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Limit Definition of the Derivative You won't have to calculate the derivative using def of derivative. You should recognize its form, then take a derivative of the function by another method. FORM A f ( x h) f ( x) Definition of the Derivative f ( x) lim h0 h Example 1: lim ( x h) 2 x 2 ? h0 h Answer: 2x. Recognize that this is definition of the derivative. The function is x2. Now take the derivative by power rule and find that d/dx of x2 is 2x. Example 2: lim tan( x h) tan( x) ? h0 h Answer: sec2 (x) f(x) = tan(x) f ' (x) = sec2 (x) Example 3: lim ( x h) 4 x 4 h0 ? h Answer: 4x3 f(x) = x4 f ' (x) = 4x3 Example 4: lim 2( x h) 2 3( x h) 2 x 3 3x ? h0 h Answer: 4x+3 f(x) = 2x2+3x f ' (x) = 4x+3 sin( 2( x h)) sin 2 x Example 5: lim ? h0 h Answer: 2 cos 2x f(x) = sin2x f ' (x) = 2 cos 2x cos( x h) 2 cos x 2 Example 6: lim ? h h0 Answer: - 2x sin(x2) f(x) = cos (x2) f ' (x) = - 2x sin(x2) FORM B Definition of the Derivative at a point (x = c) lim f ( x ) f (c ) f ' (c ) xc xc Example 7: lim x2 9 ? x3 x 3 Answer: 6 Recognize that this is definition of the derivative at a point. The function is x2 and it is evaluated at the point x=3. Now take the derivative by power rule and find that d/dx of x2 is 2x, then plug in x=3 and get 2(3) = 6. Example 8: lim x 2 52 ? x5 x 5 Answer: 10 f(x) = x2 x =5 is the x value desired f ' (x) = 2x plug in x=5 f’(5) = 10 trickier version of example 8 Example 9: lim x 2 25 x5 ? x5 Answer: 10 This time the 52 has been given in a different form so it appears more difficult but it is the same as Example 8. f(x) = x2 x =5 is the x value desired f ' (x) = 2x plug in x=5 f’(5) = 10 Example 10: lim sin x sin( ) xπ/3 3 ? x 3 Answer: 1/2 f(x) = sin(x) x =pi/3 f ' (x) = cos (x). f '(pi/3) = cos(pi/3) = 1/2 trickier version of Example 10 Example 11: lim 3 xπ/3 sin x 2 ? x 3 Answer: 1/2 This time the sin(π/3) has been given in a different form so it appears more difficult but it is the same as Example 10. √(3)/2 is given in place of sin(pi/3) f(x) = sin(x) x = π/3 so f ' (x) = cos (x) f ' (π/3) = cos(π/3) = 1/2 another tricky one: Example 12: lim 2 xπ/4 cos x 2 ? x 4 cos x cos( ) Answer: - √(2)/2 Recognize this as lim 4 x π/4 x 4 f(x) = cos(x) x = π/4 so f ' (x) = - sin(x) f ' (π/4 ) = - sin (π/4 ) = - √(2)/2 (Ostebee Zorn 2nd ed. section 2.7, page 148) Example 13: lim sin x x0 ? x Answer: 1 Recognize this as lim sin x sin x sin 0 x0 x x0 since sin(0) = 0 f(x) = sin(x) x=0 f ' (x) = cos (x) f ' (0) = cos(0) = 1 Example 14: lim sin x xπ ? x Answer: -1 Recognize this as lim sin x sin x 0 sin x sin xπ x x x since sin π = 0 f(x) = sin(x) x=π f ' (x) = cos (x) f ' (π) = cos(π) = -1

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AP Calculus AB, Calculus Test, calculus AB, Sample Questions, limit definition, ap calculus ab exam, Separable Differential Equations, 1 & 2, video covers, Calculus AB Math

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posted: | 5/22/2011 |

language: | English |

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