Chapter5(0211)

# Chapter5(0211) - Ch5/MATH0211/YMC/2009-10 1 Chapter 5...

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Unformatted text preview: Ch5/MATH0211/YMC/2009-10 1 Chapter 5. Differentiation 5.1. Derivatives In this section, we introduce the concept of a derivative. There are two main interpretations of derivatives – (i) the geometric one is the slope of the tangent line to the graph of a function; and (ii) the physical one is the instantaneous rate of change (velocity is an example) in many practical problems. Geometric Interpretation of derivatives The tangent line of the graph of f ( x ) at the point P can be defined as the limit of the secant line (a line drawn between two points on the graph) as the distance between the two points goes to zero. In other words, the tangent line to the curve at P is the line through P whose slope is the limit of the secant slopes as Q → P . To find a tangent to the graph of f ( x ) at a point P = ( x ,f ( x )) , we first calculate the slope of the secant line through P and a point Q = ( x + h,f ( x + h )) , which is given by f ( x + h )- f ( x ) h . We then investigate the limit of the secant slope as h → , that is, lim h → f ( x + h )- f ( x ) h (5.1) If the limit exists, then it is the slope of the tangent line to f ( x ) at the point P . Ch5/MATH0211/YMC/2009-10 2 Suppose the limit in (5.1) exists. Then the equation for the tangent line to f ( x ) at the point ( x ,f ( x )) is given by y- f ( x ) x- x = m, or equivalently, y = f ( x ) + m ( x- x ) (5.2) where m is the limit in (5.1). Example 5.1 Let f ( x ) = 1 /x . (a) Find the slope of the tangent line of f ( x ) at x = a 6 = 0 . (b) Write down the equation of the tangent line of f ( x ) at x = 2 . Solution . (a) The slope at the point ( a, 1 /a ) can be computed by using (5.1): lim h → f ( a + h )- f ( a ) h = lim h → 1 a + h- 1 a h = lim h → 1 h a- ( a + h ) a ( a + h ) = lim h →- 1 a ( a + h ) =- 1 a 2 . (b) Using equation (5.2) and part (a), we have y = 1 a- 1 a 2 ( x- a ) Putting a = 2 gives y = 1 2- 1 4 ( x- 2) =- 1 4 x + 1 . Ch5/MATH0211/YMC/2009-10 3 Example 5.2 At what points do the graph of the function f ( x ) = x 3- 3 x have horizontal tangents? Solution . The slope at the point ( a,f ( a )) is lim h → f ( a + h )- f ( a ) h = lim h → ( a + h ) 3- 3( a + h )- a 3 + 3 a h = lim h → 1 h ( a 3 + 3 a 2 h + 3 ah 2 + h 3- 3 a- 3 h- a 3 + 3 a ) = lim h → 3 a 2 + 3 ah + h 2- 3 = 3 a 2- 3 . The tangent line at the point ( a,f ( a )) will be a horizontal tangent if its slope is zero at that point. The slope will be zero provided that 3 a 2- 3 = 0 , that is, a = ± 1 . It follows that the graph of f ( x ) has horizontal tangents at the points (1 ,f (1)) = (1 ,- 2) and (- 1 ,f (- 1)) = (- 1 , 2) . 2 The expression f ( x + h )- f ( x ) h ( h 6 = 0) is called the difference quotient of f ( x ) at the point x = x . If the difference quotient has a limit as h approaches , then the limit lim h → f ( x + h )- f ( x ) h is called the derivative of f ( x ) at the point x = x ....
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## This note was uploaded on 01/02/2012 for the course MATH 0211 taught by Professor Chan during the Spring '10 term at HKU.

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Chapter5(0211) - Ch5/MATH0211/YMC/2009-10 1 Chapter 5...

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