# 2nd and 3rd Derivatives!!!!!

• Jun 26th 2008, 01:39 PM
elocin
2nd and 3rd Derivatives!!!!!
Hi! This is my last question, I need the 2nd and 3rd derivatives of (lnx) / (8x). I have the first derivative as being (8-8lnx) / (64x^2) is that right?? I'm not sure!!! And I need the next 2 derivatives!!! THank you!!!
• Jun 26th 2008, 01:47 PM
bleesdan
OK, your first one is right on.
For the next two, just take the derivative of the derivative, then take the derivative of that!
Remember, $\displaystyle \frac{low*high'-high*low'}{low^{2}}$
• Jun 26th 2008, 01:55 PM
elocin
Okay, for the second, I got (64x^2)(-8/x)-(8-8lnx)(128x) all over 4096x^4. Is that right?
• Jun 26th 2008, 02:04 PM
bleesdan
Let's go back to the first derivative (I didn't see this the first time)
Remember to reduce! You can pull an 8 out of there to get $\displaystyle \frac{1-ln(x)}{8x^{2}}$.
I want you to simplify it, then post back with that.
After that, you should go ahead and take another derivative.
• Jun 26th 2008, 02:12 PM
elocin
Oh okay, so if you take out a 8, then you're left with 1-lnx over 8x^2? So then finding the derivative of that, I got (-8x)-(1-lnx)(16x) all over 64x^4. Is that right?
• Jun 26th 2008, 02:24 PM
bleesdan
Yep, but you still need to simplify this second derivative. What can you factor out of the top and the bottom?
• Jun 26th 2008, 02:28 PM
elocin
An 8? So I got (-x)-(1-lnx)(2x) all over 8x^4. No?
• Jun 26th 2008, 02:38 PM
bleesdan
You're pretty close.
You can get an x out too, so you get down to (-3+2ln(x)) over 8x^3
With that, you can then take its derivative and get the third derivative
• Jun 26th 2008, 02:41 PM
elocin
Shouldn't it be all over 3x^4? How did you get 8x^3?
• Jun 26th 2008, 02:43 PM
elocin
I have to go to dinner now, but I will work on it!! THank you so much for helping me!!!!
• Jun 26th 2008, 02:47 PM
bleesdan
Well, 8x^4 is 8x^3*x, so that's that.