Expert: Paul Klarreich - 2/5/2009

Question
6. Find dy/dx. y= ln (2x/(x+3))

7. Find dy/dx. y= ln (x^3 + 3x)^3

8. Find the derivative. f(x) = ln √x^2 +1) / x(2x^3 -1)^2

14. ∫ (2x^2 + x -3) / (x-2) dx

24. A population of bacteria is changing at the rate of dP/dt = 2000/ 1+0.2t where t is time in days.
The initial population is 1000.

a) Write an equation that gives the pop. at any time t.

b) find the pop. after 10 days.

Answer
Questioner:   Pag
Category:  Calculus
Private:  No
 
Subject:  Differentiating and Integrating Logs
Question:  6. Find dy/dx. y= ln (2x/(x+3))

7. Find dy/dx. y= ln (x^3 + 3x)^3

8. Find the derivative. f(x) = ln √x^2 +1) / x(2x^3 -1)^2

14. ∫ (2x^2 + x -3) / (x-2) dx

24. A population of bacteria is changing at the rate of dP/dt = 2000/ 1+0.2t where t is time in days.
The initial population is 1000.

a) Write an equation that gives the pop. at any time t.

b) find the pop. after 10 days.
 
===============================================
This is a lot, so I will just get you started on these.  [This is not a homework-doing or -checking site, but a help-you-get-started site.]

7. Find dy/dx. y= ln (x^3 + 3x)^3


y = 3 ln(x(x^2 + 3))
y = 3 (ln x + ln(x^2 + 3))

Now use the chain rule.

......................
8. Find the derivative. f(x) = ln √x^2 +1) / x(2x^3 -1)^2

You did not parenthesize clearly, so I don't know what the example is, but do the same as #7 -- use ln rules to break it apart, then the Chain Rule.



14. ∫ (2x^2 + x -3) / (x-2) dx

First use long division.  You get poly(x) + something/(x-2).  Should be easy after that.
....................



24. A population of bacteria is changing at the rate of dP/dt = 2000/ 1+0.2t where t is time in days.
The initial population is 1000.

a) Write an equation that gives the pop. at any time t.

Integrate, using u = (1 + 0.2t).  You will get a ln(...) + C.  Use P(0) = 1000 to find C.


b) find the pop. after 10 days. = p(10)