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Unformatted text preview: 7.3 THE LOGARITHM FUNCTION, PART II Differentiation and Graphing We know that for x > 0, d dx (ln x ) = 1 x . Now suppose that u is a positive, differentiable function of x . Then (7.3.1) d dx (ln u ) = 1 u du dx . PROOF By the chain rule, d dx (ln u ) = d du (ln u ) du dx = 1 u du dx . For example, d dx [ln (1 + x 2 )] = 1 1 + x 2 · 2 x = 2 x 1 + x 2 for all x , and d dx [ln(1 + 3 x )] = 1 1 + 3 x · 3 = 3 1 + 3 x for all x > − 1 3 . Example 1 Find the domain of f and find f ( x ) if f ( x ) = ln ( x 4 + x 2 ). SOLUTION For x to be in the domain of f , we must have x √ 4 + x 2 > 0, and thus we must have x > 0. The domain of f is the set of positive numbers. Before differentiating f , we make use of the special properties of the logarithm: f ( x ) = ln x 4 + x 2 = ln x + ln [(4 + x 2 ) 1 / 2 ] = ln x + 1 2 ln (4 + x 2 ). From this we have f ( x ) = 1 x + 1 2 · 1 4 + x 2 · 2 x = 1 x + x 4 + x 2 = 4 + 2 x 2 x (4 + x 2 ) . 7.3 THE LOGARITHM FUNCTION, PART II 389 Example 2 Sketch the graph of f ( x ) = ln  x  . SOLUTION The function, defined at all x = 0, is an even function: f ( − x ) = f ( x ) for all x = 0. The graph has two branches: y = ln ( − x ), x < and y = ln x , x > 0. Each branch is the mirror image of the other. (Figure 7.3.1) Example 3 ( Important ) Show that y = In (– x ) y = In x x y y = In x Figure 7.3.1 (7.3.2) d dx (ln  x  ) = 1 x for all x = 0. SOLUTION For x > 0, d dx (ln  x  ) = d dx (ln x ) = 1 x . For x < 0, we have  x  = − x > 0, so that d dx (ln  x  ) = d dx [ln ( − x )] = 1 − x d dx ( − x ) = 1 − x ( − 1) = 1 x . It follows that if u is a differentiable function of x , then where u ( x ) = 0, (7.3.3) d dx (ln  u  ) = 1 u du dx . PROOF d dx (ln  u  ) = d du (ln  u  ) du dx = 1 u du dx . Here are two examples: d dx (ln  1 − x 3  ) = 1 1 − x 3 d dx (1 − x 3 ) = − 3 x 2 1 − x 3 = 3 x 2 x 3 − 1 . d dx ln x − 1 x − 2 = d dx (ln  x − 1  ) − d dx (ln  x − 2  ) = 1 x − 1 − 1 x − 2 . Example 4 Let f ( x ) = x ln x . (a) Specify the domain of f and find the intercepts, if any. (b) On what intervals does f increase? Decrease? (c) Find the extreme values of f . (d) Determine the concavity of the graph and find the points of inflection. (e) Sketch the graph of f . SOLUTION Since the logarithm function is defined only for positive numbers, the domain of f is (0, ∞ ) and there is no yintercept. Since f (1) = 1 · ln 1 = 0, 1 is an xintercept. 390 CHAPTER 7 THE TRANSCENDENTAL FUNCTIONS Differentiating f , we have f ( x ) = x · 1 x + ln x = 1 + ln x . To find the critical numbers of f , we set f ( x ) = 0: 1 + ln x = 0, ln x = − 1, x = 1 e (verify this). Recalling that the logarithm function is increasing on (0, ∞ ), and that ln x → −∞ as x → + and ln x → ∞ as x → ∞ , we have 1 e f : f ' : 1 2 x + + + + + + + + + + + + + + + + + + + + – – – – – – – decreases increases Therefore, f decreases on (0, 1 / e ] and increases on [1...
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 Spring '10
 SMITH
 Derivative, Intermediate Value Theorem, Logarithm

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