{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

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

Info icon This preview shows page 1. Sign up to view the full content.

View Full Document Right Arrow Icon
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
Image of page 1
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: ( x ) = e 2 x · 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 πt ). 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 ′ ( t ): V ′ ( t ) = 115(-sin(270 πt )) · 270 π. Since v is in volts and t is in units of seconds, V ′ ( 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 πt )) · 270 π · 60 volts/minute....
View Full Document

{[ snackBarMessage ]}

What students are saying

  • Left Quote Icon

    As a current student on this bumpy collegiate pathway, I stumbled upon Course Hero, where I can find study resources for nearly all my courses, get online help from tutors 24/7, and even share my old projects, papers, and lecture notes with other students.

    Student Picture

    Kiran Temple University Fox School of Business ‘17, Course Hero Intern

  • Left Quote Icon

    I cannot even describe how much Course Hero helped me this summer. It’s truly become something I can always rely on and help me. In the end, I was not only able to survive summer classes, but I was able to thrive thanks to Course Hero.

    Student Picture

    Dana University of Pennsylvania ‘17, Course Hero Intern

  • Left Quote Icon

    The ability to access any university’s resources through Course Hero proved invaluable in my case. I was behind on Tulane coursework and actually used UCLA’s materials to help me move forward and get everything together on time.

    Student Picture

    Jill Tulane University ‘16, Course Hero Intern