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quiz5-sol

# quiz5-sol - x = e 2 x Â 2 23 6 The voltage V in volts in an...

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AMS 151 (Fall, 2009) Joe Mitchell Applied Calculus I Quiz # 5, Thursday, November 5 – Solution Notes 1. Let f ( x ) = 1 x 3 2 x + e x + ln( x 2 ). Find f ( x ). f ( x ) = (2 x + e x )( - 3 x 2 ) - (1 - x 3 )(2 + e x ) (2 x + e x ) 2 + 1 x 2 · (2 x ) . 2. Let f ( x ) = cos x - 3tan(3 x ). Find f ( x ). f ( x ) = - sin x - 3(sec 2 (3 x )) · 3 . 3. Let f ( x ) = x e x . Find f ( x ). Take ln to get: ln f ( x ) = ln( x e x ) = e x ln x, and then take the derivative with respect to x : 1 f ( x ) f ( x ) = e x 1 x + e x ln x, which implies that f ( x ) = ( e x 1 x + e x ln x ) f ( x ) = ( e x 1 x + e x ln x ) x e x . 4. Suppose that h ( a ) = 2 ab + g ( a 2 ) and that g ( u ) = 3 u 2 . Find h ( a ). We compute h ( a ), the derivative of the function h (which is a function of a ; we treat b as constant). h ( a ) = 2 b + g ( a 2 ) · 2 a. Now, since g ( u ) = 3 u 2 , we know that g ( a 2 ) = 3( a 2 ) 2 = 3 a 4 . Thus, h ( a ) = 2 b + 3 a 4 · 2 a = 2 b + 6 a 5 . 5. Let h ( x ) = e 2 x . Find the 23rd derivative, h (23) ( x ). We compute: h ( x ) = e 2 x · 2 , h ′′ ( x ) = e 2 x · 2 2 , h ′′′ ( x ) = e 2 x · 2 3 , and then, by induction, we get h
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Unformatted text preview: ( x ) = e 2 x Â· 2 23 . 6. The voltage, V , in volts, in an electrical outlet is given as a function of time, t , in seconds, by the function V = 115 cos(270 Ï€t ). Give the expression for the rate of change of voltage with respect to time in units of volts per minute. We compute the rate of change, V â€² ( t ): V â€² ( t ) = 115(-sin(270 Ï€t )) Â· 270 Ï€. Since v is in volts and t is in units of seconds, V â€² ( t ) is in units of volts/second. To get volts/minute, we multiply by 1 = (60 seconds)/(1 minute), and the desired rate is 115(-sin(270 Ï€t )) Â· 270 Ï€ Â· 60 volts/minute....
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