# Thread: Calculus (Differentiating & Integrating Logs)

1. ## 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)

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

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

2. Well, this seems to be homework, so, why don't you show us what you've done on each problem?

3. Originally Posted by hmoob-chic
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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