# Derivative of a log function:

1. Jul 30, 2008

### xxclaymanxx

1. Given Y = ln [ (x+1)^3/((x^2)-1)^(1/2), find y'

2. I came out with the following answer to this question:

(3x-4)/((x^2)-1)

How ever, I typed the question into an online derivative calculator (to hopefully check my asnwer as I have no answer key, and want to make sure I'm on the right path), but it came up with a completely different answer:

(2x-3)/((x^2)-1)

Could anyone point me in the right direction...my answer worked out nicely: factored, cancelled etc. but I'm worried its not correct.

Thanks for the check!

2. Jul 30, 2008

### CompuChip

The latter is correct, unfortunately for you :)
You're not completely off, though, as
$$\frac{2x - 3}{x^2 - 1} = \frac{3 x - 4}{x^2 - 1} + \frac{x - 1}{x^2 - 1} = \frac{3 x - 4}{x^2 - 1} - \frac{1}{1 + x}$$
so it looks like you're just missing a term or you've got a sign wrong.

Also unfortunately, it is hard for us to tell you what went wrong without showing us your work. It's basically just calculating: d/du log(u) = 1/u, using the chain rule with u = (x+1)^3/((x^2)-1)^(1/2).

3. Jul 30, 2008

### xxclaymanxx

Thank you for your help! it allowed me to go back into my work, and figure out where I wen't wrong. Basically all I did, was i forgot to write an X, and instead wrote a 1...so when I was multiplying both sides by a common demonator, my numbers came out funny.

Anyways, I found the error, corrected the following calculations, and VOILA! got it.

Thanks again!

4. Jul 30, 2008

### HallsofIvy

Staff Emeritus
May I point out that Y= 3ln(x+1)-(1/2)ln(x2+ 1). Surely that is simpler to differentiate!