# ch08 - Chapter 8 Derivatives of Exponential and Logarithmic...

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Exercise 8.1 4. c. f ( x ) } e x x 3 } f' ( x )= } –3 x 2 e x 3 x ( 2 x )– e x 3 } d. s = } e t 3 2 t 2 } } d d s t } =6 te 3 t 2 ( t 2 )–2 t ( e 3 t 2 ) = } 2 e 3 t 2 [3 t t 3 2 –1] } h. g ( t )= } 1+ e 2 t e 2 t } g' ( t )= = } (1 2 + e e 2 t 2 t ) 2 } 5. a. f' ( x )= } 1 3 } (3 e 3 x – 3 e –3 x ) = e 3 x e –3 x f' (1) = e 3 e –3 c. h' ( z )=2 z (1 + e z )+ z 2 (– e z ) h' (–1) = 2(–1)(1 + e )+(–1) 2 (– e 1 ) = –2 – 2 e e = –2 – 3 e 7. y = e x Slope of the tangent is } d d y x } = e x . Slope of the given line is –3. Slope of the perpendicular line is } 1 3 } . Therefore, e x = } 1 3 } : x ln e = ln 1 – ln 3 x = –ln 3 = ˙ –1.099. 2 e 2 t (1 + e 2 t ) – 2 e 2 t ( e 2 t ) } (1 + e 2 t ) 2 186 Chapter 8: Derivatives of Exponential and Logarithmic Functions Chapter 8 • Derivatives of Exponential and Logarithmic Functions Review of Prerequisite Skills 4. a. log 2 32 Since 32 = 2 5 ,log 2 32=5 . b. log 10 0.0001 Since 0.0001 = 10 –4 ,log 10 0.0001 = –4. c. log 10 20+log 10 5 =log 10 (20 3 5) =log 10 100 =log 10 10 2 =2 d. log 2 20 – log 2 5 =log 2 1 } 2 5 0 } 2 =log 2 4 =2 e. 3 2log 3 5 =(3 log 3 5 ) 2 =5 2 =25 f. log 3 (5 3 9 –3 25 } 3 2 } ) =log 3 5 3 +log 3 9 –3 +log 3 25 } 3 2 } =log 3 5 3 +log 3 3 –6 +log 3 5 –3 = 3log 3 5 – 6 – 3log 3 5 =–6 5. a. log 2 80, b = e = } l l o o g g e 8 e 2 0 } = } l l n n 8 2 0 } = ˙ 6.322 b. 3log 5 22 – 2log 5 15, b =10 =3 1 } l l o o g g 10 1 2 0 5 2 } 2 – 2 1 } l l o o g g 10 1 1 0 5 5 } 2 = } 3log 10 2 l 2 og 10 2 5 log 10 15 } =

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The point where the tangent meets the curve has x = –ln 3 and y =3 –ln3 = } 1 3 } . The equation of the tangent is y } 1 3 } = } 1 3 } ( x + ln 3) or y = 0.3 x + 0.6995. 8. The slope of the tangent line at any point is given by } d d y x } = (1)( e x )+ x (– e x ) = e x (1 – x ). At the point (1, e –1 ), the slope is e –1 (0) = 0. The equation of the tangent line at the point A is y e –1 =0( x – 1) or y = } 1 e } . 9. The slope of the tangent line at any point on the curve is } d d y x } =2 xe x + x 2 (– e x ) =(2 x x 2 )( e x ) = } 2 x e x x 2 } . Horizontal lines have slope equal to 0. We solve } d d y x } =0 } x (2 e x x ) } =0 . Since e x >0 for all x , the solutions are x =0 and x = 2. The points on the curve at which the tangents are horizontal are (0, 0) and 1 2, } e 4 2 } 2 . 10. If y = } 5 2 } ( e } 5 x } + e } 5 x } ), then y' = } 5 2 } 1 } 1 5 } e } 5 x } } 1 5 } e } 5 x } 2 , and y'' = } 5 2 } 1 } 2 1 5 } e } 5 x } + } 2 1 5 } e } 5 x } 2 = } 2 1 5 } 3 } 5 2 } ( e } 5 x } + e } 5 x } ) 4 = } 2 1 5 }
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## This note was uploaded on 02/10/2010 for the course MATHS MCV-01 taught by Professor Mr.m during the Spring '10 term at Seneca.

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ch08 - Chapter 8 Derivatives of Exponential and Logarithmic...

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