# ws14 - WORKSHEET 14 Fall 1995 1 Reiview properties of...

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

WORKSHEET 14 - Fall 1995 1. Reiview properties of logs (if necessary) and then solve these for x : a) x = log 4 2 b) log 4 x = 5 / 2 c) log x 1 8 = 3 2 d) log 4 ( x 2) log 4 (2 x + 3) = 0 e) log 10 x + log 10 ( x 15) = 2 f) 3 log x 4 = 2 g) log 2 x = log 4 5 + 3 log 2 3 h) log x (1 x ) = 2 i) log 8 (log 4 (log 2 x )) = 0 2. Show that log a b = 1 log b a . 3. Differentiation formulae. a) Let f ( x ) = xe x . Find a formula for f ( n ) ( x ). b) Do the same for f ( x ) = x 2 e x and f ( x ) = x 3 e x . c) Let f ( x ) = e x sin x . Find a formula for f ( n ) ( x ). d) Let f ( x ) = ue x where u is a function of x . Find a formula for f ( n ) ( x ) in terms of the derivatives of u . 4. Find the derivatives of each of the following. Simplify your answer as much as possible. a) y = sin( e x 2 + x 1 ) c) y = a x x a , a > 0 a, constant b) y = 1 + e x d) y = 2 (3 x ) . 5. a) Suppose that on some interval, the function f satisfies f ( x ) = cf ( x ) for some number c . Show that f ( x ) = Ke cx for some K by considering the function g ( x ) = f ( x ) e cx . b) Suppose that f ( x ) = f ( x ) g ( x ) for some g . Show that f ( x ) = Ke g ( x ) for some K . 6. For a given disease (rumor) the rate at which it spreads is proportional to the probability that a nondiseased, but susceptible, person meets a susceptible person, which is proportional to the product
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}

### What students are saying

• 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.

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

• 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.

Dana University of Pennsylvania ‘17, Course Hero Intern

• 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.

Jill Tulane University ‘16, Course Hero Intern