little bit of calc help? update- little more help needed

bleeb

Lifer
Feb 3, 2000
10,868
0
0
this is a chain rule, and product rule...


f(x) = x*ln(3*x)

(product rule)

f'(x) = 1*ln(3*x) + ln(3*x)'*x

(u-substitution * chain rule) u = 3x, du = 3

f'(x) = ln(3x) + [ 1/u*du/dx ]*x

f'(x) = ln(3x) + [1/3x*3]*x

f'(x) = ln(3x) + 1

CORRECT me if i'm wrong, it's been a while since i've done differentiation.
 

Woodchuck2000

Golden Member
Jan 20, 2002
1,632
1
0
As Bleeb says, you need to split it in two using the product rule. You can then differentiate the half with ln(3x) in by substituting t=3x into the equation.
 

Reel

Diamond Member
Jul 14, 2001
4,484
0
76
Originally posted by: TuxDave
Originally posted by: DeafeningSilence
Looks right to me.

triple checked by me... looks good

<edit> wait a sec... yeah... definitely good

Quick calculator check. Didn't feel like thinking about u substitution. :p
 

bleeb

Lifer
Feb 3, 2000
10,868
0
0
Originally posted by: Woodchuck2000
It's right, I just thought it was better to give him a hint than to give him the answer straight off ;)

hmmm... you're probably right.. but i'm sure there will be tougher solutions.. and i assumed he was stuck and had been working on this problem for a long time... so i figure seeing how it was solved would be better than giving small hints.
 

Siva

Diamond Member
Mar 8, 2001
5,472
0
71
wow, thanks guys. i forgot about the chain rule, when you get to stuff like logs the simpler stuff just flies right out. you guys rock, that was a big help! :D
 

Siva

Diamond Member
Mar 8, 2001
5,472
0
71
ok one more for the big calc wizards

integral of x/(2x+1)^(1/2)

i ended up with [(2x+1)^(3/2)] /6 - [(2x+1)^(1/2)] /2 but my ti89 disagrees
 

silverpig

Lifer
Jul 29, 2001
27,703
12
81
Try doing a u = 2x sub and maybe a trig identity will pop out?


u/sqrt(u+1) * 1/2 or something?


<---- super quick guess
 

chiwawa626

Lifer
Aug 15, 2000
12,013
0
0
Helps to get a Ti-89 in calc since u can use it to *check* your answers. I got lucky and picked one up on ebay for $55 shipped in mint condition :)
 

Dudd

Platinum Member
Aug 3, 2001
2,865
0
0
Originally posted by: Siva
ok one more for the big calc wizards

integral of x/(2x+1)^(1/2)

i ended up with [(2x+1)^(3/2)] /6 - [(2x+1)^(1/2)] /2 but my ti89 disagrees

u= 2x+1
du=2dx
x=(u-1)/2

thus we get to take the integral of

1/2 integral symbol ((u-1)/2)/u^.5du

I work that down to

(1/6)u^(3/2)- u^(.5)

Substituting, we get ((2x+1)^(3/2)/6)-(2x+1)^(.5)

Is that what you get?

 

Dudd

Platinum Member
Aug 3, 2001
2,865
0
0
Is that what the Ti89 got, or what you got? You had the final term over 2, wheras I did not.
 

Dudd

Platinum Member
Aug 3, 2001
2,865
0
0
Originally posted by: Siva
wait, isn't u^.5 over 2?

Not according to my math. My 2nd last step before substituting looked like this: (1/2)((u^(3/2)/3)-2u^.5) The 1/2 and the 2 multiply together, creating 1.

Edit: I could be wrong, I've been skiing all day and then did some coding, so my brain is shot. Let me look my math over and reply again.
 

Siva

Diamond Member
Mar 8, 2001
5,472
0
71
but when you divide (u-1)/2 by u^(.5) you get (u-1)/(2u^.5) and you can pull that 2 out to make it 1/4 integral of (u-1)/u^.5 can't you? or am i totally off
 

Dudd

Platinum Member
Aug 3, 2001
2,865
0
0
I'm wrong, you're right. When I had (u-1)/2u^.5, I didn't divide the 2 into both, so I orginally integrated u^.5/2-u^-.5 instead of u^-.5/2. So, it appears you are right after all.