Lecture 3Section 7.3 The Logarithm Function, Part II

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1 Lectre 3Section 7.3 The Logarithm Fnction, Part II Jiwen He Section 7.2: Highlights

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3 Properties of the Log Fnction ln = t t, ln = 0, ln e =. (ln ) = > 0. ln(y) = ln + ln y, ln(/y) = ln ln y. ln ( r) = r ln, ln ( e r) = r. omain = (0, ), range = (, ). lim 0 + ln =, lim ln =. ln lim r = 0, lim r ln = Limits Differentiation an Graphing. Chain Rle Differentiation: Chain Rle Theorem. ( ) ln () = () ( ) (), for s.t. () > 0. ( ) ( ) Proof. By the chain rle, ln = ln =. Eamples 2. ( ln( + 2 ) ) = ( + 2 ) + 2 = 2 2 = , for all + 2 > 0. ( ) ( ) ln(+3) = +3 = = 3 + 3, for all > > 0. Eample Eample 3. Fin the omain of f an fin f () if f() = ln ( 4 + 2). Soltion For omain(f), we nee > 0, ths > 0. 3

4 Before ifferentiating f, simplify it: f() = ln ( 4 + 2) = ln + ln ( ) = ln + 2 ln( 4 + 2). Ths f () = + ( ) = = Graphing Eample 4

5 Eample 4. Let ( 4 f() = ln ). Specify the omain of f. On what intervals oes f increase? Decrease? Fin the etrem vales of f.determine the concavity an inflection points. Skectch the graph, specifying the asymptotes. Soltion 4 For > 0, we nee >, ths omain(f) = (, ). Simplify f() = 4 ln ln ( ). Then f () = 4 = 3 4 ( ). Ths f on (, 4 3 ) an on ( 4 3, ). At = 4 3, f () = 0. Ths f( 4 3 ) = 4 ln 4 3 ln is the (only) local an absolte minimm. From f () = 4, we have f () = )(3 2) = ( ( ) 2 2 ( ) 2. At = 2, f () = 0 ( 2 3 / omain(f) is ignore). Then, the graph is concave p on (, 2), concave own on (2, ). The point (2, f(2)) = (2, 4 ln 2) (2, 2.77) is the only point of inflection. From f () = 4, we have lim f () =, + lim f () = 0. From f() = 4 ln ln( ), we have lim f() =, + lim f() =. The line = is a vertical asymptote. 5

6 Eample Eample 5. Let f() = ln. Specify the omain of f an fin the intercepts. On what intervals oes f increase? Decrease? Fin the etrem vales of f.determine the concavity an inflection points. Skectch the graph. Soltion ln is efine only for > 0, ths omain(f) = (0, ). 6

7 There is no y-intercept. Since = is the only -intercept. f() = ln = 0, We have f () = + ln = + ln. For f () = 0, we have + ln = 0 ln = = e. Ths f on (0, e ) an on ( e, ). At = e, f () = 0. Ths is the (only) local an absolte minimm. f( e ) = e ln e = e From f () = +ln, we have f () = > 0, for all > 0. Then, the graph is concave p on (0, ). There is no point of inflection. From f () = + ln, we have lim f () =, 0 + lim f () =. From f() = ln, we have lim f() = 0, 0 + lim f() =. Qiz Qiz. ln =? : (a), (b) 0, (c). 2. ln e =? : (a) 0, (b), (c) e. 2 ln 2. Properties f() = ln, 0 7

8 Graph The graph has two branches: y = ln( ), < 0 an y = ln, > 0, each is the mirror image of the other. Theorem 6. Proof. For > 0, For < 0, Power Rle: ( ) ln = = ln + C ( ) ( ) ln = ln =. ( ) ( ) ln = ln( ) = n ( ) =. 8

9 Power Rle n = n + n+ + C, if n, ln + C, if n =. Eample 7. ( + 2 = + ) 2 = ln + C. 2.2 Chain Rle Differentiation: Chain Rle Theorem 8. ( ) ln () = () ( ) (), for s.t. () 0. ( ) ( ) Proof. By the chain rle, ln = ln =. Eamples 9. ( ln 3 ) = ( 3 ) 3 = ( ) ln 2 = (ln ) (ln 2 ) = Logarithmic Differentiation Logarithmic Differentiation Theorem 0. Let g() = g () g 2 () g n (). Then ( ) g g () = g() () g () + g 2() g 2 () + + g n(). g n () Proof. First write ln g() = ln ( g () g 2 () g n () ) = ln g () + ln g 2 () + + ln g n (). Then ifferentiate g () g() = g () g () + g 2() g 2 () + + g n() g n (). Mltiplying by g() gives the reslt. 9

10 Eamples Eamples. Fin f () if g() = ( )( 2)( 3). g() = (2 + ) 3 (2 5) 2 ( 2 + 5) 2. Soltion ln g() = ln + ln + ln 2 + ln 3. g () g() = ( g () = ( )( 2)( 3) ). 3 ln g() = 3 ln ln ln g () g() = g () = (2 + ) 3 (2 5) 2 ( 2 + 5) 2 ( ) Qiz (cont.) Qiz (cont.) 3. lim ln =? : (a), (b) 0, (c) lim ln =? : (a), (b) 0, (c). 3 Integration an Trigonometric Fnctions 3. -Sbstittion Integration: -Sbstittion Theorem 2. g () = ln g() + C, 0. g() 0

11 Proof. Let = g(), ths = g (), then g () g() = Eample 3. Calclate then = 2 = ln + C = ln g() + C. 2 Eamples: -Sbstittion ln Eamples 4... ( + ) Soltion Set = ln, 4 3. Let = 43, ths = 2 2, = 2 ln + C = 2 ln 43 + C. =. Then ln = = C = 2 (ln )2 + C. Set = +, = 2. Then = 2 ( + ) = 2 ln + C = 2 ln ( + ) + C. Set = 3 + +, = (3 2 + ). At =, = 3; at = 2, =. Then = 2 = 2 [ln ] 3 = 2 (ln ln 3). 3 Natral log arises (only) when integrating a qotient whose nmerator is the erivative of its enominator (or a constant mltiple of it). g () g() = = ln + C = ln g() + C.

12 3.2 Trigonometric Fnctions Integration of Trigonometric Fnctions Recall that [e] cos = sin + C sin = cos. sin = cos + C cos = sin. [e] sec 2 = tan + C tan = sec2. csc 2 = cot + C cot = csc2. [e] sec tan = sec + C sec = sec tan. csc cot = csc + C csc = csc cot. New Integration Formlas Integration of Trigonometric Fnctions tan = ln cos + C. cot = ln sin + C. ln sec + tan + C. csc = ln csc cot + C. Proof. Set = cos, = sin, then tan = sin cos = = ln cos + C. Set = sin, = cos, then cos cot = sin = = ln sin + C. = ln + C = ln + C sec = Set = sec + tan, = (sec tan + sec 2 ), then sec = sec = sec + tan sec tan + sec 2 sec + tan = sec + tan = ln + C = ln sec + tan + C. Set = csc cot, = ( csc cot + csc 2 ), then csc = csc = csc cot csc cot + csc 2 csc cot = csc cot = ln + C = ln csc cot + C. 2

13 Eamples: Eamples 5. sec 2. sec tan 3. tan(ln ). Soltion Set = + tan 3, = 3 sec 2 3 : sec tan 3 = 3 = 3 ln + C = ln + tan 3 + C. 3 Set = 2, = 2 : sec 2 = 2 sec = ln sec + tan + C 2 = 2 ln sec 2 + tan 2 + C. Set = ln, = : tan(ln ) = tan = ln sec + C = ln sec(ln ) + C. Otline Contents Differentiation 3. Chain Rle Graphing ln 7 2. Properties Chain Rle Log Differentiation Integration Sbstittion Trigonometric Fnctions