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 September 29th, 2017, 03:05 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 07:54 AM.
 September 29th, 2017, 03:47 AM #2 Senior Member   Joined: Feb 2016 From: Australia Posts: 1,790 Thanks: 629 Math Focus: Yet to find out. You haven't posted the question.
September 29th, 2017, 04:43 AM   #3
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Quote:
 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 07:56 AM.

September 30th, 2017, 03:49 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, 05:17 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 07:24 AM.

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