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Math Help - Calculus (Differentiating & Integrating Logs)

  1. #1
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    Calculus (Differentiating & Integrating Logs)

    6. Find dy/dx. y= ln (2x/(x+3))
    ln 2x - ln x+3
    1/2x - 1/x+3
    (x+3)/2x(x+3) - (2x)/2x(x+3)
    = -x+3/2x(x+3)

    Answer is: 3/x(x+3)

    7. Find dy/dx. y= ln (x^3 + 3x)^3
    3ln(x^3+3x)^2 (3x^2+3)
    = (9x^2+9)ln(3x^2+3)^2

    Answer is: 9(x^2 +1)/x(x^2+3)

    8. Find the derivative. f(x) = ln √x^2 +1) / x(2x^3 -1)^2 Note: radical is only over the numerator.
    this is what i have so far but i'm not getting the answer:

    ln (x^2 +1)^1/2 - ln x(2x^3 -2)^2
    1/2 ln(x^2+1) - 2 ln x(2x^3-1)
    1/2 n(x^2+1) - 2 ln x + ln (2x^3-1)
    dy/dx = 1/2 x (1/x^2+1)2x - (2 (1/x) + (1/2x^3-1)6x^2)
    = (x/x^2 +1) - (2/x) + (6x^2/2x^3-1)

    but the answer is: (x/x^2 +1) - (1/x) - (12x^2/2x^3-1)

    14. ∫ (2x^2 + x -3) / (x-2) dx
    I did long division and got 2x-3-(3/x-2)
    ∫2x-3-(3/x-2)dx
    2x^2/2 - 3x - (3∫1/x-2)dx
    =x^2 - 3x - 3ln |x-2| +c

    Answer is: x^2 + 5x + 7ln |x-2| +c

    22. ∫ (sinx^2 - cosx^2) / sinx dx

    Answer is: -2cosx+ln |cscx + cotx| +c

    16. ∫ ln√x)) /x dx
    ∫ln (x)^1/2 / x
    ∫1/2 ln (x) / x

    Let u= lnx
    du= 1/x dx

    1/2 ∫u du

    Answer is: 1/4 (ln x)^2 +c
    Last edited by hmoob-chic; February 5th 2009 at 03:56 PM.
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  2. #2
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    Krizalid's Avatar
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    Well, this seems to be homework, so, why don't you show us what you've done on each problem?
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  3. #3
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    Quote Originally Posted by hmoob-chic View Post
    6. Find dy/dx. y= ln (2x/(x+3))
    Let u= 2x/(x+3) and use the chain rule: dy/dx= (dy/du)(du/dx)

    7. Find dy/dx. y= ln (x^3 + 3x)^3
    An immediate simplification: y= 3 ln(x^3+ 3x). Now let u= x^3+ 3x and use the chain rule.

    8. Find the derivative. f(x) = ln √x^2 +1) / x(2x^3 -1)^2
    Is the entire fraction inside the logarithm or is the logarithm in the numerator of the fraction?

    14. ∫ (2x^2 + x -3) / (x-2) dx
    If you let u= x- 2, x= u+ 2 so 2x^2+ x- 3= 2(u+2)^2+ (u+ 2)- 3= 2u^2+ 8u+ 8+ u- 1= 2u^2+ 9u+ 7 and (2x^2+x-3)/(x-2)= (2u^2+ 9u+ 7)/u= 2u+ 9+ 7/u. And, of course, dx= du

    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.
    Let u= 1+ 0.2t so du= 0.2dt and dt= du/0.2. dP= (2000/u)(du/0.2)= 10000/u du. Integrating that will give you a "constant of integration" and you can use P(0)= 1000 to determine that.

    b) find the pop. after 10 days.
    Set t= 10 in the formula you get in (a).
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