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**Unformatted text preview: **MATH0201 BASIC CALCULUS MATH0201 BASIC CALCULUS Differentiation I Dr. WONG Chi Wing Department of Mathematics, HKU MATH0201 BASIC CALCULUS Instantaneous Rate of Change and Tangent Problem Differentiable Functions Derivatives from the First Principle Continuity, and Mean Value Theorem Basic Differentiation Rules Constant Function Rule Power Rule Constant Multiple Rule Sum and Difference Rule Product Rule Quotient Rule Antiderivatives Reference § 2.1–3,5; § 4.3–4; § 5.1 of the textbook. MATH0201 BASIC CALCULUS Instantaneous Rate of Change and Tangent Problem Suppose that a vehicle is moving along a straight line. Its displacement from a fixed reference point at time t is given by x ( t ) = 4 t 2 : A plot of the displacement against time elapsed is shown below What is the velocity of the vehicle? MATH0201 BASIC CALCULUS Instantaneous Rate of Change and Tangent Problem The average velocity over the time period [ 2 ; 3 ] is f ( 3 ) f ( 2 ) 3 2 = 4 3 2 4 2 2 3 2 = 20 : The average velocity over the time period [ 2 ; 2 : 5 ] is f ( 2 : 5 ) f ( 2 ) 2 : 5 2 = 4 2 : 5 2 4 2 2 2 : 5 2 = 18 : Similarly, the average velocity over the time period [ 2 ; 2 : 1 ] is 16 : 4; the average velocity over the time period [ 2 ; 2 : 01 ] is 16 : 04. In general, the average velocity over the time period [ 2 ; 2 + h ] is 4 ( 4 + h ) . MATH0201 BASIC CALCULUS Instantaneous Rate of Change and Tangent Problem Definition 1 (Instantaneous Rate of Change (p.105)) The instantaneous rate of change of f ( x ) at x = a is lim h ! f ( a + h ) f ( a ) h if the limit exists. MATH0201 BASIC CALCULUS Instantaneous Rate of Change and Tangent Problem Example 2 A small steel ball dropped from a tower will fall a distance of y feet in x seconds, as given approximately by y = f ( x ) = 16 x 2 . I The velocity is NOT uniform. I Average velocity from x = 2 to x = 2 + h , h 6 = 0, seconds is f ( 2 + h ) f ( 2 ) ( 2 + h ) 2 = 64 + 16 h feet per second. I The instantaneous velocity of the ball at x = 2 seconds is lim h ! f ( 2 + h ) f ( 2 ) h = 64 feet per second. Read also Examples 2.1.1–2. MATH0201 BASIC CALCULUS Instantaneous Rate of Change and Tangent Problem Definition 3 (Secant Line (p.105)) A line through two points of the graph of a function is called a secant line . The slope of the secant line passing through ( x ; f ( x )) and ( x + h ; f ( x + h )) is f ( x + h ) f ( x ) ( x + h ) x = f ( x + h ) f ( x ) h : MATH0201 BASIC CALCULUS Instantaneous Rate of Change and Tangent Problem Definition 4 (Slope of Graph (p.106)) Given y = f ( x ) , the slope of the graph at the point ( a ; f ( a )) is given by lim h ! f ( a + h ) f ( a ) h if the limit exists. The slope of the graph is also the slope of the tangent line at the point ( a ; f ( a )) . MATH0201 BASIC CALCULUS Instantaneous Rate of Change and Tangent Problem Example 5 Let f ( x ) = x 2 and consider its graph y = x 2 ....

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