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

[Siva]

i need to find the derivative of xln3x

any takers?

 

[bleeb]

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]

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.

 

[bleeb]

Can anyone confirm

 

[hdeck]

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

*edit* confirmed for bleeb.

 

[DeafeningSilence]

Looks right to me.

 

[TuxDave]

Originally posted by: DeafeningSilence
Looks right to me.

triple checked by me... looks good

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

 

[Woodchuck2000]

It's right, I just thought it was better to give him a hint than to give him the answer straight off

 

[Reel]

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.

 

[bleeb]

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]

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!

 

[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

 

[silverpig]

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

 

[Siva]

i did a u sub for 2x+1 and then solved for x=(u-1)/2

 

[chiwawa626]

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]

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?

 

[Siva]

wait, isn't u^.5 over 2?

 

[Dudd]

Is that what the Ti89 got, or what you got? You had the final term over 2, wheras I did not.

 

[Siva]

its what i got, the ti89 got something different altogether

 

[Dudd]

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]

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]

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.