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# chapter2 - Chapter 2 Dierentiation 2.1 2.1.1 Derivative...

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Chapter 2. Differentiation 2.1 Derivative 2.1.1 Instantaneous Speed Motion can be very complicated to describe; yet in “microscopic scale” it can be comprehended by the very simple and elegant Newton second law: F = ma . Here a is the acceleration, which is related to change in speed. Let the distance traveled by a car be given by a function f ( t ), where t denotes time. Then we all know that the average speed of the car in the time period from t = a to t = b is average speed = distance time = f ( b ) - f ( a ) b - a .

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2 MA1505 Chapter 2. Differentiation If we let the length of the time interval [ a, b ] shrink, i.e., let b get closer and closer to a , then we obtain the average speed in shorter and shorter time intervals. As b tends to a , in the limit, we will get the instanta- neous speed. This is the reading on the speedometer. Mathematically, this instantaneous rate of change is called the derivative of the function f ( t ) and is de- noted by f 0 ( t ). Using the concept of limit, one can give the following definition. 2.1.2 Derivative Let f ( x ) be a given function. The derivative of f at the point a , denoted by f 0 ( a ), is defined to be f 0 ( a ) = lim x a f ( x ) - f ( a ) x - a ( * ) 2
3 MA1505 Chapter 2. Differentiation provided the limit exists. An equivalent formulation of ( * ) is f 0 ( a ) = lim h 0 f ( a + h ) - f ( a ) h . If we use y as the dependent variable, i.e., y = f ( x ), then we also use the notation dy dx fl fl fl fl x = a = dy dx ( a ) = f 0 ( a ) . 2.1.3 Differentiable functions If the derivative f 0 ( a ) exists, we say that the function f is differentiable at the point a . If a function is differentiable at every point in its domain, we say that the function is differentiable. 3

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4 MA1505 Chapter 2. Differentiation 2.1.4 Example Let f ( x ) = x 2 . (i) Find f 0 (1). (ii) Show that f ( x ) is a differentiable function and f 0 ( x ) = 2 x . Solution . (i) Using the definition, f 0 (1) = lim h 0 f (1 + h ) - f (1) h = lim h 0 (1 + h ) 2 - 1 2 h = lim h 0 1 + 2 h + h 2 - 1 h = lim h 0 h (2 + h ) h = lim h 0 (2 + h ) = 2 . (ii) By doing the same calculation as above for a gen- eral x (instead of 1), we see that f 0 ( x ) = 2 x for every x . Hence the function f ( x ) = x 2 is differentiable. 4
5 MA1505 Chapter 2. Differentiation 2.1.5 Example Let f ( x ) = | x | . f is differentiable for x 6 = 0 but has no derivative at x = 0. Solution . We show f has no derivative at x = 0. lim h 0 + f ( h ) - f (0) h = lim h 0 + | h | h = lim h 0 + h h = 1 , lim h 0 - f ( h ) - f (0) h = lim h 0 - | h | h = lim h 0 - - h h = - 1 . Hence the limit lim h 0 f ( h ) - f (0) h does not exist. 2.1.6 Geometrical Meaning Let us start with the graph of a function f which is differentiable at a (see figure below). Then f ( b ) - f ( a ) b - a is the slope of the straight line joining the two points P = ( a, f ( a )) and Q = ( b, f ( b )) (such a line is called 5

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6 MA1505 Chapter 2. Differentiation a secant to the graph). As b tends to a (so Q ap- proaches P ), the secant becomes the tangent, and thus, geometrically, the derivative is just the slope of the tangent to the graph.
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