parenrightbigg 2 = 7 32 = . 21875 Note area A satisfies . 21875 A . 46875
General calculation using right end points Area definition using right end points Definition The area A of the region S that lies under the graph of the continuous function f is the limit of the sum of the areas of approximating rectangles: A = lim n →∞ R n = lim n →∞ [ f ( x 1 x + f ( x 2 x + ··· + f ( x n x ]

  • Left Quote Icon

    Student Picture

  • Left Quote Icon

    Student Picture

  • Left Quote Icon

    Student Picture