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03 DIFFERENTIATION HOMEWORK 14 Date: Definition of Derivative I A. Use the definition of derivative to find f'(x) for the following functions. Example 1. f(x)

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03 DIFFERENTIATION HOMEWORK 14 Date: Definition of Derivative I A. Use the definition of derivative to find f'(x) for the following functions. Example 1. f(x) = 3x2 + x f ( x ) = 2 1 - X f'(x) = lim f ( x + h)-f(x) h - h 2 2 = lim 1 - ( x + h ) 1 - x h = lim 2(1-x ) - 2[1 - (x + h)] x 5 / - h - [1-(x + h)](1-x) = lim 2-2x - 2 + 2(x + h) h-> o h[1- (x + h)](1-x) = lim 2-2x - 2+ 2x +2h h- o h[1-(x + h)](1-x) 2h = lim h- o h[1- (x + h)](1-x) 2 = lim h-> 0 [1 - (x + h) ]( 1-x) 2 [1-(x + 0)](1-x) 2 (1-x)(1-x) 2 ( 1 - x ) 2 2. f ( x ) =- X-2 x + 3 14HOMEWORK 15 Date: Definition of Derivative II A. Use the definition of derivative to find f'(x) for the following functions. Example 1 . f ( x ) = - 3 V2x - 1 f( x) = 2Vx+1 f'(x) = lim f ( x + h ) - f (x ) h -> 0 h = lim 2\\ (x + h)+ 1-2Vx+1 h - 0 h = 2 lim (x + h) + 1-vx+1 \\(x+h)+ 1+ vx+1 h - 0 ( x + h) + 1+vx+1 = 2 lim [ (x + h) + 1]-(x+1) h-oh ( x + h ) + 1+ vx+ 1) xth + 1-x-1 = 2 lim h - 0 h (x + h)+ 1+vx+1) h = 2 lim no h ( x + h ) + 1 + V x + 1 ) = 2 lim ho ( x + h ) + 1 + v x+1 = 2 ( V(x + 0) +1+Vx+1 2 = Vx + 1 + Vx+1 2 = 2Vx + 1 = Vx + 1 2. f ( x ) = x2 - Jx 15Use the definition Of derivative to find f'(x) for the function f(x) = \\/X 5x. CLONE Mar 04 Use the first principle of differentiation to 1 =3x -. X2 find f'(x) for the function f(x) CLONE Nov 05 Use the first principle of differentiation to find f'(x) for the function f(x) = \\/12x . CLONE Apr 07 Use the first principle of differentiation to . . 1 find f'x for the function fx = V . CLONE Apr 08 2 Given that f(x) = 2X . Use the 1 + 3x definition of derivative to nd f '(x). CLONE Apr 09 Use the definition of derivative to find 5 (X +3)? CLONE Apr 10 f '(X) for the function f(x) = Given that f(x) = 5x2 + 2x. Use the definition Of derivative to find f'(1). CLONE Apr 11 Find the derivative of f(x) = _ using x + 1 the definition of derivative. CLONE Sep 13 16 9. 10. 11. 12. 13. 14. 15. 16. 17. Use the denition of derivative to find f'(x) for the function x) = 5 4x + 3x2. CLONE Mar 15 Use the denition of derivative to find f'(X) for f(x) = 5\\/; . CLONE Mar 16 Given the function f(x) = x/2x3 + 6. Use the definition of derivative to find CLONE f'(x) for the function. Oct 16 Given the function f(x) = x/Xz'l' 9. Use the definition of derivative to find f'(x) for CLONE the function. Ma, 17 . Find f'(x) by using Given f(x) = 4 2x + 1 CLONE definition of derivative. Jan 13 2 Given f(x) = X . Find f'(x) using _ CLONE definition of derivative. '. Jun 18 Given f(x) =' V4x+3 5. Find f'(x) CLONE using the definition of derivative. Dec 18 Given f(x) = x2 + 5x + 3. Find rm using CLONE the definition of derivative. Jun 19 Use definition of derivative to differentiate 1 CLONE y = 3x . Dec 19 x Derivatives of Polynomial Functions A. Differentiate each of the following with respect to x. =9/Ex2 [wit f(x) x(x 6) = x(x2 12x + 36) = x3 12x2 + 36x f'(x) = 3x2 24x + 36 17 HOMEWORK 18 Date: Derivatives of Composite Functions A. Differentiate each of the following with respect to x. Example 1. y = 4( x2+ 5x) f( x ) = ( 1 - 2 x 2 )5 f' (x) = 5(1-2x2) (0-4x) = 5(1-2x2) (-4x) = -20x(1-2x2) 4 Example 2 . f( x ) = 3 / ( x 2 - 1 ) 2 y = V6x - 5 1 = (6x -5)2 dy = 1( 6x -5) 2(6 - 0) = = ( 6x -5) -2(6) = 3(6x - 5) 2 3 = V6x -5 Example 3. y= - 3 VX + 7 f ( x ) = 5 ( x2 + 1) 1 = 5(x2+ 1) f' ( x ) = - 20 ( x2 + 1) ( 2 x + 0 ) = -20(x2+ 1) (2x) = -40x (x2 + 1) 40x = ( x 2 + 1) 5 18HOMEWORK 19 Date: Product Rule A. Differentiate each of the following with respect to x. Example 1. f ( x ) = x2 (5 x2 - 2) 4 y = 8x3(1 - 2x)2 dy =u- dv du dx dx dx U = 8x3 V = (1 - 2x)2 du = 24x2 dv = 2(1 - 2x)1(0-2) dx dx = -4(1 - 2x) dy = 8x3[- 4(1-2x)] + (1-2x)2(24x2) dx = -32x3(1- 2x) + 24x2(1 - 2x)2 = 8x2(1 - 2x) [- 4x + 3(1-2x)] = 8x2(1 - 2x) [- 4x + 3-6x] = 8x2(1 - 2x)(3 - 10x) 2. y = (x+ 1)2(x-2)3 3. y = 3x2 1-x 19HOMEWORK 20 Date: Quotient Rule A. Differentiate each of the following with respect to x. Example 1. f ( x ) = _ 1-2x 4x + 3 3x + 2 x2 - 1 du - udv dy - dx dx dx V2 U = 3x + 2 V = X2 - 1 du =3 + 0 dv = 2x - 0 dx dx = 3 = 2x dy - (x2 -1) (3) - (3x + 2) (2x) dx ( x2 -1) = 3x2 -3-6x2 - 4x ( x 2 - 1) 2 -3x2 - 4x -3 ( x 2 - 1 ) 2 *2 2. y= 3. f ( x) = - X 3x + 1 Vx2 - 4 20

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