# Natural logs and derivatives

1. Oct 14, 2009

### vorcil

Q1)

If (x^5)-x = 8
then
x = e^(solve for this)

not sure, x*x*x*x*x-x = 8

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q2)
ln(3x)/x^2
dy/dx = [ ]-[ ]ln(3x) / x^3
or solve for dy/dx

I get

(x^2)*(dy/dx(ln(3x))) - ln(3x)*(dy/dx(x^2)) / x^2*x^2
=(x^2)*(1/x) - ln(3x)*2x / x^4
= [x^2]-[2x]*(ln(3x)) / x^3
(is this right?)

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q3)

if dy/dx = 3e^x
and y = 5 when x=0

then y = [ ]

I get
I think i have to integrate 3e^x
and it's just 3e^x
y = 3e^0 + c
e^0 = 1
so 3e^x + 2 = 5

y = 3e^x + 2
and I got it wrong???

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q4)

evaluate
the integral of (1/3 * x-1)dx between 6 and 3

integrate it = 1/3*x^2/2 - x
= 1/6x^2 - x
(1/6 * 6^2 - 6 ) - (1/6 * 3^2 - 3) = 1/6*36-6 - 1/6*9-3
= 0 - 1.5-3
= -4.5
and I got it wrong
although I think during the test I may have made the mistake of doing 1.5 - 3 instead of -1.5 - 3...

2. Oct 14, 2009

### Mentallic

Q1) I don't see any simple solution to the problem. You can try approximate the solution, but I wouldn't know how to get any nice analytical solution if there is one.

Q2) You were right with the derivatives but didn't simplify correctly.

$$\frac{x^2*\frac{1}{x} - ln(3x)*2x}{x^4}$$

$$=\frac{x-2xln(3x)}{x^4}$$

$$=\frac{1-2ln(3x)}{x^3}$$

Q3) Uh it looks right...

Q4) Again, it was the algebra that got you.

$$0 - (9/6 - 3)= -9/6 + 3 = 1.5$$

3. Oct 14, 2009

thanks