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 September 29th, 2017, 04:05 AM #1 Newbie   Joined: Sep 2017 From: Latvia/Denmark Posts: 21 Thanks: 0 derivatives and natural logarithms I don't really understand (b) and (c). Thanks beforehand. Last edited by skipjack; September 29th, 2017 at 08:54 AM. September 29th, 2017, 04:47 AM #2 Senior Member   Joined: Feb 2016 From: Australia Posts: 1,847 Thanks: 661 Math Focus: Yet to find out. You haven't posted the question. September 29th, 2017, 05:43 AM   #3
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 Originally Posted by vizzy22  I don't really understand (b) and (c). Thanks beforehand.
Okay, do you know what the "laws of logarithms" are?

$\displaystyle \ln(ab)= \ln(a)+ \ln(b)$
$\displaystyle \ln(a/b)= \ln(a)- \ln(b)$ and
$\displaystyle \ln(a^b)= b \ln(a)$

For example, (c) can be written as $\displaystyle y= \ln\left(\left(\frac{x+ 1}{x- 1}\right)^{1/2}\right)= \frac{1}{2} \ln\left(\frac{x+ 1}{x- 1}\right)= \frac{1}{2}\left(\ln(x+ 1)- \ln(x- 1)\right)$.

Now, can you differentiate that?

You try the others.

Last edited by skipjack; September 29th, 2017 at 08:56 AM. September 30th, 2017, 04:49 AM   #4
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 Originally Posted by Country Boy Okay, do you know what the "laws of logarithms" are? $\displaystyle \ln(ab)= \ln(a)+ \ln(b)$ $\displaystyle \ln(a/b)= \ln(a)- \ln(b)$ and $\displaystyle \ln(a^b)= b \ln(a)$ For example, (c) can be written as $\displaystyle y= \ln\left(\left(\frac{x+ 1}{x- 1}\right)^{1/2}\right)= \frac{1}{2} \ln\left(\frac{x+ 1}{x- 1}\right)= \frac{1}{2}\left(\ln(x+ 1)- \ln(x- 1)\right)$. Now, can you differentiate that? You try the others.
I differenced that, but the thing I don't really understand is - how did you take ^1/2 and made it into 1/2ln.
I tried expanding B, but did not do it correctly, as the answers don't match. September 30th, 2017, 06:17 AM #5 Math Team   Joined: Jan 2015 From: Alabama Posts: 3,264 Thanks: 902 Then you need to go back and review logarithms! One of the basic "laws of logarithms", as I said before is "$\displaystyle \ln(a^b)= b \ln(a)$". If b = 1/2, that says that $\displaystyle \ln(\sqrt{a})= \ln(a^{1/2})= (1/2) \ln(a)$. Problem (b) is $\displaystyle \ln\left(x\sqrt{3x- 1}\right)$. That can be written as $\displaystyle \ln\left(x(3x- 1)^{1/2}\right)$ and, by the "laws of logarithms", $\displaystyle \ln(x)+ \ln((3x- 1)^{1/2})= \ln(x)+ (1/2)\ln(3x- 1)$. Of course to do problems like this, you have to know that the derivative of ln(x) is 1/x. And you need to know the "chain rule": the derivative of ln(3x- 1) is $\displaystyle \frac{1}{3x- 1}$ times the derivative of 3x- 1 which is 3. Putting all of those together the derivative of $\displaystyle \ln\left(x\sqrt{3x-1}\right)$, which is the same as the derivative of $\displaystyle \ln(x)+ (1/2)\ln(3x- 1)$, is $\displaystyle \frac{1}{x}+ \frac{3}{2(3x- 1)}$. Last edited by skipjack; September 30th, 2017 at 08:24 AM. Tags derivatives, derrivatives, logarithms, natural Thread Tools Show Printable Version Email this Page Display Modes Linear Mode Switch to Hybrid Mode Switch to Threaded Mode Similar Threads Thread Thread Starter Forum Replies Last Post mickapoo Calculus 7 September 5th, 2016 08:07 PM zendetax Algebra 1 September 23rd, 2015 07:37 AM BlackSnowMarine Algebra 8 February 13th, 2014 03:17 AM johngalt47 Calculus 2 November 23rd, 2013 12:44 PM jinjouk Number Theory 12 June 3rd, 2008 07:11 AM

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