# Differentiating natural logs

• Jun 19th 2010, 07:38 PM
Celestiax
Differentiating natural logs
I'm having a little trouble differentiating these two functions, if anyone could help I would greatlyyy appreciate it.

g(x) = ln (x * sqrt(x^(2)-1))

and

f(u) = ln u / (1+ln(2u))

Whatever I do that makes sense to me, I end up with (1/x) + ((x) / (x^(2) - 1)) for g(x) and for the second, I used the quotient rule and can't end up with anything similar to the answers for either. Thank you very much in advance!
• Jun 19th 2010, 07:46 PM
Sudharaka
Quote:

Originally Posted by Celestiax
I'm having a little trouble differentiating these two functions, if anyone could help I would greatlyyy appreciate it.

g(x) = ln (x * sqrt(x^(2)-1))

and

f(u) = ln u / (1+ln(2u))

Whatever I do that makes sense to me, I end up with (1/x) + ((x) / (x^(2) - 1)) for g(x) and for the second, I used the quotient rule and can't end up with anything similar to the answers for either. Thank you very much in advance!

Dear Celestiax,

For the first one you should use the chain rule and the product rule of differentiation.

That is, $\frac{dy}{dx}=\frac{dy}{du}.\frac{du}{dx}~where~u~ and~y~are~fuctions~of~x.$

$\frac{d}{dx}(v.u)=v.\frac{du}{dx}+u.\frac{dv}{dx}$

Also you should know that, $\frac{d}{dx}\ln{x}=\frac{1}{x}$

• Jun 19th 2010, 07:48 PM
Ackbeet
I'll do the first step on each one:

$g'(x)=\frac{1}{x\sqrt{x^{2}-1}}\,\frac{d}{dx}(x\sqrt{x^{2}-1})$, and

$f'(u)=\frac{(1+\ln(2u))\frac{d}{du}\,\ln(u)-\ln(u)\frac{d}{du}(1+\ln(2u))}{(1+\ln(2u))^{2}}$. Can you continue?
• Jun 19th 2010, 07:55 PM
TheCoffeeMachine
Quote:

Originally Posted by Celestiax
Whatever I do that makes sense to me, I end up with (1/x) + ((x) / (x^(2) - 1))

Correct.
Quote:

... and for the second, I used the quotient rule and can't end up with anything similar to the answers for either
Post your answer/working. The chances are it's probably correct as well but it just isn't on the form in which it's in the book.
• Jun 19th 2010, 07:59 PM
Celestiax
Thank you to both of you, that is what I did for f(u), but I don't understand how you simplify the numerator to only (1 + ln 2), and how the 2u in the denominator cancels out? And for g(x), I cannot make it g(x) = lnx + (1/2) ln (x^(2) - 1)?
• Jun 19th 2010, 08:03 PM
Celestiax
The answers in the book are: g(x) = (2x^(2) - 1) / (x (x^(2) -1) and f(u) = (1 + ln 2) / ( (u) (1 + ln(2u))^(2) I guess I just didn't combine the terms for g(x)?
• Jun 19th 2010, 08:05 PM
Ackbeet
I must confess, I don't see how an ln(2u) can become an ln(2). Are you sure there's no u there? Be careful with parentheses!
• Jun 19th 2010, 08:25 PM
TheCoffeeMachine
Quote:

Originally Posted by Celestiax
The answers in the book are: g(x) = (2x^(2) - 1) / (x (x^(2) -1) I guess I just didn't combine the terms for g(x)?

Yup!

Quote:

[LaTeX ERROR: Compile failed]
Differentiate, either by writing it as [LaTeX ERROR: Compile failed] and using the chain rule, or using the quotient rule (as laid out earlier in the thread), and we will just work out how what we get is equivalent to the book's answer... don't worry about the simplification for now.
• Jun 19th 2010, 08:32 PM
Celestiax
Ah, thank you very much! I worked it all out now and I believe I have it correct...I didn't know ln(2u) - ln(u) could just be ln(2) for f(u). Although I'm not sure where the 2u in the denominator goes...but I guess I'll figure it out haha. For f(u) I got the same as in the book, but with a 2u in the denominator as well.
• Jun 19th 2010, 08:55 PM
TheCoffeeMachine
Quote:

Originally Posted by Celestiax
Ah, thank you very much! I worked it all out now and I believe I have it correct...I didn't know ln(2u) - ln(u) could just be ln(2) for f(u). Although I'm not sure where the 2u in the denominator goes...but I guess I'll figure it out haha. For f(u) I got the same as in the book, but with a 2u in the denominator as well.

Isn't the book's answer [LaTeX ERROR: Compile failed] If so, then you were obviously meant to have the [LaTeX ERROR: Compile failed] in the denominator. If it's [LaTeX ERROR: Compile failed] , then it's because [LaTeX ERROR: Compile failed] . It isn't important to get the answer in the exact form as in the book anyway... it's alright as far as it's correct and sufficiently simplified.
• Jun 19th 2010, 08:57 PM
Celestiax
Thank you to everyone that helped! I think I have it all worked out now.